Chapter 12 Exponents and Powers (Additional Questions)

To express $2^{-3}$ as a positive exponent, we use the rule for negative exponents, which states that for any non-zero number $a$ and any positive integer $n$, $a^{-n} = \frac{1}{a^n}$.

Here, the base is $a=2$ and the exponent is $n=3$. So, applying the rule:

$2^{-3} = \frac{1}{2^3}$

Comparing this result with the given options:

(A) $-2^3 = -8$

(B) $\frac{1}{2^3} = \frac{1}{8}$

(C) $\frac{1}{2^{-3}} = 2^3 = 8$

(D) $-6$

The expression $2^{-3}$ is equal to $\frac{1}{2^3}$.

The correct option is (B).

To find the value of $3^0$, we use the rule for zero exponents.

The rule states that for any non-zero number $a$, $a^0 = 1$.

In this case, the base is $a=3$, which is a non-zero number.

Therefore, applying the rule:

$3^0 = 1$

Comparing this result with the given options:

(A) 0

(B) 1

(C) 3

(D) Undefined

The value of $3^0$ is 1.

The correct option is (B).

To simplify the expression $a^m \times a^n$, we use the product rule of exponents.

The product rule states that when multiplying exponential expressions with the same base, we keep the base and add the exponents.

Mathematically, the rule is expressed as:

$a^m \times a^n = a^{m+n}$

In the given expression, the base is $a$, and the exponents are $m$ and $n$. Applying the product rule:

$a^m \times a^n = a^{m+n}$

Comparing this result with the given options:

(A) $a^{m+n}$

(B) $a^{m-n}$

(C) $a^{mn}$

(D) $a^{m/n}$

The simplified form of $a^m \times a^n$ is $a^{m+n}$.

The correct option is (A).

To simplify the expression $(a^m)^n$, we use the power of a power rule of exponents.

The power of a power rule states that when raising an exponential expression to another power, we keep the base and multiply the exponents.

Mathematically, the rule is expressed as:

$(a^m)^n = a^{m \times n} = a^{mn}$

In the given expression, the base is $a$, the inner exponent is $m$, and the outer exponent is $n$. Applying the power of a power rule:

$(a^m)^n = a^{m \times n} = a^{mn}$

Comparing this result with the given options:

(A) $a^{m+n}$

(B) $a^{m-n}$

(C) $a^{mn}$

(D) $a^{m/n}$

The simplified form of $(a^m)^n$ is $a^{mn}$.

The correct option is (C).

To evaluate the expression $\frac{5^7}{5^4}$, we use the quotient rule of exponents.

The quotient rule states that when dividing exponential expressions with the same base, we keep the base and subtract the exponent of the denominator from the exponent of the numerator.

Mathematically, the rule is expressed as:

$\frac{a^m}{a^n} = a^{m-n}$

In the given expression, the base is $a=5$, the exponent in the numerator is $m=7$, and the exponent in the denominator is $n=4$. Applying the quotient rule:

$\frac{5^7}{5^4} = 5^{7-4} = 5^3$

Comparing this result with the given options:

(A) $5^3$

(B) $5^{11}$

(C) $5^{-3}$

(D) $5^{28}$

The value of $\frac{5^7}{5^4}$ is $5^3$.

The correct option is (A).

To simplify the expression $(2 \times 3)^4$, we use the power of a product rule of exponents.

The power of a product rule states that when raising a product to a power, we raise each factor to that power and multiply the results.

Mathematically, the rule is expressed as:

$(ab)^n = a^n b^n$

In the given expression, $a=2$, $b=3$, and $n=4$. Applying the power of a product rule:

$(2 \times 3)^4 = 2^4 \times 3^4$

Comparing this result with the given options:

(A) $2^4 + 3^4$

(B) $2^4 \times 3^4$

(C) $6^8$ (Note: $2 \times 3 = 6$, so $(2 \times 3)^4 = 6^4$, not $6^8$)

(D) $5^4$

The simplified form of $(2 \times 3)^4$ is $2^4 \times 3^4$.

The correct option is (B).

The standard form (or scientific notation) of a number is expressed as a number between 1 and 10 (inclusive of 1, but exclusive of 10) multiplied by a power of 10.

The given number is $45000$. To write this in standard form, we need to move the decimal point so that there is only one non-zero digit before the decimal point.

The decimal point in $45000$ is initially after the last zero, i.e., $45000.$

We move the decimal point to the left:

$4\underbrace{5000}_{4 \text{ places to the left}}.$

Moving the decimal point 4 places to the left gives us $4.5$.

Since we moved the decimal point 4 places to the left, we multiply the resulting number by $10^4$.

So, the standard form of $45000$ is $4.5 \times 10^4$.

Let's check the given options:

(A) $45 \times 10^3 = 45 \times 1000 = 45000$. This is not standard form as $45$ is not between 1 and 10.

(B) $4.5 \times 10^4 = 4.5 \times 10000 = 45000$. This is standard form as $4.5$ is between 1 and 10.

(C) $0.45 \times 10^5 = 0.45 \times 100000 = 45000$. This is not standard form as $0.45$ is not between 1 and 10.

(D) $4.5 \times 10^{-4} = 4.5 \times 0.0001 = 0.00045$. This is not equal to 45000.

The correct standard form of $45000$ is $4.5 \times 10^4$.

The correct option is (B).

The standard form (or scientific notation) of a number is expressed as a number between 1 and 10 (inclusive of 1, but exclusive of 10) multiplied by a power of 10.

The given number is $0.000007$. To write this in standard form, we need to move the decimal point so that there is only one non-zero digit before the decimal point.

The decimal point is currently before the leading zeros. We need to move it to the right until it is after the digit 7.

$0.\underbrace{000007}_{6 \text{ places to the right}}$

Moving the decimal point 6 places to the right gives us $7$.

Since we moved the decimal point 6 places to the right, we multiply the resulting number by $10^{-6}$.

So, the standard form of $0.000007$ is $7 \times 10^{-6}$.

Let's check the given options:

(A) $7 \times 10^6 = 7,000,000$. Incorrect.

(B) $7 \times 10^{-6} = 0.000007$. This is standard form as $7$ is between 1 and 10.

(C) $0.7 \times 10^{-5} = 0.000007$. Correct value, but not standard form because $0.7$ is not between 1 and 10.

(D) $70 \times 10^{-7} = 0.000007$. Correct value, but not standard form because $70$ is not between 1 and 10.

The correct standard form of $0.000007$ is $7 \times 10^{-6}$.

The correct option is (B).

To convert a number from standard form to the usual form, we examine the power of 10.

The given number is $3.02 \times 10^{-5}$. The power of 10 is $-5$. A negative exponent means we need to move the decimal point to the left.

We need to move the decimal point 5 places to the left in the number $3.02$.

Starting with $3.02$, move the decimal point 5 places to the left:

First move: $0.302$

Second move: $0.0302$

Third move: $0.00302$

Fourth move: $0.000302$

Fifth move: $0.0000302$

The usual form of $3.02 \times 10^{-5}$ is $0.0000302$.

Comparing this result with the given options:

(A) $302000$

(B) $0.0000302$

(C) $0.000302$

(D) $0.00302$

The correct option is (B).

To evaluate the expression $(1/3)^{-2}$, we use the rule for negative exponents, which states that for any non-zero rational number $\frac{a}{b}$ and any positive integer $n$, $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$.

In this expression, $a=1$, $b=3$, and $n=2$. Applying the rule:

$(\frac{1}{3})^{-2} = (\frac{3}{1})^2$

Now, we evaluate $(\frac{3}{1})^2$:

$(\frac{3}{1})^2 = 3^2$

Calculate $3^2$:

$3^2 = 3 \times 3 = 9$

So, the value of $(1/3)^{-2}$ is 9.

Comparing this result with the given options:

(A) 9

(B) $-9$

(C) $\frac{1}{9}$

(D) $-\frac{1}{9}$

The correct option is (A).

To simplify the expression $(\frac{2}{5})^3 \times (\frac{2}{5})^{-5}$, we use the product rule of exponents. The product rule states that when multiplying exponential expressions with the same base, we keep the base and add the exponents.

The rule is $a^m \times a^n = a^{m+n}$.

In this case, the base is $a = \frac{2}{5}$, the first exponent is $m=3$, and the second exponent is $n=-5$.

Applying the product rule:

$(\frac{2}{5})^3 \times (\frac{2}{5})^{-5} = (\frac{2}{5})^{3 + (-5)}$

Simplify the exponent:

$3 + (-5) = 3 - 5 = -2$

So, the expression becomes:

$(\frac{2}{5})^{-2}$

Now, we express the result with a positive exponent using the rule $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$.

Applying this rule:

$(\frac{2}{5})^{-2} = (\frac{5}{2})^2$

Finally, we evaluate the square:

$(\frac{5}{2})^2 = \frac{5^2}{2^2} = \frac{25}{4}$

The simplified expression is $\frac{25}{4}$. Looking at the options, option (A) shows the steps leading to this result.

The correct option is (A).

To evaluate the expression $\frac{10^2 \times 10^5}{10^3}$, we can use the rules of exponents.

First, simplify the numerator using the product rule of exponents, which states that $a^m \times a^n = a^{m+n}$ when the bases are the same.

The numerator is $10^2 \times 10^5$. Applying the product rule:

$10^2 \times 10^5 = 10^{2+5} = 10^7$

Now the expression becomes:

$\frac{10^7}{10^3}$

Next, simplify this expression using the quotient rule of exponents, which states that $\frac{a^m}{a^n} = a^{m-n}$ when the bases are the same.

Applying the quotient rule:

$\frac{10^7}{10^3} = 10^{7-3} = 10^4$

The value of the expression is $10^4$.

Comparing this result with the given options:

(A) $10^4$

(B) $10^0$

(C) $10^{10}$

(D) $10^{-6}$

The correct option is (A).

To determine which expression is equivalent to $\frac{a^m}{b^m}$, we can use the rules of exponents.

The expression is a quotient where both the numerator and the denominator are raised to the same power, $m$. This matches the form of the power of a quotient rule.

The power of a quotient rule states that for any non-zero numbers $a$ and $b$, and any integer $m$:

$(\frac{a}{b})^m = \frac{a^m}{b^m}$

Using this rule, we can see that $\frac{a^m}{b^m}$ is equivalent to $(\frac{a}{b})^m$.

Let's verify this with an example. Let $a=2$, $b=3$, and $m=2$.

$\frac{a^m}{b^m} = \frac{2^2}{3^2} = \frac{4}{9}$

$(\frac{a}{b})^m = (\frac{2}{3})^2 = \frac{2^2}{3^2} = \frac{4}{9}$

The values are equal.

Comparing this result with the given options:

(A) $(\frac{a}{b})^m$

(B) $(ab)^m = a^m b^m$

(C) $a/b$

(D) $a^{m-m} = a^0 = 1$ (assuming $a \neq 0$)

The expression equivalent to $\frac{a^m}{b^m}$ is $(\frac{a}{b})^m$.

The correct option is (A).

To convert a number from standard form to the usual form, we look at the exponent of 10.

The given number is $6.02 \times 10^{11}$. The exponent of 10 is $11$. Since the exponent is positive, we move the decimal point to the right.

We need to move the decimal point 11 places to the right in the number $6.02$.

Starting with $6.02$, the decimal point is between 6 and 0.

Moving the decimal point 2 places to the right gives $602$. We need to move it a total of 11 places. The remaining number of places to move is $11 - 2 = 9$.

We add 9 zeros after the digit 2:

$602\underbrace{000000000}_{\text{9 zeros}}$

This number is $602,000,000,000$.

Comparing this result with the given options:

(A) 60200000000 (10 zeros after 602)

(B) 602000000000 (9 zeros after 602)

(C) 6020000000000 (10 zeros after 602)

(D) 602 followed by 9 zeroes

The usual form of $6.02 \times 10^{11}$ is $602,000,000,000$, which is 602 followed by 9 zeros.

The correct option is (B), which represents 602 followed by 9 zeroes.

To compare the masses of the Earth and the Moon given in standard form, we examine the exponents of the power of 10.

Mass of Earth $= 5.97 \times 10^{24} \text{ kg}$

Mass of Moon $= 7.35 \times 10^{22} \text{ kg}$

In standard form $a \times 10^n$, the value of the number is primarily determined by the exponent $n$. A larger exponent of 10 indicates a larger number.

The exponent for the mass of the Earth is $24$.

The exponent for the mass of the Moon is $22$.

Comparing the exponents, we have $24 > 22$.

Since the exponent for the Earth's mass ($24$) is greater than the exponent for the Moon's mass ($22$), the mass of the Earth is greater than the mass of the Moon.

Specifically, the Earth's mass is approximately $10^{24}/10^{22} = 10^{24-22} = 10^2 = 100$ times larger than the Moon's mass (more accurately, $(5.97/7.35) \times 10^2 \approx 0.812 \times 100 \approx 81.2$ times larger).

Therefore, the Earth has more mass.

The correct option is (A).

We need to evaluate each statement to determine if it is true based on the rules of exponents.

Statement (A): $a^{-m} = \frac{1}{a^m}$

This is the definition of a negative exponent. It states that any non-zero base raised to a negative exponent is equal to the reciprocal of the base raised to the positive exponent.

This statement is TRUE.

Statement (B): $a^0 = 1$ (for $a \neq 0$)

This is the definition of the zero exponent. Any non-zero base raised to the power of zero is equal to 1. The condition $a \neq 0$ is important because $0^0$ is typically considered undefined.

This statement is TRUE.

