Chapter 13 Exponents and Powers (Additional Questions)

In a mathematical expression written in the form $a^b$, where $a$ is raised to the power of $b$, $a$ is called the base and $b$ is called the exponent or power.

The expression $a^b$ represents the multiplication of the base $a$ by itself $b$ times.

The given expression is $5^3$.

Comparing this with the general form $a^b$, we have $a = 5$ and $b = 3$.

Therefore, in the expression $5^3$, the base is 5 and the exponent is 3.

Looking at the given options:

(A) 5

(B) 3

(C) $5^3$

(D) 15

The base is 5, which corresponds to option (A).

The final answer is (A) 5.

The expression $2^4$ represents the base 2 multiplied by itself 4 times. The exponent is 4.

The calculation is as follows:

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

$2 \times 2 = 4$

$4 \times 2 = 8$

$8 \times 2 = 16$

So, the value of $2^4$ is 16.

Comparing the calculated value with the given options:

(A) 6

(B) 8

(C) 16

(D) 4

The value 16 matches option (C).

The final answer is (C) 16.

To express a number in exponential form with a given base, we need to find the power to which the base must be raised to equal the number.

In this case, we want to express 81 with base 3. This means we need to find an exponent $x$ such that $3^x = 81$.

We can find the exponent by repeatedly multiplying the base 3:

$3^1 = 3$

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

$3^3 = 3 \times 3 \times 3 = 9 \times 3 = 27$

$3^4 = 3 \times 3 \times 3 \times 3 = 27 \times 3 = 81$

So, $81$ can be written as $3^4$.

Comparing this result with the given options:

(A) $3^2 = 9$

(B) $3^3 = 27$

(C) $3^4 = 81$

(D) $3^5 = 243$

The exponential form of 81 with base 3 is $3^4$, which matches option (C).

The final answer is (C) $3^4$.

The expression $(-1)^5$ means the base -1 is multiplied by itself 5 times. The exponent is 5.

The calculation is as follows:

$(-1)^5 = (-1) \times (-1) \times (-1) \times (-1) \times (-1)$

We can group the terms:

$((-1) \times (-1)) \times ((-1) \times (-1)) \times (-1)$

$1 \times 1 \times (-1)$

$1 \times (-1)$

$-1$

So, the value of $(-1)^5$ is -1.

Alternatively, we know that raising -1 to an odd power results in -1, and raising -1 to an even power results in 1.

Since the exponent is 5, which is an odd number, $(-1)^5 = -1$.

Comparing the calculated value with the given options:

(A) 1

(B) -1

(C) 0

(D) 5

The value -1 matches option (B).

The final answer is (B) -1.

This question asks for the simplification of the expression $a^m \times a^n$. This involves the laws of exponents.

One of the fundamental laws of exponents states that when multiplying two exponential expressions with the same base, you keep the base the same and add the exponents.

Mathematically, this law is expressed as:

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

Let's verify this with an example. Suppose $a=2$, $m=3$, and $n=4$.

$2^3 \times 2^4 = (2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2) = 8 \times 16 = 128$

Using the rule, $2^{3+4} = 2^7$.

$2^7 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 128$.

The results match, confirming the rule.

Comparing the simplified form $a^{m+n}$ with the given options:

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

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

(C) $a^{mn}$

(D) $(a^m)^n$

The simplified form $a^{m+n}$ matches option (A).

The final answer is (A) $a^{m+n}$.

To simplify the expression $5^2 \times 5^3$, we use the law of exponents that states when multiplying powers with the same base, we add the exponents. The rule is $a^m \times a^n = a^{m+n}$.

In the given expression, the base is 5, the first exponent is $m=2$, and the second exponent is $n=3$.

Applying the rule:

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

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

So, the simplified form is $5^5$.

Comparing the result with the given options:

(A) $5^5$

(B) $5^6$

(C) $25^5$

(D) $5^1$

The simplified form $5^5$ matches option (A).

The final answer is (A) $5^5$.

To simplify the expression $a^m \div a^n$, we use a fundamental law of exponents that deals with the division of powers with the same base.

The law states that when dividing two exponential expressions with the same base, you keep the base the same and subtract the exponent of the denominator from the exponent of the numerator.

Mathematically, this law is expressed as:

$a^m \div a^n = a^{m-n}$, where $a \neq 0$.

The condition $m > n$ mentioned in the question ensures that the resulting exponent $m-n$ is positive, which is a specific case of this general rule.

Let's illustrate this with an example. Suppose $a=2$, $m=5$, and $n=2$. Here, $m > n$ is satisfied.

$2^5 \div 2^2 = (2 \times 2 \times 2 \times 2 \times 2) \div (2 \times 2) = 32 \div 4 = 8$

Using the rule, $2^{5-2} = 2^3$.

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

The results match, confirming the rule $a^m \div a^n = a^{m-n}$.

Comparing the simplified form $a^{m-n}$ with the given options:

(A) $a^{m+n}$ (This is for multiplication: $a^m \times a^n$)

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

(C) $a^{mn}$ (This is for power of a power: $(a^m)^n$)

(D) $a^{n-m}$ (This would be the result of $a^n \div a^m$)

The simplified form $a^{m-n}$ matches option (B).

The final answer is (B) $a^{m-n}$.

To simplify the expression $7^6 \div 7^2$, we use the law of exponents for division of powers with the same base. The rule states that for any non-zero base $a$ and integers $m$ and $n$, $a^m \div a^n = a^{m-n}$.

In the given expression, the base is $a=7$. The exponent of the numerator is $m=6$, and the exponent of the denominator is $n=2$.

Applying the rule, we subtract the exponents:

$7^6 \div 7^2 = 7^{6-2}$

Now, we calculate the difference in the exponents:

$6 - 2 = 4$

So, the simplified form of the expression is $7^4$.