Statement (C): $(a^m)^n = a^{m+n}$

This statement describes the power of a power rule. The correct rule is that when raising an exponential expression to another power, you multiply the exponents, i.e., $(a^m)^n = a^{m \times n} = a^{mn}$. The statement given adds the exponents ($m+n$) instead of multiplying them.

For example, let $a=2, m=2, n=3$. $(2^2)^3 = 4^3 = 64$. $2^{2+3} = 2^5 = 32$. $64 \neq 32$.

This statement is FALSE.

Statement (D): $a^m \times b^m = (ab)^m$

This statement describes the power of a product rule, applied in reverse. It states that if you have a product of bases raised to the same exponent, you can multiply the bases first and then raise the result to that exponent. This rule is true for any bases $a$ and $b$ and any integer exponent $m$.

For example, let $a=2, b=3, m=2$. $2^2 \times 3^2 = 4 \times 9 = 36$. $(2 \times 3)^2 = 6^2 = 36$. The values are equal.

This statement is TRUE.

The statements that are TRUE are (A), (B), and (D).

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): "The standard form of a number is written as $k \times 10^n$, where $k$ is a number between 1 and 10 (inclusive of 1 but exclusive of 10) and $n$ is an integer."

This statement provides the definition of standard form (also known as scientific notation). The conditions on $k$ ($1 \leq k < 10$) and $n$ ($n$ is an integer) are correct for the standard definition of scientific notation used for representing numbers.

Thus, Assertion (A) is TRUE.

Reason (R): "This form helps in representing very large and very small numbers concisely."

The standard form was developed precisely for this purpose. Numbers like the mass of the Earth ($5.97 \times 10^{24} \text{ kg}$) or the size of an atom are either extremely large or extremely small. Writing them out fully in usual form would involve many zeros, making them difficult to read, write, and calculate with. The standard form $k \times 10^n$ provides a concise way to represent these numbers.

Thus, Reason (R) is TRUE.

Now let's consider if Reason (R) is the correct explanation of Assertion (A).

Assertion (A) defines the standard form. Reason (R) explains *why* this specific form (with the condition $1 \leq k < 10$) is useful and widely adopted. The structure of the standard form defined in A (especially the constraint on $k$) is what makes it concise and practical for handling very large and small numbers. Therefore, the reason (R) correctly explains the significance and utility of the definition given in Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A).

To match the expressions with their simplified forms, we need to simplify each expression individually.

(i) Simplify $7^{-1} \times 7^3$:

Using the product rule $a^m \times a^n = a^{m+n}$:

$7^{-1} \times 7^3 = 7^{-1+3} = 7^2$

Evaluating $7^2$:

$7^2 = 7 \times 7 = 49$

So, (i) simplifies to $49$. This matches option (a).

(ii) Simplify $(5^2)^0$:

Using the power of a power rule $(a^m)^n = a^{mn}$:

$(5^2)^0 = 5^{2 \times 0} = 5^0$

Using the zero exponent rule $a^0 = 1$ (for $a \neq 0$):

$5^0 = 1$

So, (ii) simplifies to $1$. This matches option (d).

(iii) Simplify $(\frac{2}{3})^{-1}$:

Using the negative exponent rule for fractions $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$:

$(\frac{2}{3})^{-1} = (\frac{3}{2})^1 = \frac{3}{2}$

So, (iii) simplifies to $\frac{3}{2}$. This matches option (c).

(iv) Simplify $\frac{4^5}{4^7}$:

Using the quotient rule $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{4^5}{4^7} = 4^{5-7} = 4^{-2}$

Using the negative exponent rule $a^{-n} = \frac{1}{a^n}$:

$4^{-2} = \frac{1}{4^2} = \frac{1}{16}$

So, (iv) simplifies to $\frac{1}{16}$. This matches option (b).

Based on the simplifications, the matches are:

(i) - (a)

(ii) - (d)

(iii) - (c)

(iv) - (b)

Comparing these matches with the given options, we find that option (A) correctly lists these matches.

The correct option is (A).

To express the diameter of the virus in standard form, we need to write the number $0.00000001$ in the form $k \times 10^n$, where $1 \leq k < 10$ and $n$ is an integer.

The diameter is given as $0.00000001 \text{ meters}$.

We need to move the decimal point to the right until the number is between 1 and 10.

Starting from $0.00000001$, we move the decimal point past the non-zero digit (which is 1).

$0.\underbrace{00000001}_{8 \text{ places to the right}}$

Moving the decimal point 8 places to the right gives us the number 1.

So, the value of $k$ is $1$. Since $1$ is between 1 and 10 (inclusive of 1), this satisfies the condition $1 \leq k < 10$.

The exponent $n$ is determined by the number of places the decimal point was moved and the direction. Since we moved the decimal point 8 places to the right (to make the number larger), the exponent $n$ is negative and equal to the number of places moved.

So, $n = -8$.

Therefore, the standard form of $0.00000001$ is $1 \times 10^{-8}$.

Adding the unit, the diameter of the virus in standard form is $1 \times 10^{-8} \text{ meters}$.

Comparing this result with the given options:

(A) $1 \times 10^8 \text{ m}$ (Incorrect exponent)

(B) $1 \times 10^{-8} \text{ m}$ (Matches our result)

(C) $10 \times 10^{-9} \text{ m}$ (This equals $1 \times 10 \times 10^{-9} = 1 \times 10^{1-9} = 1 \times 10^{-8}$, but $k=10$ is not in the range $1 \leq k < 10$, so it's not standard form)

(D) $10 \times 10^7 \text{ m}$ (Incorrect value and not standard form)

The correct standard form is $1 \times 10^{-8} \text{ m}$.

The correct option is (B).

To compare the diameter of the virus and the distance to the Sun, we should express both quantities in a comparable form, such as standard form.

From the Case Study:

Diameter of virus $\approx 0.00000001 \text{ meters}$

Distance from Earth to Sun $\approx 1.496 \times 10^{11} \text{ meters}$

First, let's express the diameter of the virus in standard form. As determined in Question 19, $0.00000001 = 1 \times 10^{-8}$.

So, Diameter of virus $= 1 \times 10^{-8} \text{ meters}$.

The distance from Earth to Sun is already given in standard form:

Distance to Sun $= 1.496 \times 10^{11} \text{ meters}$.

Now we compare the two numbers in standard form:

Virus Diameter: $1 \times 10^{-8}$

Distance to Sun: $1.496 \times 10^{11}$

When comparing numbers in standard form ($k \times 10^n$), the magnitude is primarily determined by the exponent $n$. The larger the exponent, the larger the number.

Comparing the exponents:

Exponent for Virus Diameter $= -8$

Exponent for Distance to Sun $= 11$

Since $11 > -8$, the number multiplied by $10^{11}$ is much larger than the number multiplied by $10^{-8}$.

Therefore, the distance from the Earth to the Sun is much larger than the diameter of the virus.

Comparing our finding with the given options:

(A) Diameter of the virus

(B) Distance to the Sun

(C) They are comparable.

(D) Cannot compare due to different units (although both are in meters)

The correct option is (B).

To find out how many times larger the distance from the Earth to the Sun is than the diameter of the virus, we need to divide the distance to the Sun by the diameter of the virus.

From the Case Study (Question 19):

Distance to Sun $= 1.496 \times 10^{11} \text{ meters}$

Diameter of virus $= 0.00000001 \text{ meters}$

First, express the diameter of the virus in standard form. As found in Question 19, the standard form of $0.00000001$ is $1 \times 10^{-8}$.

Diameter of virus $= 1 \times 10^{-8} \text{ meters}$.

Now, calculate the ratio:

Ratio $= \frac{\text{Distance to Sun}}{\text{Diameter of Virus}}$

Ratio $= \frac{1.496 \times 10^{11}}{1 \times 10^{-8}}$

We can perform the division by separating the numerical parts and the powers of 10:

Ratio $= \left(\frac{1.496}{1}\right) \times \left(\frac{10^{11}}{10^{-8}}\right)$

Simplify the numerical part:

$\frac{1.496}{1} = 1.496$

Simplify the powers of 10 using the quotient rule $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{10^{11}}{10^{-8}} = 10^{11 - (-8)} = 10^{11 + 8} = 10^{19}$

Combine the results:

Ratio $= 1.496 \times 10^{19}$

So, the distance from the Earth to the Sun is approximately $1.496 \times 10^{19}$ times larger than the diameter of the virus.

Comparing this result with the given options:

(A) $1.496 \times 10^{19}$ times

(B) $1.496 \times 10^3$ times

(C) $1.496 \times 10^{-19}$ times

(D) $1.496 \times 10^{20}$ times

The calculated ratio matches option (A).

The correct option is (A).

To find the value of $(-1)^{-1}$, we use the rule for negative exponents.

The rule states that for any non-zero number $a$ and any integer $n$, $a^{-n} = \frac{1}{a^n}$.

In this expression, the base is $a = -1$ and the exponent is $-1$. Applying the rule with $n=1$:

$(-1)^{-1} = \frac{1}{(-1)^1}$

Now, evaluate the denominator:

$(-1)^1 = -1$

Substitute this back into the expression:

$(-1)^{-1} = \frac{1}{-1}$

Simplify the fraction:

$\frac{1}{-1} = -1$

So, the value of $(-1)^{-1}$ is $-1$.

Comparing this result with the given options:

(A) 1

(B) -1

(C) Undefined

(D) 0

The correct option is (B).

The question asks to complete the definition of standard form (scientific notation).

Standard form is defined as writing a number as the product of a number $k$ and a power of 10, where $k$ is a number such that $1 \leq k < 10$. The general form is $k \times 10^n$, where $n$ is an integer.

The sentence provided is: "When a number is written in standard form, it is expressed as the product of a number between 1 and 10 (inclusive of 1, exclusive of 10) and a power of _________."

Comparing this with the standard form $k \times 10^n$, the blank needs to be filled with the base of the power, which is 10.

The completed sentence is: "When a number is written in standard form, it is expressed as the product of a number between 1 and 10 (inclusive of 1, exclusive of 10) and a power of 10."

Comparing the required word with the given options:

(A) $e$ (The base of the natural logarithm, not used in standard form)

(B) 10 (Correct base for standard form)

(C) 100 (Incorrect base)

(D) 0 (Incorrect base)

The correct option is (B).

To simplify the expression $\frac{(2^3 \times 3^4 \times 4)}{(3^2 \times 32)}$, we will express all numbers as powers of their prime factors and then use the rules of exponents.

First, express $4$ and $32$ as powers of 2:

$4 = 2 \times 2 = 2^2$

$32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5$

Substitute these into the expression:

$\frac{(2^3 \times 3^4 \times 2^2)}{(3^2 \times 2^5)}$

Rearrange the terms in the numerator and denominator to group terms with the same base:

$\frac{(2^3 \times 2^2 \times 3^4)}{(2^5 \times 3^2)}$

Use the product rule of exponents ($a^m \times a^n = a^{m+n}$) in the numerator:

$2^3 \times 2^2 = 2^{3+2} = 2^5$

The expression becomes:

$\frac{(2^5 \times 3^4)}{(2^5 \times 3^2)}$

Now, we can use the quotient rule of exponents ($\frac{a^m}{a^n} = a^{m-n}$) for each base:

For base 2: $\frac{2^5}{2^5} = 2^{5-5} = 2^0$

For base 3: $\frac{3^4}{3^2} = 3^{4-2} = 3^2$

Substitute these simplified terms back into the expression:

$2^0 \times 3^2$

Evaluate the terms:

$2^0 = 1$ (using the zero exponent rule $a^0 = 1$ for $a \neq 0$)

$3^2 = 3 \times 3 = 9$

Multiply the results:

$1 \times 9 = 9$

The simplified value of the expression is 9.

Comparing this result with the given options:

(A) 9

(B) 12

(C) 18

(D) 24

The correct option is (A).

We are given the equation $3^x = 243$.

To find the value of $x$, we need to express both sides of the equation with the same base. The base on the left side is 3. So, we need to express 243 as a power of 3.

Let's find the prime factorization of 243 by repeatedly dividing by 3:

$243 \div 3 = 81$

$81 \div 3 = 27$

$27 \div 3 = 9$

$9 \div 3 = 3$

$3 \div 3 = 1$

So, $243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5$.

Now, substitute $3^5$ for 243 in the original equation:

$3^x = 3^5$

When two exponential expressions with the same non-zero, non-one base are equal, their exponents must be equal. That is, if $a^x = a^y$ and $a \neq 0, 1, -1$, then $x=y$.

In our equation, the base is 3, which is not 0, 1, or -1. Therefore, we can equate the exponents:

$x = 5$

The value of $x$ is 5.

Comparing this result with the given options:

(A) 3

(B) 4

(C) 5

(D) 6

The correct option is (C).

To find the reciprocal of a number, we take 1 and divide it by the number. The reciprocal of $a$ is $\frac{1}{a}$.

The given number is $5^{-2}$. Its reciprocal is $\frac{1}{5^{-2}}$.

To simplify $\frac{1}{5^{-2}}$, we can use the rule for negative exponents which states that $\frac{1}{a^{-n}} = a^n$.

Applying this rule, with $a=5$ and $n=2$:

$\frac{1}{5^{-2}} = 5^2$

Now, evaluate $5^2$:

$5^2 = 5 \times 5 = 25$

So, the reciprocal of $5^{-2}$ is $5^2$ or 25.

Comparing our simplified form $5^2$ with the given options:

(A) $5^2$

(B) $-5^2$

(C) $\frac{1}{5^2}$

(D) $-\frac{1}{5^2}$

The correct option is (A).

To evaluate the expression $((\frac{1}{2})^{-2})^{-1}$, we can simplify it step by step using the rules of exponents.

First, simplify the inner part of the expression: $(\frac{1}{2})^{-2}$.

Using the rule for negative exponents for fractions $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$:

$(\frac{1}{2})^{-2} = (\frac{2}{1})^2 = 2^2$

Now, substitute this result back into the original expression:

$(2^2)^{-1}$

Next, simplify $(2^2)^{-1}$ using the power of a power rule $(a^m)^n = a^{mn}$:

$(2^2)^{-1} = 2^{2 \times (-1)} = 2^{-2}$

Finally, evaluate $2^{-2}$ using the rule for negative exponents $a^{-n} = \frac{1}{a^n}$:

$2^{-2} = \frac{1}{2^2} = \frac{1}{4}$

The value of the expression is $\frac{1}{4}$.

Alternatively, we can use the power of a power rule first on the outer exponents: $((a^m)^n)^p = a^{m \times n \times p}$.

In this case, the base is $\frac{1}{2}$, the exponents are $-2$ and $-1$.

$((\frac{1}{2})^{-2})^{-1} = (\frac{1}{2})^{(-2) \times (-1)}$

Calculate the product of the exponents: $(-2) \times (-1) = 2$.

So the expression becomes:

$(\frac{1}{2})^2$

Evaluate the square:

$(\frac{1}{2})^2 = \frac{1^2}{2^2} = \frac{1}{4}$

Both methods yield the same result, $\frac{1}{4}$.

Comparing this result with the given options:

(A) 4

(B) $-4$

(C) $\frac{1}{4}$

(D) $-\frac{1}{4}$

The correct option is (C).

To convert a number from standard form to the usual form, we look at the exponent of 10.

The given number is the speed of light, approximately $3 \times 10^8 \text{ m/s}$. The exponent of 10 is $8$. Since the exponent is positive, we move the decimal point to the right.

The number part is 3, which can be written as $3.0$. We need to move the decimal point 8 places to the right.

$3.\underbrace{00000000}_{\text{8 places}}$

This means we place 8 zeros after the digit 3.

The resulting number is $300,000,000$.

The usual form of $3 \times 10^8 \text{ m/s}$ is $300,000,000 \text{ m/s}$.

Comparing this result with the given options:

(A) 300,000,000 m/s (3 followed by 8 zeros)

(B) 30,000,000 m/s (3 followed by 7 zeros)

(C) 3,000,000 m/s (3 followed by 6 zeros)

(D) 3,00,00,000 m/s (This format is sometimes used, but it represents 30,000,000)

The number $300,000,000$ corresponds to option (A).

The correct option is (A).

To express the number $0.00000051$ in standard form, we need to write it in the form $k \times 10^n$, where $1 \leq k < 10$ and $n$ is an integer.

The given number is $0.00000051$. We need to move the decimal point to the right until the number is between 1 and 10. The non-zero digits are 5 and 1.

Starting from $0.00000051$, we move the decimal point to the right until it is after the first non-zero digit (which is 5).

$0.\underbrace{0000005}_{7 \text{ places to the right}}1$

Moving the decimal point 7 places to the right gives us the number $5.1$.

So, the value of $k$ is $5.1$. Since $5.1$ is between 1 and 10, it satisfies the condition $1 \leq k < 10$.

The exponent $n$ is determined by the number of places the decimal point was moved and the direction. Since we moved the decimal point 7 places to the right (to make the number larger), the exponent $n$ is negative and equal to the number of places moved.

So, $n = -7$.

Therefore, the standard form of $0.00000051$ is $5.1 \times 10^{-7}$.

Comparing this result with the given options:

(A) $5.1 \times 10^7$ (Incorrect exponent sign and value)

(B) $5.1 \times 10^{-7}$ (Matches our result)

(C) $51 \times 10^{-8}$ (This equals $5.1 \times 10 \times 10^{-8} = 5.1 \times 10^{1-8} = 5.1 \times 10^{-7}$, but $k=51$ is not in the range $1 \leq k < 10$)

(D) $5.1 \times 10^{-8}$ (Incorrect exponent value)

The correct standard form is $5.1 \times 10^{-7}$.

The correct option is (B).

To simplify the expression $\frac{1}{2^{-3} \times 5^{-3}}$, we can use the properties of exponents. Let's analyze the different ways shown in the options.

The given expression is:

$\frac{1}{2^{-3} \times 5^{-3}}$

Method 1 (Matching Option A):

We can use the property $a^m \times b^m = (ab)^m$ in the denominator:

$2^{-3} \times 5^{-3} = (2 \times 5)^{-3} = 10^{-3}$

So, the expression becomes:

$\frac{1}{10^{-3}}$

Now, use the property $\frac{1}{a^{-n}} = a^n$:

$\frac{1}{10^{-3}} = 10^3$

Evaluate $10^3$:

$10^3 = 10 \times 10 \times 10 = 1000$

This matches the steps shown in option (A): $\frac{1}{(10)^{-3}} = 10^3 = 1000$. So, option (A) shows a correct way to evaluate the expression.

Method 2 (Matching Option B):

Start with the original expression and use the property $\frac{1}{a^{-n}} = a^n$ or $a^{-n} = \frac{1}{a^n}$ on the terms in the denominator.

$\frac{1}{2^{-3} \times 5^{-3}} = \frac{1}{\frac{1}{2^3} \times \frac{1}{5^3}} = \frac{1}{\frac{1}{2^3 \times 5^3}}$

Dividing by a fraction is equivalent to multiplying by its reciprocal:

$\frac{1}{\frac{1}{2^3 \times 5^3}} = 1 \times (2^3 \times 5^3) = 2^3 \times 5^3$

Now, use the property $a^m \times b^m = (ab)^m$:

$2^3 \times 5^3 = (2 \times 5)^3 = 10^3$

Evaluate $10^3$:

$10^3 = 10 \times 10 \times 10 = 1000$

This matches the steps shown in option (B): $2^3 \times 5^3 = (10)^3 = 1000$. So, option (B) shows a correct way to evaluate the expression.

Method 3 (Matching Option C):

Continue from the step $2^3 \times 5^3$ derived in Method 2. Instead of using the power of a product rule, evaluate each power individually and then multiply.

$2^3 = 2 \times 2 \times 2 = 8$

$5^3 = 5 \times 5 \times 5 = 125$

Now, multiply the results:

$8 \times 125 = 1000$

This matches the steps shown in option (C): $2^3 \times 5^3 = 8 \times 125 = 1000$. So, option (C) shows a correct way to evaluate the expression.

Since options (A), (B), and (C) all demonstrate valid mathematical steps leading to the correct value (1000), all of these methods are correct ways to evaluate the expression.

The correct option is (D).

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): "$2^5 > 5^2$."

To check if this statement is true, we need to evaluate both sides:

$2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$

$5^2 = 5 \times 5 = 25$

Now, compare the values: $32 > 25$.

Since $32$ is indeed greater than $25$, the assertion "$2^5 > 5^2$" is TRUE.

Reason (R): "$2^5 = 32$ and $5^2 = 25$, and $32 > 25$."

This statement calculates the values of $2^5$ and $5^2$ correctly ($32$ and $25$ respectively) and then correctly states that $32 > 25$. These calculations are accurate.

Thus, Reason (R) is TRUE.

Now let's consider if Reason (R) is the correct explanation of Assertion (A).

Assertion (A) makes a claim about the inequality between $2^5$ and $5^2$. Reason (R) provides the exact values of $2^5$ and $5^2$ and shows the comparison that directly supports and proves the claim made in Assertion (A). The reason gives the specific calculation and comparison that justifies why $2^5$ is greater than $5^2$.

Therefore, the reason (R) is a correct explanation of Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A).

The size of the dust particle is given as $10^{-6} \text{ m}$. We need to express this number in usual form, which means without using exponents.

Using the rule for negative exponents, $a^{-n} = \frac{1}{a^n}$, we can rewrite $10^{-6}$:

$10^{-6} = \frac{1}{10^6}$

Now, calculate the value of $10^6$:

$10^6 = 1,000,000$

Substitute this back into the expression:

$10^{-6} = \frac{1}{1,000,000}$

To convert the fraction $\frac{1}{1,000,000}$ to a decimal, we divide 1 by 1,000,000. This results in a decimal with 6 places after the decimal point, with the digit 1 in the sixth place.

$\frac{1}{1,000,000} = 0.000001$

So, the usual form of $10^{-6} \text{ m}$ is $0.000001 \text{ m}$.

Comparing this result with the given options:

(A) 0.000001 m (Matches our result)

(B) 0.0000001 m ($10^{-7}$)

(C) 1000000 m ($10^6$)

(D) 0.00001 m ($10^{-5}$)

The correct option is (A).

To find out how many times larger the distance to Proxima Centauri is than the size of the dust particle, we need to divide the distance to Proxima Centauri by the size of the dust particle.

From the Case Study (Question 32):

Size of dust particle $= 10^{-6} \text{ m}$

Distance to Proxima Centauri $= 4.24 \times 10^{16} \text{ m}$

We need to calculate the ratio:

Ratio $= \frac{\text{Distance to Proxima Centauri}}{\text{Size of dust particle}}$

Ratio $= \frac{4.24 \times 10^{16}}{10^{-6}}$

We can rewrite the denominator as $1 \times 10^{-6}$ to make the division clearer:

Ratio $= \frac{4.24 \times 10^{16}}{1 \times 10^{-6}}$

Now, we perform the division by separating the numerical parts and the powers of 10:

Ratio $= \left(\frac{4.24}{1}\right) \times \left(\frac{10^{16}}{10^{-6}}\right)$

Simplify the numerical part:

$\frac{4.24}{1} = 4.24$

Simplify the powers of 10 using the quotient rule of exponents $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{10^{16}}{10^{-6}} = 10^{16 - (-6)} = 10^{16 + 6} = 10^{22}$

Combine the results:

Ratio $= 4.24 \times 10^{22}$

So, the distance to Proxima Centauri is approximately $4.24 \times 10^{22}$ times larger than the size of the dust particle.

Comparing this result with the given options:

(A) $4.24 \times 10^{10}$ times

(B) $4.24 \times 10^{22}$ times

(C) $4.24 \times 10^{-22}$ times

(D) $4.24 \times 10^{16} - 10^{-6}$ times

The calculated ratio matches option (B).

The correct option is (B).

To evaluate the expression $((\frac{-1}{2})^{-3})^{-2}$, we can use the rules of exponents.

We apply the power of a power rule, which states that $(a^m)^n = a^{mn}$. In our case, the base is $a = \frac{-1}{2}$, the inner exponent is $m = -3$, and the outer exponent is $n = -2$.

Multiplying the exponents:

$(-3) \times (-2) = 6$

So, the expression simplifies to:

$(\frac{-1}{2})^6$

Now, we evaluate $(\frac{-1}{2})^6$. Using the rule $(\frac{a}{b})^n = \frac{a^n}{b^n}$:

$(\frac{-1}{2})^6 = \frac{(-1)^6}{2^6}$

Calculate the powers:

$(-1)^6 = 1$ (A negative number raised to an even power is positive)

$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$

Substitute the values back into the fraction:

$\frac{1}{64}$

So, the value of the expression $((\frac{-1}{2})^{-3})^{-2}$ is $\frac{1}{64}$.

Let's compare this result and the steps with the given options:

(A) $(-8)^{-2} = \frac{1}{(-8)^2} = \frac{1}{64}$. This path evaluates the inner exponent first: $(\frac{-1}{2})^{-3} = (\frac{2}{-1})^3 = (-2)^3 = -8$. Then it evaluates $(-8)^{-2}$. This is a correct evaluation path and result.

(B) $(-\frac{1}{2})^6 = \frac{1}{64}$. This path applies the power of a power rule first, resulting in $(-\frac{1}{2})^6$, and then evaluates it. This is also a correct evaluation path and result.

(C) $(-\frac{1}{2})^5 = -\frac{1}{32}$. This is incorrect as the resulting exponent is 5, not 6.

(D) $64$. This is incorrect; the value is $\frac{1}{64}$.

Both options (A) and (B) show correct methods of evaluation leading to the correct answer $\frac{1}{64}$. In typical multiple-choice questions with a single correct answer, option (B) which directly applies the power of a power rule $(a^m)^n = a^{mn}$ first is often considered the intended or primary simplification method. However, (A) is also perfectly valid.

Based on the direct application of the power of a power rule, the evaluation $(-\frac{1}{2})^6 = \frac{1}{64}$ is shown in option (B).

The correct option is (B).

To simplify the expression $(5^0 \times 6^0 \times 7^0)$, we use the rule for zero exponents.

The rule states that for any non-zero number $a$, $a^0 = 1$.

Applying this rule to each term in the expression:

$5^0 = 1$ (since 5 is non-zero)

$6^0 = 1$ (since 6 is non-zero)

$7^0 = 1$ (since 7 is non-zero)

Now substitute these values back into the expression:

$(5^0 \times 6^0 \times 7^0) = 1 \times 1 \times 1$

Calculate the product:

$1 \times 1 \times 1 = 1$

So, the value of the expression is 1.

Comparing this result with the given options:

(A) 0

(B) 1

(C) 210 ($5 \times 6 \times 7$)

(D) $18^0 = 1$ (While $18^0$ is also 1, the direct simplification of the expression using the property is shown in the steps)

The simplified value is 1.

The correct option is (B).