Comparing the result with the given options:

(A) $7^3$

(B) $7^4$

(C) $7^8$ (This would be $7^6 \times 7^2$)

(D) $7^{12}$ (This would be $(7^6)^2$ or $(7^2)^6$)

The simplified form $7^4$ matches option (B).

The final answer is (B) $7^4$.

To simplify the expression $(a^m)^n$, we use the law of exponents for raising a power to another power.

The rule states that when an exponential expression $a^m$ is raised to another power $n$, we keep the base $a$ the same and multiply the exponents $m$ and $n$.

Mathematically, this law is expressed as:

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

Let's consider an example. Suppose $a=3$, $m=2$, and $n=3$.

$(3^2)^3 = (3 \times 3)^3 = 9^3 = 9 \times 9 \times 9 = 81 \times 9 = 729$

Using the rule, $(3^2)^3 = 3^{2 \times 3} = 3^6$.

$3^6 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 9 \times 9 = 729$.

The results match, confirming the rule $(a^m)^n = a^{mn}$.

Comparing the simplified form $a^{mn}$ with the given options:

(A) $a^{m+n}$ (This is for multiplication: $a^m \times a^n$)

(B) $a^{m-n}$ (This is for division: $a^m \div a^n$)

(C) $a^{mn}$

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

The simplified form $a^{mn}$ matches option (C).

The final answer is (C) $a^{mn}$.

To simplify the expression $(3^2)^3$, we use the law of exponents which states that when raising a power to another power, we multiply the exponents. The rule is $(a^m)^n = a^{mn}$.

In the given expression, the base is $a=3$. The inner exponent is $m=2$, and the outer exponent is $n=3$.

Applying the rule $(a^m)^n = a^{mn}$, we have:

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

Now, we calculate the product of the exponents:

$2 \times 3 = 6$

So, the simplified form of the expression is $3^6$.

Comparing the result with the given options:

(A) $3^5$

(B) $3^6$

(C) $3^9$ (This would be $3^{3 \times 3}$ or $(3^3)^3$)

(D) $9^3$ (Note that $9 = 3^2$, so $9^3 = (3^2)^3$, which is the original expression, not a further simplified form in base 3)

The simplified form $3^6$ matches option (B).

The final answer is (B) $3^6$.

This question asks for the value of any non-zero number raised to the power of zero. This is a standard rule in the study of exponents.

For any non-zero number $a$, the value of $a^0$ is defined to be 1.

Mathematically:

$a^0 = 1$, for $a \neq 0$.

We can understand this rule by considering the division of powers with the same base. According to the division rule of exponents, $a^m \div a^n = a^{m-n}$ for $a \neq 0$.

If we take the specific case where the exponents are equal, i.e., $m=n$, then the division becomes $a^m \div a^m$.

Using the division rule:

$a^m \div a^m = a^{m-m} = a^0$

Also, any non-zero number divided by itself is equal to 1.

$a^m \div a^m = 1$ (since $a \neq 0$, $a^m \neq 0$)

Equating the two results, we get:

$a^0 = 1$

This demonstrates why any non-zero number raised to the power of zero is 1.

Comparing the value 1 with the given options:

(A) The number itself

(B) 0

(C) 1

(D) Undefined

The value 1 matches option (C).

The final answer is (C) 1.

To simplify the expression $(a \times b)^m$, we use the law of exponents for the power of a product.

The rule states that when a product of two or more factors is raised to a power, each factor is raised to that power, and the resulting powers are multiplied.

Mathematically, this law is expressed as:

$(a \times b)^m = a^m \times b^m$

or simply

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

We can illustrate this with an example. Let $a=2$, $b=3$, and $m=2$.

$(2 \times 3)^2 = 6^2 = 6 \times 6 = 36$

Using the rule, $2^2 \times 3^2 = (2 \times 2) \times (3 \times 3) = 4 \times 9 = 36$.

The results match, confirming the rule $(a \times b)^m = a^m \times b^m$.

Comparing the simplified form $a^m b^m$ with the given options:

(A) $a^m + b^m$

(B) $a^m - b^m$

(C) $a^m b^m$

(D) $a^m / b^m$

The simplified form $a^m b^m$ matches option (C).

The final answer is (C) $a^m b^m$.

To simplify the expression $(2 \times 3)^4$, we can approach this in two ways:

Method 1: Calculate the product first, then apply the exponent.

Inside the parentheses, we have the product $2 \times 3$.

$2 \times 3 = 6$

So, the expression becomes $6^4$.

This matches option (A).

Method 2: Apply the power of a product rule.

The power of a product rule states that $(a \times b)^m = a^m \times b^m$.

In the given expression $(2 \times 3)^4$, we have $a=2$, $b=3$, and $m=4$.

Applying the rule:

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

This matches option (C).

Since both $6^4$ and $2^4 \times 3^4$ are equivalent ways to express the simplified form of $(2 \times 3)^4$, both options (A) and (C) are correct simplifications according to the properties of exponents.

Comparing our findings with the given options:

(A) $6^4$ (Matches Method 1)

(B) $2^4 + 3^4$ (Incorrect, the exponent applies to the multiplication, not addition)

(C) $2^4 \times 3^4$ (Matches Method 2)

(D) Both (A) and (C) (Since both (A) and (C) are correct)

Option (D) correctly identifies that both (A) and (C) are valid simplifications.

The final answer is (D) Both (A) and (C).

To express 1024 as a power of 2, we need to find the exponent $x$ such that $2^x = 1024$. We can do this by finding how many times 2 must be multiplied by itself to get 1024.

Method 1: Repeated Multiplication

We multiply 2 by itself repeatedly until we reach 1024:

$2^1 = 2$

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

$2^3 = 4 \times 2 = 8$

$2^4 = 8 \times 2 = 16$

$2^5 = 16 \times 2 = 32$

$2^6 = 32 \times 2 = 64$

$2^7 = 64 \times 2 = 128$

$2^8 = 128 \times 2 = 256$

$2^9 = 256 \times 2 = 512$

$2^{10} = 512 \times 2 = 1024$

From this, we see that $2^{10} = 1024$.

Method 2: Prime Factorization

We can find the prime factorization of 1024 by repeatedly dividing by 2:

By counting the number of times 2 appears in the factorization, we find that $1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{10}$.

Both methods show that 1024 as a power of 2 is $2^{10}$.

Comparing our result with the given options:

(A) $2^8$

(B) $2^9$

(C) $2^{10}$

(D) $2^{11}$

The power of 2 that equals 1024 is $2^{10}$, which matches option (C).

The final answer is (C) $2^{10}$.

Let's simplify each expression using the laws of exponents:

(i) Simplify $x^5 \times x^3$:

Using the rule $a^m \times a^n = a^{m+n}$ (product of powers with the same base), we add the exponents:

$x^5 \times x^3 = x^{5+3} = x^8$

This matches option (b) $x^8$.

(ii) Simplify $y^7 \div y^4$:

Using the rule $a^m \div a^n = a^{m-n}$ (division of powers with the same base), we subtract the exponents:

$y^7 \div y^4 = y^{7-4} = y^3$

This matches option (a) $y^3$.

(iii) Simplify $(m^2)^4$:

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

$(m^2)^4 = m^{2 \times 4} = m^8$

This matches option (d) $m^8$.

(iv) Simplify $(2a)^3$:

Using the rule $(ab)^m = a^m b^m$ (power of a product), we raise each factor in the base to the power:

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

Calculate the value of $2^3$:

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

So, $(2a)^3 = 8a^3$

This matches option (c) $8a^3$.

Summary of the matches:

(i) $x^5 \times x^3 \rightarrow x^8$ (b)

(ii) $y^7 \div y^4 \rightarrow y^3$ (a)

(iii) $(m^2)^4 \rightarrow m^8$ (d)

(iv) $(2a)^3 \rightarrow 8a^3$ (c)

Now let's compare these matches with the given options (A), (B), (C), and (D):

(A) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)

(B) (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c)

(C) (i)-(a), (ii)-(b), (iii)-(d), (iv)-(c)

(D) (i)-(b), (ii)-(a), (iii)-(c), (iv)-(d)

The matches we found correspond exactly to option (A).

The final answer is (A) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c).

Let's evaluate the truthfulness of the Assertion (A) and the Reason (R), and then determine if R is the correct explanation for A.

Assertion (A): $(-3)^4$ is a positive number.

We calculate the value of $(-3)^4$:

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

We multiply step by step:

$(-3) \times (-3) = 9$ (The product of two negative numbers is positive).

$9 \times (-3) = -27$ (The product of a positive and a negative number is negative).

$-27 \times (-3) = 81$ (The product of two negative numbers is positive).

The value of $(-3)^4$ is 81. Since 81 is greater than 0, it is a positive number.

Thus, Assertion (A) is true.

Reason (R): A negative base raised to an even exponent results in a positive value.

Consider a negative base, say $-a$ where $a > 0$. When this base is raised to an even exponent, say $n=2k$ where $k$ is a positive integer, the expression is $(-a)^{2k}$.

$(-a)^{2k} = \underbrace{(-a) \times (-a) \times \dots \times (-a)}_{2k \text{ times}}$

We can pair the negative terms: $2k$ is an even number of terms, so there are $k$ pairs of $(-a) \times (-a)$.

Each pair $(-a) \times (-a) = a^2$, which is a positive value (since $a > 0$).

So, $(-a)^{2k} = (a^2)^k = a^{2k}$. Since $a > 0$ and $2k$ is a positive integer, $a^{2k}$ is positive.

If the even exponent is 0, $(-a)^0 = 1$ (for $a \neq 0$), which is also positive.

Thus, Reason (R) is a correct statement and is true.

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

Assertion (A) is about the specific expression $(-3)^4$. In this expression, the base is -3 (which is a negative number) and the exponent is 4 (which is an even number).

Reason (R) provides the general rule that applies precisely to the situation in Assertion (A): a negative base raised to an even exponent gives a positive result.

Therefore, Reason (R) directly explains why $(-3)^4$ is a positive number.

Based on our analysis:

- Assertion (A) is true.

- Reason (R) is true.

- Reason (R) is the correct explanation for Assertion (A).

This matches the description in option (A).

The final answer is (A) Both A and R are true, and R is the correct explanation of A.

Let's analyze the truthfulness of Assertion (A) and Reason (R) and whether R explains A.

Assertion (A): $10^5$ is equal to 100000.

The expression $10^5$ means 10 multiplied by itself 5 times:

$10^5 = 10 \times 10 \times 10 \times 10 \times 10$

Calculating the value:

$10 \times 10 = 100$

$100 \times 10 = 1000$

$1000 \times 10 = 10000$

$10000 \times 10 = 100000$

So, $10^5 = 100000$. Assertion (A) is true.

Reason (R): $10^n$ represents 1 followed by $n$ zeros.

Let's check this statement for a few positive integer values of $n$:

For $n=1$, $10^1 = 10$, which is 1 followed by 1 zero.

For $n=2$, $10^2 = 100$, which is 1 followed by 2 zeros.

For $n=3$, $10^3 = 1000$, which is 1 followed by 3 zeros.

This pattern holds true for any positive integer $n$. The reason is that multiplying by 10 simply shifts the digits one place to the left and adds a zero at the end. Starting with $1 = 10^0$ (conceptually, before any zeros), each multiplication by 10 adds one zero.

Thus, Reason (R) is a correct property of powers of 10 and is true.

Now, let's determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) is the statement $10^5 = 100000$. According to Reason (R), for $n=5$, $10^5$ should be 1 followed by 5 zeros. 1 followed by 5 zeros is indeed 100000.