To express the number $153600000$ in standard form, we need to write it in the form $k \times 10^n$, where $1 \leq k < 10$ and $n$ is an integer.

The given number is $153600000$. The decimal point is implicitly at the end of the number.

$153600000.$

We need to move the decimal point to the left until there is only one non-zero digit before the decimal point. The first non-zero digit is 1.

Move the decimal point past the digits $5, 3, 6, 0, 0, 0, 0, 0$. There are 8 digits after the first digit 1.

$1\underbrace{53600000.}_{8 \text{ places}}$

Moving the decimal point 8 places to the left gives us the number $1.536$. This is the value of $k$. It satisfies the condition $1 \leq 1.536 < 10$.

The exponent $n$ is equal to the number of places the decimal point was moved. Since we moved the decimal point 8 places to the left to make the number smaller, the exponent $n$ is positive.

So, $n = 8$.

Therefore, the standard form of $153600000$ is $1.536 \times 10^8$.

Comparing this result with the given options:

(A) $1.536 \times 10^8$

(B) $1.536 \times 10^7$

(C) $15.36 \times 10^7$

(D) $0.1536 \times 10^9$

The correct standard form is $1.536 \times 10^8$.

The correct option is (A).

We are given that $m = -2$. We need to find the value of $m^m$.

Substitute the value of $m$ into the expression $m^m$:

$m^m = (-2)^{-2}$

Now, we evaluate $(-2)^{-2}$ using the rule for negative exponents, $a^{-n} = \frac{1}{a^n}$.

In this case, the base is $a = -2$ and the exponent is $n = 2$.

$(-2)^{-2} = \frac{1}{(-2)^2}$

Evaluate the denominator: $(-2)^2$.

$(-2)^2 = (-2) \times (-2) = 4$ (A negative number raised to an even power is positive).

Substitute the value of the denominator back into the fraction:

$\frac{1}{4}$

So, the value of $m^m$ when $m=-2$ is $\frac{1}{4}$.

Comparing this result and the steps with the given options:

(A) $(-2)^{-2} = \frac{1}{(-2)^2} = \frac{1}{4}$. This option shows the correct steps and the final result.

(B) 4. Incorrect.

(C) -4. Incorrect.

(D) $\frac{1}{2}$. Incorrect.

The correct option is (A).

We need to find which of the given options is equal to the expression $\frac{1}{(2/3)^4}$.

Let's analyze the given expression and each option using the rules of exponents.

The given expression is $\frac{1}{(\frac{2}{3})^4}$.

Consider Option (A): $(2/3)^{-4}$.

Using the rule for negative exponents, $a^{-n} = \frac{1}{a^n}$, we can rewrite $(2/3)^{-4}$ as:

$(\frac{2}{3})^{-4} = \frac{1}{(\frac{2}{3})^4}$

This matches the original expression exactly. So, Option (A) is equal to the given expression.

Consider Option (B): $(3/2)^4$.

We can relate negative exponents of fractions to positive exponents of the reciprocal using the rule $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$.

From Option (A), we know $\frac{1}{(\frac{2}{3})^4} = (\frac{2}{3})^{-4}$.

Applying the rule $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$ to $(\frac{2}{3})^{-4}$, with $a=2$, $b=3$, and $n=4$, we get:

$(\frac{2}{3})^{-4} = (\frac{3}{2})^4$

This shows that $(\frac{3}{2})^4$ is also equal to $(\frac{2}{3})^{-4}$, which is equal to the original expression. So, Option (B) is equal to the given expression.

Consider Option (C): $\frac{3^4}{2^4}$.

Using the power of a quotient rule, $(\frac{a}{b})^n = \frac{a^n}{b^n}$, we can simplify Option (B):

$(\frac{3}{2})^4 = \frac{3^4}{2^4}$

Since $(\frac{3}{2})^4$ is equal to the original expression (as shown for Option B), and $(\frac{3}{2})^4$ is equal to $\frac{3^4}{2^4}$, it follows that $\frac{3^4}{2^4}$ is also equal to the original expression. So, Option (C) is equal to the given expression.

Since Options (A), (B), and (C) are all equivalent to the original expression $\frac{1}{(2/3)^4}$, the statement "All of the above" is correct.

The correct option is (D).

To write expressions with a positive exponent, we use the rules for negative exponents.

(a) Write $5^{-2}$ with a positive exponent.

We use the rule $a^{-n} = \frac{1}{a^n}$ for any non-zero number $a$ and positive integer $n$.

Here, $a=5$ and $n=2$.

Applying the rule:

$5^{-2} = \frac{1}{5^2}$

The expression $5^{-2}$ written with a positive exponent is $\frac{1}{5^2}$.

(b) Write $(1/3)^{-4}$ with a positive exponent.

We use the rule $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$ for any non-zero rational number $\frac{a}{b}$ and positive integer $n$.

Here, $a=1$, $b=3$, and $n=4$.

Applying the rule:

$(\frac{1}{3})^{-4} = (\frac{3}{1})^4$

Simplifying the base $\frac{3}{1} = 3$:

$(\frac{3}{1})^4 = 3^4$

The expression $(1/3)^{-4}$ written with a positive exponent is $3^4$.

To evaluate the given expressions, we use the rule for negative exponents, which states that for any non-zero number $a$ and positive integer $n$, $a^{-n} = \frac{1}{a^n}$.

(a) Evaluate $3^{-2}$.

Here, $a=3$ and $n=2$. Applying the rule $a^{-n} = \frac{1}{a^n}$:

$3^{-2} = \frac{1}{3^2}$

Now, we evaluate the power in the denominator:

$3^2 = 3 \times 3 = 9$

Substitute this value back into the fraction:

$\frac{1}{3^2} = \frac{1}{9}$

So, the value of $3^{-2}$ is $\frac{1}{9}$.

(b) Evaluate $(-2)^{-3}$.

Here, $a=-2$ and $n=3$. Applying the rule $a^{-n} = \frac{1}{a^n}$:

$(-2)^{-3} = \frac{1}{(-2)^3}$

Now, we evaluate the power in the denominator:

$(-2)^3 = (-2) \times (-2) \times (-2)$

$(-2) \times (-2) = 4$

$4 \times (-2) = -8$

So, $(-2)^3 = -8$.

Substitute this value back into the fraction:

$\frac{1}{(-2)^3} = \frac{1}{-8}$

This can also be written as $-\frac{1}{8}$.

So, the value of $(-2)^{-3}$ is $-\frac{1}{8}$.

To simplify the expression $5^3 \times 5^{-7}$, we use the product rule of exponents.

The product rule states that when multiplying exponential expressions with the same base, we keep the base and add the exponents. Mathematically, $a^m \times a^n = a^{m+n}$.

In this expression, the base is $a=5$, the first exponent is $m=3$, and the second exponent is $n=-7$.

Applying the product rule:

$5^3 \times 5^{-7} = 5^{3 + (-7)}$

Simplify the exponent:

$3 + (-7) = 3 - 7 = -4$

So, the simplified expression is $5^{-4}$.

If we need to write the answer with a positive exponent, we can use the rule $a^{-n} = \frac{1}{a^n}$:

$5^{-4} = \frac{1}{5^4}$

The simplified form using laws of exponents is $5^{-4}$ (or $\frac{1}{5^4}$ if a positive exponent is required, though the question only asks to simplify).

Given Expression:

$a^{-5} \times a^{10}$

Law of Exponents Used:

When multiplying exponents with the same base, we add the powers. The general form is:

$a^m \times a^n = a^{m+n}$

Simplification:

Using the law $a^m \times a^n = a^{m+n}$ with $m = -5$ and $n = 10$, we get:

$a^{-5} \times a^{10} = a^{-5 + 10}$

$a^{-5 + 10} = a^{5}$

Final Answer:

The simplified form of $a^{-5} \times a^{10}$ is $a^{5}$.

$a^{-5} \times a^{10} = a^{5}$

Given Expression:

$7^{-5} \div 7^{-2}$

Law of Exponents Used:

When dividing exponents with the same base, we subtract the powers. The general form is:

$a^m \div a^n = a^{m-n}$

Simplification:

Using the law $a^m \div a^n = a^{m-n}$ with $a = 7$, $m = -5$, and $n = -2$, we get:

$7^{-5} \div 7^{-2} = 7^{-5 - (-2)}$

$7^{-5 - (-2)} = 7^{-5 + 2}$

$7^{-5 + 2} = 7^{-3}$

Alternatively, we can write $7^{-3}$ as $\frac{1}{7^3}$.

$\frac{1}{7^3} = \frac{1}{7 \times 7 \times 7} = \frac{1}{343}$

Final Answer:

The simplified form of $7^{-5} \div 7^{-2}$ is $7^{-3}$ or $\frac{1}{343}$.

$7^{-5} \div 7^{-2} = 7^{-3} = \frac{1}{343}$

Given Expression:

$(3^{-2})^4$

Law of Exponents Used:

When raising a power to another power, we multiply the exponents. The general form is:

$(a^m)^n = a^{m \times n}$

Simplification:

Using the law $(a^m)^n = a^{m \times n}$ with $a = 3$, $m = -2$, and $n = 4$, we get:

$(3^{-2})^4 = 3^{-2 \times 4}$

$3^{-2 \times 4} = 3^{-8}$

Alternatively, we can write $3^{-8}$ with a positive exponent using the rule $a^{-n} = \frac{1}{a^n}$:

$3^{-8} = \frac{1}{3^8}$

We can calculate the value of $3^8$:

$3^8 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 9 \times 9 \times 9 = 81 \times 81 = 6561$

So, $\frac{1}{3^8} = \frac{1}{6561}$.

Final Answer:

The simplified form of $(3^{-2})^4$ is $3^{-8}$ or $\frac{1}{6561}$.

$(3^{-2})^4 = 3^{-8} = \frac{1}{6561}$

Given Expression:

$(2 \times 3)^{-3}$

Laws of Exponents Used:

1. When a product is raised to a power, each factor is raised to that power: $(ab)^n = a^n b^n$.

2. A term with a negative exponent can be written with a positive exponent in the denominator: $a^{-n} = \frac{1}{a^n}$.

Simplification using Law 1:

Using the law $(ab)^n = a^n b^n$ with $a=2$, $b=3$, and $n=-3$, we get:

$(2 \times 3)^{-3} = 2^{-3} \times 3^{-3}$

Now, using the law $a^{-n} = \frac{1}{a^n}$ for each term:

$2^{-3} = \frac{1}{2^3}$

$3^{-3} = \frac{1}{3^3}$

So, $2^{-3} \times 3^{-3} = \frac{1}{2^3} \times \frac{1}{3^3}$

Calculate the values of the powers:

$2^3 = 2 \times 2 \times 2 = 8$

$3^3 = 3 \times 3 \times 3 = 27$

Substitute these values back:

$\frac{1}{8} \times \frac{1}{27} = \frac{1 \times 1}{8 \times 27} = \frac{1}{216}$

Alternate Simplification (calculating the base first):

First, calculate the value inside the parenthesis:

$2 \times 3 = 6$

So the expression becomes $6^{-3}$.

Now, use the law $a^{-n} = \frac{1}{a^n}$ with $a=6$ and $n=3$:

$6^{-3} = \frac{1}{6^3}$

Calculate the value of the power:

$6^3 = 6 \times 6 \times 6 = 36 \times 6 = 216$

So, $\frac{1}{6^3} = \frac{1}{216}$

Final Answer:

The simplified form of $(2 \times 3)^{-3}$ is $\frac{1}{216}$.

$(2 \times 3)^{-3} = \frac{1}{216}$

Given Expression:

$(\frac{2}{5})^{-2}$

Laws of Exponents Used:

1. $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$

2. $(\frac{a}{b})^n = \frac{a^n}{b^n}$

Simplification:

Using the law $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$ with $a=2$, $b=5$, and $n=2$, we invert the fraction and change the sign of the exponent:

$(\frac{2}{5})^{-2} = (\frac{5}{2})^2$

Now, using the law $(\frac{a}{b})^n = \frac{a^n}{b^n}$:

$(\frac{5}{2})^2 = \frac{5^2}{2^2}$

Calculate the powers:

$5^2 = 5 \times 5 = 25$

$2^2 = 2 \times 2 = 4$

Substitute the calculated values:

$\frac{5^2}{2^2} = \frac{25}{4}$

Final Answer:

The simplified form of $(\frac{2}{5})^{-2}$ is $\frac{25}{4}$.

$(\frac{2}{5})^{-2} = \frac{25}{4}$

Given Expression:

$(\frac{1}{2})^{-2} + (\frac{1}{3})^{-2} + (\frac{1}{4})^{-2}$

Law of Exponents Used:

We use the law for negative exponents of fractions: $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$.

Evaluation of Each Term:

Apply the law to the first term:

$(\frac{1}{2})^{-2} = (\frac{2}{1})^2 = 2^2 = 4$

Apply the law to the second term:

$(\frac{1}{3})^{-2} = (\frac{3}{1})^2 = 3^2 = 9$

Apply the law to the third term:

$(\frac{1}{4})^{-2} = (\frac{4}{1})^2 = 4^2 = 16$

Adding the Results:

Now, we add the evaluated terms:

$4 + 9 + 16$

Adding the numbers:

$4 + 9 = 13$

$13 + 16 = 29$

Final Answer:

The value of the expression $(\frac{1}{2})^{-2} + (\frac{1}{3})^{-2} + (\frac{1}{4})^{-2}$ is 29.