Reason (R) provides the general principle that explains how to determine the value of any power of 10, including the specific case $10^5$. Therefore, Reason (R) is the correct explanation for Assertion (A).

Based on the analysis:

- Assertion (A) is true.

- Reason (R) is true.

- Reason (R) is the correct explanation for Assertion (A).

This matches option (A).

The final answer is (A) Both A and R are true, and R is the correct explanation of A.

The mass of the Earth is given as $5.97 \times 10^{24}\text{ kg}$. We need to check which of the given options represents the same value.

The original value is $5.97 \times 10^{24}$. This is in scientific notation because the coefficient (5.97) is between 1 and 10.

Let's examine each option:

(A) $597 \times 10^{22}\text{ kg}$

We can rewrite the coefficient 597 in scientific notation: $597 = 5.97 \times 100 = 5.97 \times 10^2$.

Substitute this back into the expression:

$597 \times 10^{22} = (5.97 \times 10^2) \times 10^{22}$

Using the law of exponents $a^m \times a^n = a^{m+n}$:

$5.97 \times 10^2 \times 10^{22} = 5.97 \times 10^{2+22} = 5.97 \times 10^{24}$

So, $597 \times 10^{22}\text{ kg}$ is equivalent to $5.97 \times 10^{24}\text{ kg}$.

(B) $59.7 \times 10^{23}\text{ kg}$

We can rewrite the coefficient 59.7 in scientific notation: $59.7 = 5.97 \times 10 = 5.97 \times 10^1$.

Substitute this back into the expression:

$59.7 \times 10^{23} = (5.97 \times 10^1) \times 10^{23}$

Using the law of exponents $a^m \times a^n = a^{m+n}$:

$5.97 \times 10^1 \times 10^{23} = 5.97 \times 10^{1+23} = 5.97 \times 10^{24}$

So, $59.7 \times 10^{23}\text{ kg}$ is equivalent to $5.97 \times 10^{24}\text{ kg}$.

(C) $0.597 \times 10^{25}\text{ kg}$

We can rewrite the coefficient 0.597 in scientific notation: $0.597 = 5.97 \div 10 = 5.97 \times 10^{-1}$.

Substitute this back into the expression:

$0.597 \times 10^{25} = (5.97 \times 10^{-1}) \times 10^{25}$

Using the law of exponents $a^m \times a^n = a^{m+n}$:

$5.97 \times 10^{-1} \times 10^{25} = 5.97 \times 10^{-1+25} = 5.97 \times 10^{24}$

So, $0.597 \times 10^{25}\text{ kg}$ is equivalent to $5.97 \times 10^{24}\text{ kg}$.

Since options (A), (B), and (C) all represent the same value as the given mass of the Earth ($5.97 \times 10^{24}\text{ kg}$), the statement that all of the above are equivalent representations is true.

The final answer is (D) All of the above are equivalent representations.

To find out how many times the Earth's mass is greater than the Moon's mass, we need to divide the mass of the Earth by the mass of the Moon.

Given:

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

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

The ratio of Earth's mass to Moon's mass is:

Ratio = $\frac{\text{Mass of Earth}}{\text{Mass of Moon}} = \frac{5.97 \times 10^{24}}{7.35 \times 10^{22}}$

We can separate the division into two parts: the division of the decimal coefficients and the division of the powers of 10.

Ratio = $\frac{5.97}{7.35} \times \frac{10^{24}}{10^{22}}$

First, calculate the division of the powers of 10 using the rule $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{10^{24}}{10^{22}} = 10^{24-22} = 10^2 = 100$

Next, calculate the division of the decimal coefficients $\frac{5.97}{7.35}$.

$\frac{5.97}{7.35} \approx 0.8122$ (using a calculator or approximate calculation)

Now, multiply these two results:

Ratio $\approx 0.8122 \times 100$

Ratio $\approx 81.22$

The Earth's mass is approximately 81.22 times greater than the Moon's mass.

Now, let's compare this result with the given options:

(A) 8 times

(B) 80 times

(C) 100 times

(D) 1000 times

Our calculated value of approximately 81.22 is closest to 80.

The final answer is (B) 80 times.

To express a number as a power of a prime number, we need to find the prime factors of the number and determine how many times each prime factor appears. Since the question asks for a power of *a* prime number, we expect only one unique prime factor.

We find the prime factorization of 128 by repeatedly dividing by the smallest prime number, which is 2.

The prime factorization of 128 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$.

We can express this product in exponential form with the base 2 (since it's the only prime factor) and the exponent being the number of times 2 appears in the factorization.

Counting the factors of 2, we have 7 factors.

So, $128 = 2^7$.

Now, let's compare this result with the given options:

(A) $2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$

(B) $2^7 = 128$ (2 is a prime number, and 7 is the correct exponent)

(C) $4^3 = 4 \times 4 \times 4 = 64$ (4 is not a prime number)

(D) $128^1 = 128$ (128 is not a prime number)

The expression $2^7$ correctly represents 128 as a power of a prime number (which is 2).

The final answer is (B) $2^7$.

To find the equivalent expression for $(\frac{a}{b})^m$, we use the law of exponents for the power of a quotient.

The rule states that when a quotient (a fraction) $\frac{a}{b}$ is raised to a power $m$, we raise both the numerator $a$ and the denominator $b$ to the power $m$.

Mathematically, this law is expressed as:

$(\frac{a}{b})^m = \frac{a^m}{b^m}$, provided that $b \neq 0$.

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

$(\frac{2}{3})^2 = \frac{2}{3} \times \frac{2}{3} = \frac{2 \times 2}{3 \times 3} = \frac{4}{9}$

Using the rule $(\frac{a}{b})^m = \frac{a^m}{b^m}$:

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

The results from both methods are the same, confirming the rule.