$(\frac{1}{2})^{-2} + (\frac{1}{3})^{-2} + (\frac{1}{4})^{-2} = 4 + 9 + 16 = 29$

Given Expression:

$(5^0 + 6^0)^{-1}$

Laws of Exponents Used:

1. Any non-zero number raised to the power of zero is 1: $a^0 = 1$ (where $a \neq 0$).

2. A term with a negative exponent can be written with a positive exponent in the denominator: $a^{-n} = \frac{1}{a^n}$.

Simplification Steps:

First, evaluate the terms inside the parenthesis using the law $a^0 = 1$:

$5^0 = 1$

$6^0 = 1$

Substitute these values back into the expression:

$(1 + 1)^{-1}$

Now, perform the addition inside the parenthesis:

$1 + 1 = 2$

The expression becomes:

$(2)^{-1}$

Finally, apply the law for negative exponents $a^{-n} = \frac{1}{a^n}$ with $a=2$ and $n=1$:

$2^{-1} = \frac{1}{2^1} = \frac{1}{2}$

Final Answer:

The simplified form of $(5^0 + 6^0)^{-1}$ is $\frac{1}{2}$.

$(5^0 + 6^0)^{-1} = \frac{1}{2}$

Standard Form (Scientific Notation):

A number is written in standard form if it is expressed as $m \times 10^n$, where $1 \leq m < 10$ and $n$ is an integer.

(a) Express $0.0000056$ in Standard Form:

The given number is $0.0000056$.

To write this in standard form, we need to move the decimal point to the right of the first non-zero digit, which is 5.

The number becomes $5.6$.

We moved the decimal point 6 places to the right (from its original position after 0 to between 5 and 6).

Since the original number is less than 1, the exponent of 10 will be negative.

The exponent is -6.

So, $0.0000056 = 5.6 \times 10^{-6}$.

(b) Express $0.00001234$ in Standard Form:

The given number is $0.00001234$.

To write this in standard form, we need to move the decimal point to the right of the first non-zero digit, which is 1.

The number becomes $1.234$.

We moved the decimal point 5 places to the right (from its original position after 0 to between 1 and 2).

Since the original number is less than 1, the exponent of 10 will be negative.

The exponent is -5.

So, $0.00001234 = 1.234 \times 10^{-5}$.

Final Answers:

(a) $0.0000056 = 5.6 \times 10^{-6}$

(b) $0.00001234 = 1.234 \times 10^{-5}$

Converting Standard Form to Usual Form:

To convert a number from standard form ($m \times 10^n$) to usual form:

If $n$ is positive, move the decimal point $n$ places to the right.

If $n$ is negative, move the decimal point $|n|$ places to the left.

(a) Express $3.02 \times 10^{-6}$ in Usual Form:

The given number is $3.02 \times 10^{-6}$.

The exponent is -6. This means we need to move the decimal point in $3.02$ six places to the left.

Starting with $3.02$, move the decimal point 6 places left:

$3.02 \xrightarrow{\text{1 place left}} 0.302 \xrightarrow{\text{2 places left}} 0.0302 \xrightarrow{\text{3 places left}} 0.00302 \xrightarrow{\text{4 places left}} 0.000302 \xrightarrow{\text{5 places left}} 0.0000302 \xrightarrow{\text{6 places left}} 0.00000302$

So, $3.02 \times 10^{-6} = 0.00000302$.

(b) Express $5.8 \times 10^{-4}$ in Usual Form:

The given number is $5.8 \times 10^{-4}$.

The exponent is -4. This means we need to move the decimal point in $5.8$ four places to the left.

Starting with $5.8$, move the decimal point 4 places left:

$5.8 \xrightarrow{\text{1 place left}} 0.58 \xrightarrow{\text{2 places left}} 0.058 \xrightarrow{\text{3 places left}} 0.0058 \xrightarrow{\text{4 places left}} 0.00058$

So, $5.8 \times 10^{-4} = 0.00058$.

Final Answers:

(a) $3.02 \times 10^{-6} = 0.00000302$

(b) $5.8 \times 10^{-4} = 0.00058$

Given Size:

Size of a plant cell = $0.00001275$ meters.

Standard Form (Scientific Notation):

A number in standard form is written as $m \times 10^n$, where $1 \leq m < 10$ and $n$ is an integer.

Converting to Standard Form:

The given number is $0.00001275$.

To express this number in standard form, we need to move the decimal point so that there is only one non-zero digit before the decimal point.

The first non-zero digit is 1. We move the decimal point from its current position to after the digit 1.

Original position: $0.00001275$

New position: $1.275$

We count the number of places the decimal point has moved to the right. It moved 5 places to the right.

Since the original number is less than 1, the exponent of 10 will be negative, and its value will be the number of places the decimal point was moved.

Number of places moved = 5.

Exponent of 10 = -5.

Therefore, the number in standard form is $1.275 \times 10^{-5}$.

Include the unit: $1.275 \times 10^{-5}$ meters.

Final Answer:

The size of a plant cell in standard form is $1.275 \times 10^{-5}$ meters.

$0.00001275 \text{ m} = 1.275 \times 10^{-5} \text{ m}$

Given Thickness:

Thickness of a piece of paper = $0.0016$ cm.

Standard Form (Scientific Notation):

A number in standard form is written as $m \times 10^n$, where $1 \leq m < 10$ and $n$ is an integer.

Converting to Standard Form:

The given number is $0.0016$.

To express this number in standard form, we need to move the decimal point so that there is only one non-zero digit before the decimal point.

The first non-zero digit is 1. We move the decimal point from its current position to after the digit 1.

Original position: $0.0016$

New position: $1.6$

We count the number of places the decimal point has moved to the right. It moved 3 places to the right.

Since the original number is less than 1, the exponent of 10 will be negative, and its value will be the number of places the decimal point was moved.

Number of places moved = 3.

Exponent of 10 = -3.

Therefore, the number in standard form is $1.6 \times 10^{-3}$.

Include the unit: $1.6 \times 10^{-3}$ cm.

Final Answer:

The thickness of a piece of paper in standard form is $1.6 \times 10^{-3}$ cm.

$0.0016 \text{ cm} = 1.6 \times 10^{-3} \text{ cm}$

Given Numbers:

Number 1: $1.5 \times 10^8$

Number 2: $2.3 \times 10^7$

Comparison Method:

To compare numbers in standard form ($m \times 10^n$), we first compare the exponents of 10 ($n$).

The number with the larger positive exponent of 10 is the greater number, assuming the coefficient $m$ is within the standard range ($1 \leq m < 10$).

Comparing Exponents:

For the first number, the exponent of 10 is 8.

For the second number, the exponent of 10 is 7.

Comparing the exponents, we have $8 > 7$.

Conclusion based on Exponents:

Since the exponent of 10 in $1.5 \times 10^8$ (which is 8) is greater than the exponent of 10 in $2.3 \times 10^7$ (which is 7), the first number is greater.

Thus, $1.5 \times 10^8 > 2.3 \times 10^7$.

Verification (by converting to usual form):

$1.5 \times 10^8 = 1.5 \times 100,000,000 = 150,000,000$

$2.3 \times 10^7 = 2.3 \times 10,000,000 = 23,000,000$

Comparing $150,000,000$ and $23,000,000$, it is clear that $150,000,000$ is greater than $23,000,000$.

Final Answer:

Comparing $1.5 \times 10^8$ and $2.3 \times 10^7$, the number $1.5 \times 10^8$ is greater.

Given Numbers:

Number 1: $4.1 \times 10^{-5}$

Number 2: $3.9 \times 10^{-6}$

Comparison Method:

To compare numbers in standard form ($m \times 10^n$) with different exponents of 10, we first compare the exponents of 10 ($n$).

When the exponents are negative, the number with the exponent that is further from zero (more negative) is the smaller number.

Comparing Exponents:

For the first number, the exponent of 10 is -5.

For the second number, the exponent of 10 is -6.

Comparing the exponents, we have $-6 < -5$.

Conclusion based on Exponents:

Since the exponent of 10 in $3.9 \times 10^{-6}$ (which is -6) is less than the exponent of 10 in $4.1 \times 10^{-5}$ (which is -5), the second number is smaller.

Thus, $3.9 \times 10^{-6} < 4.1 \times 10^{-5}$.

Verification (by converting to usual form):

$4.1 \times 10^{-5} = 0.000041$

$3.9 \times 10^{-6} = 0.0000039$

Comparing $0.000041$ and $0.0000039$, it is clear that $0.0000039$ is smaller than $0.000041$.

Final Answer:

Comparing $4.1 \times 10^{-5}$ and $3.9 \times 10^{-6}$, the number $3.9 \times 10^{-6}$ is smaller.

Given Expression:

$(3^{-7} \div 3^{-10}) \times 3^{-5}$

Laws of Exponents Used:

1. Division with the same base: $a^m \div a^n = a^{m-n}$

2. Multiplication with the same base: $a^m \times a^n = a^{m+n}$

3. Negative exponent: $a^{-n} = \frac{1}{a^n}$

Simplification Steps:

First, simplify the expression inside the parenthesis using the division law ($a^m \div a^n = a^{m-n}$):

$3^{-7} \div 3^{-10} = 3^{-7 - (-10)}$

$3^{-7 - (-10)} = 3^{-7 + 10}$

$3^{-7 + 10} = 3^3$

Now, substitute this result back into the original expression:

$(3^3) \times 3^{-5}$

Next, use the multiplication law ($a^m \times a^n = a^{m+n}$):

$3^3 \times 3^{-5} = 3^{3 + (-5)}$

$3^{3 + (-5)} = 3^{3 - 5}$

$3^{3 - 5} = 3^{-2}$

The expression is now simplified to $3^{-2}$. The question asks for the answer in exponential form with a positive exponent.

Use the negative exponent law ($a^{-n} = \frac{1}{a^n}$):

$3^{-2} = \frac{1}{3^2}$

Final Answer:

The simplified expression in exponential form with a positive exponent is $\frac{1}{3^2}$.

$(3^{-7} \div 3^{-10}) \times 3^{-5} = \frac{1}{3^2}$

Given Equation:

$5^m \times 5^{-3} = 5^5$

Law of Exponents Used:

When multiplying exponents with the same base, we add the powers: $a^p \times a^q = a^{p+q}$.

Solution Steps:

Apply the multiplication law of exponents to the left side of the equation:

$5^m \times 5^{-3} = 5^{m + (-3)} = 5^{m-3}$

Now, the equation becomes:

$5^{m-3} = 5^5$

Since the bases on both sides of the equation are equal (and are not 0, 1, or -1), the exponents must be equal for the equation to hold true.

Equate the exponents:

Solve for $m$ by adding 3 to both sides of the equation:

Verification:

Substitute $m=8$ back into the original equation:

$5^8 \times 5^{-3}$

Using the multiplication law: $5^{8 + (-3)} = 5^{8 - 3} = 5^5$.

The left side equals the right side, so the value of $m=8$ is correct.

Final Answer:

The value of $m$ is 8.

Given Expression:

$(2^5 \div 2^8)^{-7} \times 2^{-5}$

Laws of Exponents Used:

1. Division with the same base: $a^m \div a^n = a^{m-n}$

2. Raising a power to another power: $(a^m)^n = a^{m \times n}$

3. Multiplication with the same base: $a^m \times a^n = a^{m+n}$

Simplification Steps:

First, simplify the expression inside the parenthesis using the division law ($a^m \div a^n = a^{m-n}$):

$2^5 \div 2^8 = 2^{5-8}$

$2^{5-8} = 2^{-3}$

Now, substitute this result back into the original expression:

$(2^{-3})^{-7} \times 2^{-5}$

Next, apply the law for raising a power to another power ($(a^m)^n = a^{m \times n}$) to the first part:

$(2^{-3})^{-7} = 2^{-3 \times -7}$

$2^{-3 \times -7} = 2^{21}$

Substitute this result back into the expression:

$2^{21} \times 2^{-5}$

Finally, use the multiplication law ($a^m \times a^n = a^{m+n}$):

$2^{21} \times 2^{-5} = 2^{21 + (-5)}$

$2^{21 + (-5)} = 2^{21 - 5}$

$2^{21 - 5} = 2^{16}$

The expression is now in the form $2^n$, where $n=16$.

Final Answer:

The simplified expression in the form $2^n$ is $2^{16}$.

$(2^5 \div 2^8)^{-7} \times 2^{-5} = 2^{16}$

Given Expression:

$(\frac{p}{q})^{-3} \times (\frac{q}{p})^{-2}$

Laws of Exponents Used:

1. $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$

2. $(\frac{a}{b})^n = \frac{a^n}{b^n}$

3. $\frac{a^m}{a^n} = a^{m-n}$

4. $a^{-n} = \frac{1}{a^n}$

Simplification Steps:

Apply the first law, $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$, to each term:

$(\frac{p}{q})^{-3} = (\frac{q}{p})^3$

$(\frac{q}{p})^{-2} = (\frac{p}{q})^2$

Substitute these back into the expression:

$(\frac{q}{p})^3 \times (\frac{p}{q})^2$

Now, apply the second law, $(\frac{a}{b})^n = \frac{a^n}{b^n}$:

$\frac{q^3}{p^3} \times \frac{p^2}{q^2}$

Multiply the fractions:

$\frac{q^3 \times p^2}{p^3 \times q^2}$

Rearrange the terms:

$\frac{q^3}{q^2} \times \frac{p^2}{p^3}$

Apply the third law, $\frac{a^m}{a^n} = a^{m-n}$, to each fraction:

$\frac{q^3}{q^2} = q^{3-2} = q^1 = q$

$\frac{p^2}{p^3} = p^{2-3} = p^{-1}$

Now multiply these simplified terms:

$q \times p^{-1}$

Using the fourth law, $a^{-n} = \frac{1}{a^n}$, rewrite $p^{-1}$:

$p^{-1} = \frac{1}{p^1} = \frac{1}{p}$

So, the expression is:

$q \times \frac{1}{p} = \frac{q}{p}$

Final Answer:

The simplified form of the expression is $\frac{q}{p}$.