Comparing the simplified form $\frac{a^m}{b^m}$ with the given options:

(A) $a^m - b^m$ (Incorrect, the exponent does not distribute over subtraction)

(B) $a^m + b^m$ (Incorrect, the exponent does not distribute over addition)

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

(D) $\frac{a}{b^m}$ (Incorrect, the numerator is not raised to the power $m$)

The simplified form $\frac{a^m}{b^m}$ matches option (C).

The final answer is (C) $\frac{a^m}{b^m}$.

To simplify the expression $(\frac{2}{3})^3$, we use the law of exponents for the power of a quotient, which states that $(\frac{a}{b})^m = \frac{a^m}{b^m}$.

In this expression, the base is the fraction $\frac{2}{3}$ and the exponent is 3.

Applying the rule, we raise both the numerator (2) and the denominator (3) to the power of 3:

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

Now, we calculate the value of the numerator $2^3$ and the denominator $3^3$:

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

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

Substitute these values back into the fraction:

$\frac{2^3}{3^3} = \frac{8}{27}$

So, the simplified form of $(\frac{2}{3})^3$ is $\frac{8}{27}$.

Comparing our result with the given options:

(A) $\frac{6}{9}$ (Incorrect)

(B) $\frac{8}{27}$ (Correct)

(C) $\frac{2}{9}$ (Incorrect)

(D) $\frac{4}{9}$ (Incorrect)

The simplified form $\frac{8}{27}$ matches option (B).

The final answer is (B) $\frac{8}{27}$.

The given value for the speed of light is $3 \times 10^8\text{ m/s}$. We need to identify the form in which this number is written.

Let's analyze the options:

(A) Expanded form: This typically refers to writing a number as the sum of its digits multiplied by their place values (e.g., $123 = 1 \times 100 + 2 \times 10 + 3 \times 1$). The given number is not in this form.

(B) Standard form: In the context of representing very large or very small numbers, standard form (also known as scientific notation) is a way of writing numbers using powers of 10. A number in standard form is written as the product of a number between 1 (inclusive) and 10 (exclusive) and an integer power of 10. The general format is $a \times 10^n$, where $1 \leq |a| < 10$ and $n$ is an integer.

(C) Exponential form: This is a general term for writing a number using a base and an exponent (e.g., $64 = 8^2 = 4^3 = 2^6$). While $3 \times 10^8$ contains a part in exponential form ($10^8$), the entire expression is a specific type of notation for large/small numbers, more precisely called standard form or scientific notation.

(D) Normal form: This term is not standard for describing this type of numerical representation.

The given number $3 \times 10^8$ consists of a coefficient, 3, which is a number between 1 and 10 ($1 \leq 3 < 10$), multiplied by a power of 10, $10^8$. This structure perfectly matches the definition of standard form (or scientific notation).

Comparing the given number's structure with the descriptions of the options, it is clear that the number is written in standard form.

The final answer is (B) Standard form.

To find the value of the expression $10^0 + 20^0 + 30^0$, we need to evaluate each term separately.

We use the rule of exponents which states that any non-zero number raised to the power of zero is equal to 1.

Mathematically, $a^0 = 1$ for any $a \neq 0$.

Applying this rule to each term in the expression:

For the first term, $10^0$: Since 10 is a non-zero number, $10^0 = 1$.

For the second term, $20^0$: Since 20 is a non-zero number, $20^0 = 1$.

For the third term, $30^0$: Since 30 is a non-zero number, $30^0 = 1$.

Now, we substitute these values back into the original expression and perform the addition:

$10^0 + 20^0 + 30^0 = 1 + 1 + 1$

$1 + 1 + 1 = 3$

The value of the expression is 3.

Comparing our result with the given options:

(A) 0

(B) 1

(C) 3

(D) 60

The calculated value 3 matches option (C).

The final answer is (C) 3.

To simplify the expression $\frac{(2^3)^2 \times 3^4}{6^2}$, we will use the laws of exponents.

First, simplify the numerator, $(2^3)^2 \times 3^4$. We use the power of a power rule, $(a^m)^n = a^{mn}$, for the term $(2^3)^2$:

$(2^3)^2 = 2^{3 \times 2} = 2^6$

So the numerator becomes $2^6 \times 3^4$.

Next, simplify the denominator, $6^2$. We recognize that $6$ is a composite number and its prime factors are 2 and 3 ($6 = 2 \times 3$). We use the power of a product rule, $(ab)^m = a^m b^m$:

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

Now, rewrite the original expression using the simplified numerator and denominator:

$\frac{(2^3)^2 \times 3^4}{6^2} = \frac{2^6 \times 3^4}{2^2 \times 3^2}$

Finally, simplify the fraction by using the division rule of exponents, $\frac{a^m}{a^n} = a^{m-n}$, for terms with the same base:

For base 2: $\frac{2^6}{2^2} = 2^{6-2} = 2^4$

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

Combine these results:

$\frac{2^6 \times 3^4}{2^2 \times 3^2} = 2^4 \times 3^2$

The simplified form of the expression is $2^4 \times 3^2$. Let's compare this with the given options:

(A) $\frac{2^6 \times 3^4}{2^2 \times 3^2} = 2^4 \times 3^2$. This option shows the correct intermediate step after simplifying the numerator and denominator and then the final simplified form.

(B) $\frac{2^5 \times 3^4}{6^2}$. This is incorrect because $(2^3)^2 = 2^6$, not $2^5$.

(C) $\frac{2^6 \times 3^4}{6^2}$. This is the expression after simplifying the numerator only. It is equivalent to the original expression but not fully simplified using prime bases in the denominator.

(D) $2^4 \times 3^6$. This is incorrect because the exponent for 3 should be $4-2=2$, not 6.

Option (A) correctly shows the division step and the final simplified form $2^4 \times 3^2$.

The final answer is (A) $\frac{2^6 \times 3^4}{2^2 \times 3^2} = 2^4 \times 3^2$.

To determine the correct comparison, we need to evaluate the value of each exponential expression in the options.

Let's calculate the values:

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

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

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

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

$4^2 = 4 \times 4 = 16$

$2^4 = 2 \times 2 \times 2 \times 2 = 16$

Now, let's check each option using these values:

(A) $2^3 > 3^2$ becomes $8 > 9$. This is false.

(B) $2^5 = 5^2$ becomes $32 = 25$. This is false.

(C) $4^2 < 2^4$ becomes $16 < 16$. This is false (as 16 is equal to 16).

(D) $2^3 < 3^2$ becomes $8 < 9$. This is true.

The only correct comparison among the given options is $2^3 < 3^2$.

The final answer is (D) $2^3 < 3^2$.

Standard form (or scientific notation) requires writing a number as the product of a number between 1 (inclusive) and 10 (exclusive) and an integer power of 10. The general form is $a \times 10^n$, where $1 \leq a < 10$ and $n$ is an integer.

The given number is 0.000007.

We need to move the decimal point so that the number is between 1 and 10. The significant digit is 7, so the coefficient $a$ will be 7.

The original position of the decimal point is after the first 0. The desired position is after the digit 7.

$0.000007$

To move the decimal point from its current position to after the 7, we need to move it 6 places to the right.

$0.000007 \rightarrow 7.0$

Since we moved the decimal point 6 places to the right to get a larger coefficient (7 from 0.000007), the power of 10 must be negative. The exponent $n$ is equal to the number of places moved, with the sign indicating the direction (negative for right movement when converting from decimal to standard form for numbers less than 1).

The number of places moved is 6. The direction is to the right. So the exponent is -6.

Therefore, $0.000007$ in standard form is $7 \times 10^{-6}$.

Comparing this result with the given options:

(A) $7 \times 10^6$ (Incorrect, this represents 7,000,000)

(B) $7 \times 10^{-6}$ (Correct)

(C) $7 \times 10^7$ (Incorrect, this represents 70,000,000)

(D) $7 \times 10^{-7}$ (Incorrect, this would be 0.0000007)

The standard form $7 \times 10^{-6}$ matches option (B).

The final answer is (B) $7 \times 10^{-6}$.

We are asked to simplify the expression $(5^0 \times 5^2) \div 5^1$. Let's simplify this expression using the laws of exponents and evaluate the result.

Method 1: Using laws of exponents directly.

The expression is $(5^0 \times 5^2) \div 5^1$.

First, simplify the part inside the parentheses, $5^0 \times 5^2$. Using the rule $a^m \times a^n = a^{m+n}$, we have:

$5^0 \times 5^2 = 5^{0+2} = 5^2$

Now the expression is $5^2 \div 5^1$. Using the rule $a^m \div a^n = a^{m-n}$, we have:

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

The value of $5^1$ is 5.

This approach is represented in option (B) which shows the combined exponent calculation $5^{0+2-1} = 5^1 = 5$. This is a correct calculation.

Method 2: Evaluating each exponential term first.

We evaluate each term: $5^0$, $5^2$, and $5^1$.

Using the rule $a^0 = 1$ (for $a \neq 0$), $5^0 = 1$.

Using the definition of exponentiation, $5^2 = 5 \times 5 = 25$.

Using the definition of exponentiation, $5^1 = 5$.

Substitute these values into the original expression:

$(5^0 \times 5^2) \div 5^1 = (1 \times 25) \div 5$

Perform the multiplication in the parentheses:

$1 \times 25 = 25$

Now perform the division:

$25 \div 5 = 5$

The value of the expression is 5.

This approach is represented in option (C) which shows the calculation $(1 \times 25) \div 5 = 25 \div 5 = 5$. This is also a correct calculation.

Both method 1 (represented in option B) and method 2 (represented in option C) lead to the same correct result, 5. Therefore, both calculations shown in options (B) and (C) are correct.

Comparing our findings with the given options:

(A) $5^0 = 1$. Incorrect, this is only the value of one term.

(B) $5^{0+2-1} = 5^1 = 5$. Correct calculation and result.

(C) $(1 \times 25) \div 5 = 25 \div 5 = 5$. Correct calculation and result.

(D) Both (B) and (C) are correct calculations. Since both (B) and (C) show valid methods leading to the correct answer, this option is the most appropriate.

The final answer is (D) Both (B) and (C) are correct calculations.

The reciprocal of a non-zero number $x$ is defined as $\frac{1}{x}$.

The given number is $5^3$. To find its reciprocal, we take 1 and divide it by $5^3$.

The reciprocal of $5^3$ is $\frac{1}{5^3}$.

This form matches option (B).

Next, let's consider the rule for negative exponents, which states that for a non-zero base $a$ and an integer exponent $n$, $a^{-n} = \frac{1}{a^n}$.

Applying this rule, $\frac{1}{5^3}$ can be written as $5^{-3}$ (here $a=5$ and $n=3$).

So, $5^{-3}$ is also a representation of the reciprocal of $5^3$.

This form matches option (A).

Finally, let's evaluate the value of $5^3$.

$5^3 = 5 \times 5 \times 5 = 25 \times 5 = 125$

Therefore, the reciprocal $\frac{1}{5^3}$ is equal to $\frac{1}{125}$.

This value matches option (C).

Since options (A), (B), and (C) all correctly represent the reciprocal of $5^3$, the statement that all of the above are correct is true.

The final answer is (D) All of the above.

To simplify the expression $\frac{a^8 b^5}{a^3 b^2}$, we can use the law of exponents for division of powers with the same base.