$(\frac{p}{q})^{-3} \times (\frac{q}{p})^{-2} = \frac{q}{p}$

Given Expression:

$\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$

Laws of Exponents Used:

1. $(ab)^n = a^n b^n$

2. $(a/b)^n = a^n / b^n$

3. $a^m \times a^n = a^{m+n}$

4. $a^m \div a^n = a^{m-n}$ or $\frac{a^m}{a^n} = a^{m-n}$

5. $a^0 = 1$ (where $a \neq 0$)

6. $a^{-n} = \frac{1}{a^n}$

Simplification Steps:

Rewrite the numbers 10, 125, and 6 in terms of their prime factors:

$10 = 2 \times 5$

$125 = 5 \times 5 \times 5 = 5^3$

$6 = 2 \times 3$

Substitute these into the expression:

$\frac{3^{-5} \times (2 \times 5)^{-5} \times 5^3}{5^{-7} \times (2 \times 3)^{-5}}$

Apply the law $(ab)^n = a^n b^n$ to $(2 \times 5)^{-5}$ and $(2 \times 3)^{-5}$:

$\frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^3}{5^{-7} \times 2^{-5} \times 3^{-5}}$

Combine terms with the same base in the numerator using $a^m \times a^n = a^{m+n}$ ($5^{-5} \times 5^3 = 5^{-5+3} = 5^{-2}$):

$\frac{3^{-5} \times 2^{-5} \times 5^{-2}}{5^{-7} \times 2^{-5} \times 3^{-5}}$

Rearrange the terms in the denominator to group by base:

$\frac{3^{-5} \times 2^{-5} \times 5^{-2}}{3^{-5} \times 2^{-5} \times 5^{-7}}$

Now, apply the law $\frac{a^m}{a^n} = a^{m-n}$ for each base:

For base 3: $\frac{3^{-5}}{3^{-5}} = 3^{-5 - (-5)} = 3^{-5 + 5} = 3^0$

For base 2: $\frac{2^{-5}}{2^{-5}} = 2^{-5 - (-5)} = 2^{-5 + 5} = 2^0$

For base 5: $\frac{5^{-2}}{5^{-7}} = 5^{-2 - (-7)} = 5^{-2 + 7} = 5^5$

Multiply the simplified terms:

$3^0 \times 2^0 \times 5^5$

Using the law $a^0 = 1$:

$1 \times 1 \times 5^5 = 5^5$

Final Answer:

The simplified form of the expression is $5^5$.

$\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}} = 5^5$

If the final value is required:

$5^5 = 5 \times 5 \times 5 \times 5 \times 5 = 3125$

Given Distances:

Distance from Earth to Sun = $1.496 \times 10^{11}$ meters

Distance from Earth to Moon = $3.84 \times 10^8$ meters

Comparing the Distances:

To compare numbers in standard form ($m \times 10^n$), we first compare the exponents of 10 ($n$).

The exponent for the Earth-Sun distance is 11.

The exponent for the Earth-Moon distance is 8.

Comparing the exponents, we have $11 > 8$.

Since the exponent of 10 for the Earth-Sun distance is greater than the exponent of 10 for the Earth-Moon distance, the Earth-Sun distance is greater.

Finding the Factor:

To find by what factor the Earth-Sun distance is greater than the Earth-Moon distance, we divide the greater distance by the smaller distance.

Factor = $\frac{\text{Distance from Earth to Sun}}{\text{Distance from Earth to Moon}}$

Factor = $\frac{1.496 \times 10^{11}}{3.84 \times 10^8}$

We can separate the coefficients and the powers of 10:

Factor = $(\frac{1.496}{3.84}) \times (\frac{10^{11}}{10^8})$

Calculate the division of the powers of 10 using the law $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{10^{11}}{10^8} = 10^{11-8} = 10^3$

Calculate the division of the coefficients:

$\frac{1.496}{3.84}$

Let's approximate this value. $1.496 \approx 1.5$ and $3.84 \approx 4$. $\frac{1.5}{4} = \frac{15}{40} = \frac{3}{8} = 0.375$.

For a more precise calculation:

$\frac{1.496}{3.84} \approx 0.3896$

So, Factor $\approx 0.3896 \times 10^3$

Convert $0.3896 \times 10^3$ to usual form by moving the decimal point 3 places to the right:

$0.3896 \times 1000 = 389.6$

Thus, the Earth-Sun distance is approximately 389.6 times the Earth-Moon distance.

Conclusion:

The distance from the Earth to the Sun ($1.496 \times 10^{11}$ m) is greater than the distance from the Earth to the Moon ($3.84 \times 10^8$ m).

The Earth-Sun distance is approximately 389.6 times greater than the Earth-Moon distance.

Final Answer:

The distance from the Earth to the Sun is greater.

It is greater by a factor of approximately 389.6.

Given Expression:

$\frac{(2^0 + 3^{-1}) \times 3^2}{4^{-2}}$

Laws of Exponents Used:

1. Any non-zero number raised to the power of zero is 1: $a^0 = 1$ (where $a \neq 0$).

2. A term with a negative exponent can be written as its reciprocal with a positive exponent: $a^{-n} = \frac{1}{a^n}$.

3. Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{A}{B/C} = A \times \frac{C}{B}$.

Simplification Steps:

Evaluate each term with an exponent:

$2^0 = 1$ (Using law 1)

$3^{-1} = \frac{1}{3^1} = \frac{1}{3}$ (Using law 2)

$3^2 = 3 \times 3 = 9$

$4^{-2} = \frac{1}{4^2} = \frac{1}{16}$ (Using law 2)

Substitute these values back into the expression:

$\frac{(1 + \frac{1}{3}) \times 9}{\frac{1}{16}}$

Simplify the sum inside the parenthesis in the numerator:

$1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{3+1}{3} = \frac{4}{3}$

Substitute the simplified parenthesis back into the expression:

$\frac{(\frac{4}{3}) \times 9}{\frac{1}{16}}$

Simplify the numerator by multiplying $\frac{4}{3}$ by 9:

$\frac{4}{\cancel{3}} \times \cancelto{3}{9} = 4 \times 3 = 12$

The expression becomes:

$\frac{12}{\frac{1}{16}}$

Now, divide the numerator by the denominator. Dividing by a fraction is equivalent to multiplying by its reciprocal (Using law 3):

$12 \div \frac{1}{16} = 12 \times \frac{16}{1}$

$12 \times 16$

Perform the multiplication:

$12 \times 16 = 192$

Final Answer:

The simplified value of the expression is 192.

$\frac{(2^0 + 3^{-1}) \times 3^2}{4^{-2}} = 192$

Given Conversions:

1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds

Calculation of Total Seconds in One Day:

Number of seconds in one day = (Number of hours in a day) $\times$ (Number of minutes in an hour) $\times$ (Number of seconds in a minute)

Number of seconds = $24 \times 60 \times 60$

First, calculate $60 \times 60$:

$60 \times 60 = 3600$

Now, multiply the result by 24:

$3600 \times 24$

$3600 \times 24 = 86400$

So, there are 86,400 seconds in one day.

Expressing in Standard Form:

Standard form is written as $m \times 10^n$, where $1 \leq m < 10$ and $n$ is an integer.

The number is 86,400.

To write this in standard form, we need to move the decimal point (which is currently after the last zero) to the position after the first non-zero digit (8).

$86400.0 \rightarrow 8.64$

We moved the decimal point 4 places to the left.

Since the original number is greater than 1, the exponent of 10 is positive, and its value is the number of places the decimal point was moved.

Number of places moved = 4.

Exponent of 10 = 4.

Therefore, 86,400 in standard form is $8.64 \times 10^4$.

Final Answer:

The number of seconds in one day in standard form is $8.64 \times 10^4$.

1 day = $8.64 \times 10^4$ seconds

Meaning of Powers with Negative Exponents:

Powers with negative exponents represent the reciprocal of the base raised to the corresponding positive exponent. They indicate that the base is in the denominator of a fraction, raised to a positive power.

Think about the pattern in powers of a number, like powers of 10:

$10^3 = 1000$

$10^2 = 100$

$10^1 = 10$

$10^0 = 1$ (Dividing by 10 each time)

Continuing this pattern, to get the next term, we again divide by 10:

$10^{-1} = 1 \div 10 = \frac{1}{10}$

$10^{-2} = \frac{1}{10} \div 10 = \frac{1}{10 \times 10} = \frac{1}{10^2}$

$10^{-3} = \frac{1}{10^2} \div 10 = \frac{1}{10^2 \times 10} = \frac{1}{10^3}$

This pattern shows that a negative exponent corresponds to taking the reciprocal.

Definition of $a^{-n}$:

For any non-zero base $a$ and any positive integer $n$, the power with a negative exponent $a^{-n}$ is defined as the reciprocal of $a^n$.

Here, $a$ is the base and $-n$ is the exponent. The condition $a \neq 0$ is necessary because division by zero is undefined.

Examples:

1. Consider the expression $3^{-2}$.

Here, the base is $a=3$ and the positive integer is $n=2$.

Using the definition, $3^{-2} = \frac{1}{3^2}$.

Calculating the value: $\frac{1}{3^2} = \frac{1}{3 \times 3} = \frac{1}{9}$.

So, $3^{-2} = \frac{1}{9}$.

2. Consider the expression $5^{-1}$.

Here, the base is $a=5$ and the positive integer is $n=1$.

Using the definition, $5^{-1} = \frac{1}{5^1}$.

Calculating the value: $\frac{1}{5^1} = \frac{1}{5}$.

So, $5^{-1} = \frac{1}{5}$.

3. Consider the expression $x^{-4}$ (where $x \neq 0$).

Here, the base is $a=x$ and the positive integer is $n=4$.

Using the definition, $x^{-4} = \frac{1}{x^4}$.

This form is typically the simplified expression for $x^{-4}$.

So, $x^{-4} = \frac{1}{x^4}$.

The laws of exponents hold true regardless of whether the exponents are positive, negative, or zero, provided the base is non-zero. We can verify these laws using the definition of negative exponents, $a^{-n} = \frac{1}{a^n}$.

Law 1: Product of Powers with the Same Base

Statement: When multiplying powers with the same non-zero base, we add the exponents.

Explanation: This law combines powers by summing their exponents. This property arises from the definition of exponents as repeated multiplication. When exponents are negative, the definition $a^{-n} = \frac{1}{a^n}$ allows us to rewrite the terms and apply the basic multiplication rules for fractions and powers.

Let's look at how it works when exponents are negative using the definition. Suppose $m$ is positive and $n$ is negative, say $n = -k$ where $k$ is positive. Then:

$a^m \times a^{-k} = a^m \times \frac{1}{a^k} = \frac{a^m}{a^k} = a^{m-k}$

Since $n = -k$, $m-k = m+(-n) = m+n$. So, $a^m \times a^n = a^{m+n}$. The law holds.

Examples for Law 1 ($a^m \times a^n = a^{m+n}$):

Example 1: Simplify $2^5 \times 2^{-3}$.

Using the law $a^m \times a^n = a^{m+n}$ with $a=2$, $m=5$, and $n=-3$:

$2^5 \times 2^{-3} = 2^{5 + (-3)}$

$= 2^{5-3}$

$= 2^2$

$= 4$

Alternatively, using the definition $a^{-n} = \frac{1}{a^n}$:

$2^5 \times 2^{-3} = 2^5 \times \frac{1}{2^3}$

$= \frac{2^5}{2^3}$

Using the division law $\frac{a^m}{a^n} = a^{m-n}$ (which we will discuss next, but is consistent):

$= 2^{5-3}$

$= 2^2 = 4$

Both methods yield the same result.

Example 2: Simplify $a^{-2} \times a^{-4}$.

Using the law $a^m \times a^n = a^{m+n}$ with $a=a$, $m=-2$, and $n=-4$:

$a^{-2} \times a^{-4} = a^{-2 + (-4)}$

$= a^{-2 - 4}$

$= a^{-6}$

We can write the result with a positive exponent if required: $a^{-6} = \frac{1}{a^6}$.

Alternatively, using the definition $a^{-n} = \frac{1}{a^n}$:

$a^{-2} \times a^{-4} = \frac{1}{a^2} \times \frac{1}{a^4}$

$= \frac{1 \times 1}{a^2 \times a^4}$

$= \frac{1}{a^{2+4}}$

$= \frac{1}{a^6}$

$= a^{-6}$

Law 2: Quotient of Powers with the Same Base

Statement: When dividing powers with the same non-zero base, we subtract the exponents.

Explanation: This law is derived from the inverse relationship between multiplication and division. When exponents are negative, the definition $a^{-n} = \frac{1}{a^n}$ helps in understanding its application. Dividing by $a^n$ is the same as multiplying by $a^{-n}$. Thus, $a^m \div a^n = a^m \times a^{-n} = a^{m+(-n)} = a^{m-n}$ (using Law 1). This confirms the law for negative exponents as well.

Examples for Law 2 ($a^m \div a^n = a^{m-n}$):

Example 1: Simplify $5^3 \div 5^{-2}$.