The expression can be rewritten as the product of two fractions:

$\frac{a^8 b^5}{a^3 b^2} = \frac{a^8}{a^3} \times \frac{b^5}{b^2}$

We apply the rule $\frac{x^m}{x^n} = x^{m-n}$ separately to each fraction:

For the terms with base $a$:

$\frac{a^8}{a^3} = a^{8-3} = a^5$

For the terms with base $b$:

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

Now, we combine the simplified terms:

$\frac{a^8 b^5}{a^3 b^2} = a^5 \times b^3 = a^5 b^3$

Comparing our simplified expression with the given options:

(A) $a^{11} b^7$ (This would be the result of $a^8 b^5 \times a^3 b^2$)

(B) $a^5 b^3$ (Correct)

(C) $a^5 b^7$ (Incorrect exponent for $b$)

(D) $a^{11} b^3$ (Incorrect exponent for $a$)

The simplified form $a^5 b^3$ matches option (B).

The final answer is (B) $a^5 b^3$.

To determine which expression is NOT equal to 1, we need to evaluate each option.

(A) $100^0$: According to the law of exponents, any non-zero number raised to the power of 0 is 1.

$100^0 = 1$

(B) $(\frac{5}{5})^2$: First, evaluate the expression inside the parentheses. $\frac{5}{5} = 1$. Then, raise the result to the power of 2.

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

(C) $0^1$: According to the definition of exponents, any number raised to the power of 1 is the number itself.

$0^1 = 0$

(D) $1^5$: According to the definition of exponents, 1 raised to any power is 1.

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

Comparing the values:

(A) $100^0 = 1$

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

(C) $0^1 = 0$

(D) $1^5 = 1$

The expression that is not equal to 1 is $0^1$, which has a value of 0.

The final answer is (C) $0^1$.

Standard form (or scientific notation) is a way of writing numbers as the product of a number between 1 (inclusive) and 10 (exclusive) and an integer power of 10. The general format is $a \times 10^n$, where $1 \leq |a| < 10$ and $n$ is an integer.

The given number is the distance of the Moon from the Earth, which is $384,000\text{ km}$.

To express this number in standard form, we need to find a number $a$ such that $1 \leq a < 10$ and an integer $n$ such that $384,000 = a \times 10^n$.

The significant digits in 384,000 are 3, 8, and 4. To get a number between 1 and 10 using these digits, we place the decimal point after the first digit: 3.84.

So, the coefficient $a$ is 3.84.

Now, we need to find the power of 10, $10^n$. The original number 384,000 can be written as 384,000. (with the decimal point at the end).

To get from 384,000. to 3.84, we need to move the decimal point to the left.

Let's count the number of places the decimal point is moved:

$384000.$

$38400.0$ (1 place left)

$3840.00$ (2 places left)

$384.000$ (3 places left)

$38.4000$ (4 places left)

$3.84000$ (5 places left)

The decimal point was moved 5 places to the left. When the decimal point is moved to the left, the exponent of 10 is positive and equal to the number of places moved.

So, the exponent $n$ is 5.

Therefore, 384,000 in standard form is $3.84 \times 10^5$.

Including the unit, the standard form is $3.84 \times 10^5\text{ km}$.

Comparing our result with the given options:

(A) $3.84 \times 10^5\text{ km}$ (Correct)

(B) $3.84 \times 10^3\text{ km}$ (Incorrect, this is 3840 km)

(C) $3.84 \times 10^6\text{ km}$ (Incorrect, this is 3,840,000 km)

(D) $384 \times 10^3\text{ km}$ (This is equal to 384,000 km, but it is not in standard form because 384 is not between 1 and 10)

The standard form representation is uniquely given by option (A).

We need to simplify the expression $(a^0)^5$. We can use the laws of exponents to do this. We assume $a \neq 0$, as $0^0$ is undefined.

Method 1: Evaluate the inner term first.

According to the law of exponents, any non-zero number raised to the power of zero is 1.

$a^0 = 1$ (assuming $a \neq 0$)

So, the expression becomes:

$(a^0)^5 = (1)^5$

Now, we evaluate $1^5$. Any power of 1 is always 1.

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

The value of the expression is 1.

Option (B) states $a^0 = 1$, which is a correct step in this method and the value of the simplified expression.

Method 2: Use the Power of a Power rule first.

The power of a power rule states that $(x^m)^n = x^{mn}$. In the expression $(a^0)^5$, the base is $a$, the inner exponent is $m=0$, and the outer exponent is $n=5$.

Applying the rule:

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

Calculate the product of the exponents:

$0 \times 5 = 0$

So the expression simplifies to $a^0$.

$a^0 = 1$ (assuming $a \neq 0$)

The value of the expression is 1.

Option (C) states $a^{0 \times 5} = a^0 = 1$, which shows the application of the power of a power rule and the final evaluation, arriving at the correct value 1.

Both methods of simplification lead to the same result, which is 1. Option (B) shows a correct intermediate step and the final value. Option (C) shows the application of exponent rules leading to the final value. Both are correct calculations performed during the simplification process.

Comparing our findings with the given options:

(A) $a^5$. Incorrect.

(B) $a^0 = 1$. This is a correct calculation ($a^0$ equals 1) and the final value of the expression.

(C) $a^{0 \times 5} = a^0 = 1$. This is a sequence of correct calculations using exponent rules that leads to the final value.

(D) Both (B) and (C) are correct calculations. Since both calculations shown in (B) and (C) are mathematically correct ways to arrive at the simplified value, option (D) is the most accurate choice.

The final answer is (D) Both (B) and (C) are correct calculations.

We are given the equation $2^x = 16$. We need to find the value of $x$ that satisfies this equation. This involves expressing 16 as a power of the same base, which is 2.

We need to find the exponent to which 2 must be raised to equal 16. We can do this by listing the powers of 2:

$2^1 = 2$

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

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

$2^4 = 2 \times 2 \times 2 \times 2 = 16$

We see that $2^4 = 16$.

Now we can rewrite the given equation by replacing 16 with $2^4$:

$2^x = 2^4$

In exponential equations where the bases are equal, the exponents must also be equal.