Using the law $a^m \div a^n = a^{m-n}$ with $a=5$, $m=3$, and $n=-2$:

$5^3 \div 5^{-2} = 5^{3 - (-2)}$

$= 5^{3 + 2}$

$= 5^5$

$= 3125$

Alternatively, using the definition $a^{-n} = \frac{1}{a^n}$:

$5^3 \div 5^{-2} = 5^3 \div \frac{1}{5^2}$

Dividing by a fraction is multiplying by its reciprocal:

$= 5^3 \times 5^2$

Using the multiplication law $a^m \times a^n = a^{m+n}$:

$= 5^{3+2}$

$= 5^5 = 3125$

Example 2: Simplify $y^{-6} \div y^{-2}$.

Using the law $a^m \div a^n = a^{m-n}$ with $a=y$, $m=-6$, and $n=-2$:

$y^{-6} \div y^{-2} = y^{-6 - (-2)}$

$= y^{-6 + 2}$

$= y^{-4}$

We can write the result with a positive exponent if required: $y^{-4} = \frac{1}{y^4}$.

Alternatively, using the definition $a^{-n} = \frac{1}{a^n}$:

$y^{-6} \div y^{-2} = \frac{1}{y^6} \div \frac{1}{y^2}$

Dividing by a fraction is multiplying by its reciprocal:

$= \frac{1}{y^6} \times \frac{y^2}{1}$

$= \frac{y^2}{y^6}$

Using the definition again: $\frac{y^2}{y^6} = \frac{1}{y^{6-2}} = \frac{1}{y^4}$

Which is $y^{-4}$.

The laws of exponents involving powers of powers and powers of products also apply when the exponents are negative integers, provided the bases are non-zero.

Law 1: Power of a Power

Statement: When raising a power to another power, we multiply the exponents.

This law holds for any non-zero base $a$ and any integers $m$ and $n$. The negative exponent rule $a^{-k} = \frac{1}{a^k}$ is consistent with this law. For example, $(a^m)^{-n} = \frac{1}{(a^m)^n} = \frac{1}{a^{mn}} = a^{-mn}$. Since $(-n) \times m = -mn$, the law $(a^m)^n = a^{mn}$ works when $n$ is negative.

Examples for Law 1 ($(a^m)^n = a^{mn}$):

Example 1: Simplify $(5^{-2})^3$.

Using the law $(a^m)^n = a^{mn}$ with $a=5$, $m=-2$, and $n=3$:

$(5^{-2})^3 = 5^{-2 \times 3}$

$= 5^{-6}$

Using the definition $a^{-k} = \frac{1}{a^k}$ to write with a positive exponent:

$5^{-6} = \frac{1}{5^6}$

Alternatively, using the definition first:

$(5^{-2})^3 = (\frac{1}{5^2})^3$

Using $(\frac{b}{c})^n = \frac{b^n}{c^n}$:

$= \frac{1^3}{(5^2)^3}$

$= \frac{1}{5^{2 \times 3}}$

$= \frac{1}{5^6}$

$= 5^{-6}$

Both methods yield the same result.

Example 2: Simplify $(y^4)^{-2}$.

Using the law $(a^m)^n = a^{mn}$ with $a=y$, $m=4$, and $n=-2$:

$(y^4)^{-2} = y^{4 \times (-2)}$

$= y^{-8}$

Using the definition $a^{-k} = \frac{1}{a^k}$ to write with a positive exponent:

$y^{-8} = \frac{1}{y^8}$

Alternatively, using the definition first:

$(y^4)^{-2} = \frac{1}{(y^4)^2}$

Using $(a^m)^n = a^{mn}$ for positive $n$ in the denominator:

$= \frac{1}{y^{4 \times 2}}$

$= \frac{1}{y^8}$

$= y^{-8}$

Law 2: Power of a Product

Statement: When a product of bases is raised to a power, each base is raised to that power.

This law holds for any non-zero bases $a$ and $b$ and any integer exponent $m$. When $m$ is negative, say $m = -k$ where $k$ is positive, the law still applies. $(ab)^{-k} = \frac{1}{(ab)^k} = \frac{1}{a^k b^k}$. Also, $a^{-k} b^{-k} = \frac{1}{a^k} \times \frac{1}{b^k} = \frac{1}{a^k b^k}$. Both results are the same, showing the law works for negative exponents.

Examples for Law 2 ($(ab)^m = a^m b^m$):

Example 1: Simplify $(3x)^{-1}$.

Using the law $(ab)^m = a^m b^m$ with $a=3$, $b=x$, and $m=-1$:

$(3x)^{-1} = 3^{-1} \times x^{-1}$

Using the definition $a^{-k} = \frac{1}{a^k}$ for each term:

$3^{-1} = \frac{1}{3^1} = \frac{1}{3}$

$x^{-1} = \frac{1}{x^1} = \frac{1}{x}$

So, $3^{-1} \times x^{-1} = \frac{1}{3} \times \frac{1}{x} = \frac{1}{3x}$.

Alternatively, using the definition first:

$(3x)^{-1} = \frac{1}{(3x)^1}$

$= \frac{1}{3x}$

Both methods yield the same result.

Example 2: Simplify $(2y^3)^{-2}$.

Using the law $(ab)^m = a^m b^m$ with $a=2$, $b=y^3$, and $m=-2$:

$(2y^3)^{-2} = 2^{-2} \times (y^3)^{-2}$

Now, use the power of a power law $(a^m)^n = a^{mn}$ on $(y^3)^{-2}$:

$(y^3)^{-2} = y^{3 \times (-2)} = y^{-6}$

So the expression is $2^{-2} \times y^{-6}$.

Using the definition $a^{-k} = \frac{1}{a^k}$ for each term:

$2^{-2} = \frac{1}{2^2} = \frac{1}{4}$

$y^{-6} = \frac{1}{y^6}$

So, $2^{-2} \times y^{-6} = \frac{1}{4} \times \frac{1}{y^6} = \frac{1}{4y^6}$.

Alternatively, using the definition first:

$(2y^3)^{-2} = \frac{1}{(2y^3)^2}$

Using $(ab)^n = a^n b^n$ in the denominator:

$= \frac{1}{2^2 \times (y^3)^2}$

Using $(a^m)^n = a^{mn}$ in the denominator:

$= \frac{1}{4 \times y^{3 \times 2}}$

$= \frac{1}{4y^6}$

Here are the explanations and examples for the two specified laws of exponents, including cases with negative exponents.

Law 1: Power of a Quotient

Statement: When a quotient of two non-zero numbers is raised to a power, the numerator and the denominator are both raised to that power separately.

This law is valid for any non-zero numbers $a$ and $b$, and any integer $m$. When $m$ is a negative integer, say $m = -k$ where $k$ is a positive integer, we can show consistency using the definition of negative exponents:

$(\frac{a}{b})^m = (\frac{a}{b})^{-k}$

Using the definition $(\frac{x}{y})^{-k} = (\frac{y}{x})^k$:

$(\frac{a}{b})^{-k} = (\frac{b}{a})^k$

Using the law for positive exponents $(\frac{y}{x})^k = \frac{y^k}{x^k}$:

$(\frac{b}{a})^k = \frac{b^k}{a^k}$

Now, let's look at $\frac{a^m}{b^m}$ for $m=-k$:

$\frac{a^m}{b^m} = \frac{a^{-k}}{b^{-k}}$

Using the definition $x^{-k} = \frac{1}{x^k}$ for the numerator and denominator:

$\frac{a^{-k}}{b^{-k}} = \frac{\frac{1}{a^k}}{\frac{1}{b^k}}$

Dividing by a fraction is multiplying by its reciprocal:

$\frac{1}{a^k} \times \frac{b^k}{1} = \frac{b^k}{a^k}$

Since $(\frac{a}{b})^{-k} = \frac{b^k}{a^k}$ and $\frac{a^{-k}}{b^{-k}} = \frac{b^k}{a^k}$, the law $(\frac{a}{b})^m = \frac{a^m}{b^m}$ holds for negative exponents $m$ as well.

Examples for Law 1:

Example 1: Simplify $(\frac{2}{3})^{-2}$.

Using the law $(\frac{a}{b})^m = \frac{a^m}{b^m}$ with $a=2$, $b=3$, and $m=-2$:

$(\frac{2}{3})^{-2} = \frac{2^{-2}}{3^{-2}}$

Using the definition $x^{-n} = \frac{1}{x^n}$ for the numerator and denominator:

$\frac{2^{-2}}{3^{-2}} = \frac{\frac{1}{2^2}}{\frac{1}{3^2}}$

Dividing by a fraction is multiplying by its reciprocal:

$= \frac{1}{2^2} \times \frac{3^2}{1}$

$= \frac{3^2}{2^2}$

$= \frac{9}{4}$

Alternatively, using the property $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$ first:

$(\frac{2}{3})^{-2} = (\frac{3}{2})^2$

Using $(\frac{a}{b})^n = \frac{a^n}{b^n}$ for positive $n$:

$= \frac{3^2}{2^2} = \frac{9}{4}$

Both methods give the same result.

Example 2: Simplify $(\frac{x}{y})^{-1}$.

Using the law $(\frac{a}{b})^m = \frac{a^m}{b^m}$ with $a=x$, $b=y$, and $m=-1$:

$(\frac{x}{y})^{-1} = \frac{x^{-1}}{y^{-1}}$

Using the definition $a^{-n} = \frac{1}{a^n}$ for the numerator and denominator:

$\frac{x^{-1}}{y^{-1}} = \frac{\frac{1}{x}}{\frac{1}{y}}$

Dividing by a fraction is multiplying by its reciprocal:

$= \frac{1}{x} \times \frac{y}{1}$

$= \frac{y}{x}$

Law 2: Zero Exponent

Statement: Any non-zero number raised to the power of zero is equal to 1.

This law applies for any base $a$ such that $a \neq 0$.

Explanation: This law can be understood using the division law of exponents. Consider the expression $\frac{a^m}{a^m}$ where $a \neq 0$.

Using the division law $a^p \div a^q = a^{p-q}$ with $p=m$ and $q=m$:

$\frac{a^m}{a^m} = a^{m-m} = a^0$

However, any non-zero quantity divided by itself is equal to 1:

$\frac{a^m}{a^m} = 1$

Therefore, equating the two results, we get $a^0 = 1$. This reasoning holds true regardless of whether $m$ is positive or negative, provided $a \neq 0$.

Examples for Law 2 ($a^0 = 1$):

Example 1: Evaluate $7^0$.

Using the law $a^0 = 1$ with $a=7$ (which is non-zero):

$7^0 = 1$

Example 2: Evaluate $(xy^2)^0$, where $x \neq 0$ and $y \neq 0$.

Using the law $(ab)^m = a^m b^m$, we have $(xy^2)^0 = x^0 (y^2)^0$.

Using the law $(a^m)^n = a^{mn}$, we have $(y^2)^0 = y^{2 \times 0} = y^0$.

So, the expression becomes $x^0 y^0$.

Using the law $a^0 = 1$ for $x \neq 0$ and $y \neq 0$:

$x^0 = 1$ and $y^0 = 1$

Therefore, $x^0 y^0 = 1 \times 1 = 1$.

Alternatively, since $x \neq 0$ and $y \neq 0$, the product $xy^2$ is also non-zero. Let $Z = xy^2$. Then the expression is $Z^0$. Since $Z \neq 0$, directly using the law $a^0 = 1$ with $a=Z$, we get:

$(xy^2)^0 = 1$

Given Expression:

$\frac{(a^5 \times a^{-2})^3}{a^{-4} \times (a^2)^{-1}}$

Laws of Exponents Used:

$a^m \times a^n = a^{m+n}$ (Product of powers)

$(a^m)^n = a^{mn}$ (Power of a power)

$\frac{a^m}{a^n} = a^{m-n}$ (Quotient of powers)

Simplifying the Numerator:

Numerator = $(a^5 \times a^{-2})^3$

First, simplify the expression inside the parenthesis using the product of powers law:

$a^5 \times a^{-2} = a^{5 + (-2)} = a^{5-2} = a^3$

So the numerator becomes $(a^3)^3$.

Now, apply the power of a power law:

$(a^3)^3 = a^{3 \times 3} = a^9$

The simplified numerator is $a^9$.

Simplifying the Denominator:

Denominator = $a^{-4} \times (a^2)^{-1}$

First, simplify $(a^2)^{-1}$ using the power of a power law:

$(a^2)^{-1} = a^{2 \times (-1)} = a^{-2}$

So the denominator becomes $a^{-4} \times a^{-2}$.

Now, use the product of powers law:

$a^{-4} \times a^{-2} = a^{-4 + (-2)} = a^{-4-2} = a^{-6}$

The simplified denominator is $a^{-6}$.

Dividing the Simplified Numerator by the Simplified Denominator:

The expression is now $\frac{a^9}{a^{-6}}$.

Use the quotient of powers law:

$\frac{a^9}{a^{-6}} = a^{9 - (-6)}$

$= a^{9 + 6}$

$= a^{15}$

Final Answer:

The simplified form of the expression is $a^{15}$.

$\frac{(a^5 \times a^{-2})^3}{a^{-4} \times (a^2)^{-1}} = a^{15}$

Exponents in Standard Form for Very Small Numbers:

Standard form, also known as scientific notation, is a way to express very large or very small numbers concisely. It is written in the form $m \times 10^n$, where $m$ is a number greater than or equal to 1 and less than 10 ($1 \leq m < 10$), and $n$ is an integer (a whole number, positive, negative, or zero).

When dealing with very small numbers (numbers between 0 and 1), the exponent $n$ in the standard form $m \times 10^n$ is a negative integer. A negative exponent $10^{-n}$ means $\frac{1}{10^n}$, which represents a very small fraction (like $\frac{1}{10}$, $\frac{1}{100}$, $\frac{1}{1000}$, etc.). Multiplying the number $m$ (which is between 1 and 10) by such a small fraction results in a very small number.

The value of the negative exponent $-n$ indicates how many places the decimal point has been moved to the right from its original position in the very small number to get the coefficient $m$.

Expressing $0.0000000085$ in Standard Form:

The given number is $0.0000000085$. This is a very small number.

We need to move the decimal point to the right until it is after the first non-zero digit. The first non-zero digit is 8.

Original number: $0.0000000085$

Move the decimal point: $0.00000000\textbf{8}.5$

The decimal point was moved 9 places to the right.

The resulting number $m$ is $8.5$, which satisfies $1 \leq 8.5 < 10$.

Since we moved the decimal point 9 places to the right, the exponent of 10 is -9.

So, $0.0000000085 = 8.5 \times 10^{-9}$.

Expressing $1.6 \times 10^{-19}$ in Usual Form:

The given number is $1.6 \times 10^{-19}$. This number is in standard form, and the negative exponent (-19) indicates it is a very small number.

To convert this to usual form, we need to move the decimal point to the left by the number indicated by the absolute value of the exponent, which is 19.

Starting with $1.6$, we need to move the decimal point 19 places to the left. We will need to add zeros to the left of 1.

$1.6 \times 10^{-19}$ means we move the decimal point 19 places left from its current position (after 1).

The first move takes the decimal point before the 1. This accounts for 1 place.

The remaining moves are $19 - 1 = 18$ places.

So, we will have a decimal point, followed by 18 zeros, followed by the digits 16.

$1.6 \times 10^{-19} = 0.00000000000000000016$

(There are 18 zeros between the decimal point and the digit 1).

Summary:

Very small numbers have negative exponents in standard form.

$0.0000000085$ in standard form is $8.5 \times 10^{-9}$.

$1.6 \times 10^{-19}$ in usual form is $0.00000000000000000016$.

Given Expression:

$(\frac{1}{4})^{-2} \times 5^2 \times 4^{-3}$

Laws of Exponents Used:

1. $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$

2. $a^{-n} = \frac{1}{a^n}$

3. $a^m \times a^n = a^{m+n}$

4. $a^0 = 1$

Simplification Steps:

Given the expression: $(\frac{1}{4})^{-2} \times 5^2 \times 4^{-3}$

Simplify the first term using the law $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$:

$(\frac{1}{4})^{-2} = (\frac{4}{1})^2 = 4^2$

The expression becomes: $4^2 \times 5^2 \times 4^{-3}$

Rearrange the terms to group those with the same base:

$(4^2 \times 4^{-3}) \times 5^2$

Combine the powers of 4 using the law $a^m \times a^n = a^{m+n}$:

$4^2 \times 4^{-3} = 4^{2 + (-3)} = 4^{2-3} = 4^{-1}$

The expression is now: $4^{-1} \times 5^2$

Evaluate the terms using the law $a^{-n} = \frac{1}{a^n}$ and by calculating the square:

$4^{-1} = \frac{1}{4^1} = \frac{1}{4}$

$5^2 = 5 \times 5 = 25$

Multiply the results:

$\frac{1}{4} \times 25 = \frac{25}{4}$

Convert the fraction to a decimal:

$\frac{25}{4} = 6.25$

Expressing in Standard Form:

The simplified value is $6.25$.

Standard form (scientific notation) is $m \times 10^n$, where $1 \leq m < 10$ and $n$ is an integer.

The number $6.25$ is already within the required range for $m$ ($1 \leq 6.25 < 10$).

This means the decimal point does not need to be moved from its current position to place it after the first non-zero digit.

Therefore, the exponent $n$ is 0.

Using the law $a^0 = 1$, we can write $10^0 = 1$.

So, $6.25 = 6.25 \times 1 = 6.25 \times 10^0$.

Final Answer in Standard Form:

The simplified expression in standard form is $6.25 \times 10^0$.

Given Masses:

Mass of a proton = $1.672 \times 10^{-27}$ kg

Mass of an electron = $9.109 \times 10^{-31}$ kg

Comparing the Masses:

Both masses are given in standard form ($m \times 10^n$). To compare them, we look at the exponents of 10.

The exponent for the proton's mass is -27.

The exponent for the electron's mass is -31.

Comparing the exponents, we see that $-27$ is greater than $-31$.

$-27 > -31$

Since the coefficient $m$ for both numbers is between 1 and 10 (1.672 and 9.109), the number with the larger exponent of 10 is the greater number.

Therefore, the mass of the proton ($1.672 \times 10^{-27}$ kg) is greater than the mass of the electron ($9.109 \times 10^{-31}$ kg).

Finding the Factor:

To find by what factor the proton's mass is greater than the electron's mass, we divide the greater mass by the smaller mass.

Factor = $\frac{\text{Mass of Proton}}{\text{Mass of Electron}}$

Factor = $\frac{1.672 \times 10^{-27}}{9.109 \times 10^{-31}}$

We can separate the division of the coefficients and the division of the powers of 10:

Factor = $(\frac{1.672}{9.109}) \times (\frac{10^{-27}}{10^{-31}})$

Using the law of exponents $\frac{a^m}{a^n} = a^{m-n}$ for the powers of 10:

$\frac{10^{-27}}{10^{-31}} = 10^{-27 - (-31)} = 10^{-27 + 31} = 10^4$

Now, perform the division of the coefficients:

$\frac{1.672}{9.109} \approx 0.1835656...$

Combining the results:

Factor $\approx 0.1835656 \times 10^4$

To express this factor as a standard number, move the decimal point 4 places to the right (as indicated by $10^4$):

$0.1835656 \times 10^4 = 1835.656$

Rounding to one decimal place, the factor is approximately 1835.7.

Conclusion:

Comparing the given masses, the mass of the proton is greater than the mass of the electron.

The proton's mass is approximately 1835.7 times greater than the electron's mass.

Final Answer:

The proton has a greater mass.

It is greater by a factor of approximately 1835.7.

Given Expression:

$\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}$

Laws of Exponents Used:

$a^m \times a^n = a^{m+n}$

$\frac{a^m}{a^n} = a^{m-n}$

$(ab)^n = a^n b^n$

$a^{-n} = \frac{1}{a^n}$

Simplification Steps:

Rewrite the numerical terms in terms of their prime factors:

$25 = 5^2$

$10 = 2 \times 5$

Substitute these into the expression:

$\frac{5^2 \times t^{-4}}{5^{-3} \times (2 \times 5) \times t^{-8}}$

Expand the term in the denominator and group terms with the same base:

$\frac{5^2 \times t^{-4}}{5^{-3} \times 2^1 \times 5^1 \times t^{-8}}$

Combine the terms with base 5 in the denominator using the product law ($5^{-3} \times 5^1 = 5^{-3+1} = 5^{-2}$):

$\frac{5^2 \times t^{-4}}{5^{-2} \times 2^1 \times t^{-8}}$

Separate the terms based on their bases (5, 2, and t):

$(\frac{5^2}{5^{-2}}) \times (\frac{t^{-4}}{t^{-8}}) \times (\frac{1}{2^1})$

Apply the quotient law ($\frac{a^m}{a^n} = a^{m-n}$) to the terms with bases 5 and t:

For base 5: $\frac{5^2}{5^{-2}} = 5^{2 - (-2)} = 5^{2+2} = 5^4$

For base t: $\frac{t^{-4}}{t^{-8}} = t^{-4 - (-8)} = t^{-4+8} = t^4$

The term with base 2 remains $\frac{1}{2^1} = \frac{1}{2}$.

Combine the simplified terms:

$5^4 \times t^4 \times \frac{1}{2}$

Calculate the value of $5^4$:

$5^4 = 5 \times 5 \times 5 \times 5 = 625$

Substitute the value back:

$625 \times t^4 \times \frac{1}{2}$

Write the expression as a single fraction:

$\frac{625 \times t^4}{2}$

The exponents in the expression ($4$ for $t$ and implicitly $1$ for $2$) are positive.

Final Answer with Positive Exponents:

The simplified expression with positive exponents is $\frac{625 t^4}{2}$.

$\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} = \frac{625 t^4}{2}$

Given Diameters:

Diameter of the Sun = $1.39 \times 10^9$ meters

Diameter of the Earth = $1.276 \times 10^7$ meters

Comparison:

Both diameters are given in standard form ($m \times 10^n$). To compare them, we look at the exponents of 10.

The exponent for the Sun's diameter is 9.

The exponent for the Earth's diameter is 7.

Comparing the exponents, we see that $9 > 7$.

Since the exponent of 10 for the Sun's diameter is greater than the exponent of 10 for the Earth's diameter, the diameter of the Sun is larger.

Finding the Factor of Difference:

To find how many times larger the diameter of the Sun is compared to the diameter of the Earth, we divide the Sun's diameter by the Earth's diameter.

Factor = $\frac{\text{Diameter of the Sun}}{\text{Diameter of the Earth}}$

Factor = $\frac{1.39 \times 10^9}{1.276 \times 10^7}$

Separate the coefficients and the powers of 10:

Factor = $(\frac{1.39}{1.276}) \times (\frac{10^9}{10^7})$

Using the law of exponents $\frac{a^m}{a^n} = a^{m-n}$ for the powers of 10:

$\frac{10^9}{10^7} = 10^{9-7} = 10^2$

Calculate the division of the coefficients:

$\frac{1.39}{1.276} \approx 1.0893417...$

Combine the results:

Factor $\approx 1.0893417 \times 10^2$

To express this factor as a standard number, multiply by $10^2$ (or 100), which means moving the decimal point 2 places to the right:

$1.0893417 \times 100 = 108.93417$

Round the factor to one decimal place. The second decimal digit is 3, which is less than 5, so we round down.

$108.93417 \approx 108.9$

Conclusion:

The diameter of the Sun is larger than the diameter of the Earth.

The diameter of the Sun is approximately 108.9 times larger than the diameter of the Earth.

Final Answer:

The diameter of the Sun is greater.

The diameter of the Sun is approximately 108.9 times larger than the diameter of the Earth.

Given Expression:

$(\frac{2}{3})^{-2} + (\frac{1}{2})^{-3} - (\frac{1}{4})^{-1}$

Law of Exponents Used:

We use the law for negative exponents of fractions: $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$.

Evaluation of Each Term:

Apply the law to the first term:

$(\frac{2}{3})^{-2} = (\frac{3}{2})^2 = \frac{3^2}{2^2} = \frac{9}{4}$

Apply the law to the second term:

$(\frac{1}{2})^{-3} = (\frac{2}{1})^3 = 2^3 = 2 \times 2 \times 2 = 8$

Apply the law to the third term:

$(\frac{1}{4})^{-1} = (\frac{4}{1})^1 = 4^1 = 4$

Combining the Results:

Substitute the evaluated terms back into the expression:

$\frac{9}{4} + 8 - 4$

Perform the addition and subtraction:

$\frac{9}{4} + 8 - 4 = \frac{9}{4} + 4$

To add the fraction and the whole number, find a common denominator:

$\frac{9}{4} + 4 = \frac{9}{4} + \frac{4 \times 4}{4} = \frac{9}{4} + \frac{16}{4}$

Add the numerators:

$\frac{9 + 16}{4} = \frac{25}{4}$

Final Answer:

The value of the expression $(\frac{2}{3})^{-2} + (\frac{1}{2})^{-3} - (\frac{1}{4})^{-1}$ is $\frac{25}{4}$.

$(\frac{2}{3})^{-2} + (\frac{1}{2})^{-3} - (\frac{1}{4})^{-1} = \frac{25}{4}$

Given:

Thickness of one page = $0.001$ cm

Number of pages in the book = $500$ pages

To Find:

The total thickness of the book in centimeters, expressed in standard form.

Solution:

The total thickness of the book is found by multiplying the thickness of one page by the total number of pages.

Total thickness = (Thickness of one page) $\times$ (Number of pages)

Total thickness = $0.001 \text{ cm/page} \times 500 \text{ pages}$

Perform the multiplication:

$0.001 \times 500$

We can write $0.001$ as $\frac{1}{1000}$.

Total thickness = $\frac{1}{1000} \times 500 = \frac{500}{1000}$

Simplify the fraction:

$\frac{500}{1000} = \frac{5}{10} = 0.5$

The total thickness of the book in usual form is $0.5$ cm.

Expressing in Standard Form:

Standard form (scientific notation) is written as $m \times 10^n$, where $1 \leq m < 10$ and $n$ is an integer.

The usual form of the thickness is $0.5$ cm.

To write $0.5$ in standard form, we need to move the decimal point to the right until it is after the first non-zero digit. The first non-zero digit is 5.

Original number: $0.5$

Move the decimal point: $0\textbf{.}5 \rightarrow 5\textbf{.}$

The decimal point was moved 1 place to the right.

The resulting number $m$ is $5$, which satisfies $1 \leq 5 < 10$.

Since we moved the decimal point 1 place to the right from a number less than 1, the exponent of 10 is negative and is equal to -1.

So, $0.5 = 5 \times 10^{-1}$.

Include the unit:

Total thickness = $5 \times 10^{-1}$ cm.

Final Answer:

The thickness of the book is $0.5$ cm in usual form.

In standard form, the thickness of the book is $5 \times 10^{-1}$ cm.