In our equation $2^x = 2^4$, the base is 2, which satisfies the condition $a \neq 0, 1, -1$.

Therefore, we can equate the exponents:

$x = 4$

Comparing our result with the given options:

(A) 2 ($2^2 = 4 \neq 16$)

(B) 4 ($2^4 = 16$, correct)

(C) 8 ($2^8 = 256 \neq 16$)

(D) 16 ($2^{16}$ is a very large number, not 16)

The value of $x$ is 4, which matches option (B).

The final answer is (B) 4.

The number given is $1,50,000$, which is 150,000 in the international numbering system.

Standard form (or scientific notation) is a way of writing numbers as the product of a number between 1 (inclusive) and 10 (exclusive) and an integer power of 10. The general format is $a \times 10^n$, where $1 \leq |a| < 10$ and $n$ is an integer.

We need to express $150,000$ in the form $a \times 10^n$, where $1 \leq a < 10$.

The significant digits in 150,000 are 1 and 5. To obtain a number between 1 and 10 using these digits, we place the decimal point after the first non-zero digit, which is 1. So, the coefficient $a$ is $1.5$.

Now, we need to determine the exponent $n$. The original number $150,000$ can be thought of as $150,000.$ (with the decimal point at the end).

We need to move the decimal point from its current position to the position after the digit 1 to get 1.5.

$150000.$

Moving the decimal point one place to the left gives $15000.0$.

Moving it two places to the left gives $1500.00$.

Moving it three places to the left gives $150.000$.

Moving it four places to the left gives $15.0000$.

Moving it five places to the left gives $1.50000$.

The decimal point has been moved 5 places to the left. When moving the decimal point to the left for a number greater than 1, the exponent of 10 is positive and is equal to the number of places moved.

Thus, the exponent $n = 5$.

The standard form of $150,000$ is $1.5 \times 10^5$.

Let's examine the given options:

(A) $1.5 \times 10^5$: The coefficient $1.5$ is between 1 and 10, and the power of 10 is $10^5$. This matches our result and is in standard form.

(B) $1.5 \times 10^4$: This equals $1.5 \times 10,000 = 15,000$. This is not $150,000$. It is also in standard form, but it represents a different number.

(C) $15 \times 10^4$: This equals $15 \times 10,000 = 150,000$. This is the correct value, but it is not in standard form because the coefficient 15 is not between 1 and 10 ($1 \leq 15 < 10$ is false).

(D) $150 \times 10^3$: This equals $150 \times 1,000 = 150,000$. This is also the correct value, but it is not in standard form because the coefficient 150 is not between 1 and 10 ($1 \leq 150 < 10$ is false).

Only option (A) represents the number $150,000$ in standard form.

The final answer is (A) $1.5 \times 10^5$.

This question asks for the value of a negative base, specifically -1, raised to an odd integer exponent.

Let's consider the result of multiplying -1 by itself multiple times:

$(-1)^1 = -1$ (The exponent 1 is odd)

$(-1)^2 = (-1) \times (-1) = 1$ (The exponent 2 is even)

$(-1)^3 = (-1) \times (-1) \times (-1) = 1 \times (-1) = -1$ (The exponent 3 is odd)

$(-1)^4 = (-1) \times (-1) \times (-1) \times (-1) = 1 \times 1 = 1$ (The exponent 4 is even)

$(-1)^5 = (-1) \times (-1) \times (-1) \times (-1) \times (-1) = 1 \times 1 \times (-1) = -1$ (The exponent 5 is odd)

From these examples, we observe a pattern:

When the exponent is an odd positive integer, the result is -1.

When the exponent is an even positive integer, the result is 1.

This rule holds true generally: A negative base raised to an odd exponent results in a negative value, and a negative base raised to an even exponent results in a positive value.

For the base -1:

$(-1)^n = -1$ if $n$ is an odd integer.

$(-1)^n = 1$ if $n$ is an even integer.

The question asks for the value of $(-1)^{\text{odd number}}$. Based on the rule, when the exponent is an odd number, the value of $(-1)$ raised to that power is always -1.

Comparing this result with the given options:

(A) 1

(B) -1

(C) 0

(D) Undefined

The value is -1, which matches option (B).

The final answer is (B) -1.

This question asks for the value of any non-zero base raised to the power of zero.

According to the definition of exponents, for any non-zero number $a$, the value of $a^0$ is equal to 1.

Mathematically:

$a^0 = 1$, for $a \neq 0$.

This rule can be understood using the division property of exponents. For any non-zero base $a$ and any positive integer $m$, we know that:

$\frac{a^m}{a^m} = a^{m-m} = a^0$

Also, any non-zero number divided by itself is equal to 1:

$\frac{a^m}{a^m} = 1$

Equating these two results gives us $a^0 = 1$ for $a \neq 0$.

Comparing the value 1 with the given options:

(A) $a$

(B) 0

(C) 1

(D) $\frac{1}{a}$

The value 1 matches option (C).

The final answer is (C) 1.

(a) The expression $5 \times 5 \times 5 \times 5$ means that the number 5 is multiplied by itself 4 times. Therefore, the exponential form is $5^4$.

(b) The expression $a \times a \times a \times a \times a$ means that the variable $a$ is multiplied by itself 5 times. Therefore, the exponential form is $a^5$.

(a) In the expression $7^3$:

The base is 7.

The exponent is 3.

(b) In the expression $(-2)^5$:

The base is $(-2)$.

The exponent is 5.

(a) To express 36 as a product of powers of its prime factors, we find the prime factorisation of 36.

$\begin{array}{c|cc} 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$.

(b) To express 108 as a product of powers of its prime factors, we find the prime factorisation of 108.

$\begin{array}{c|cc} 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^3$.

(a) To evaluate $2^6$, we multiply the base 2 by itself 6 times.

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

(b) To evaluate $(-3)^4$, we multiply the base $(-3)$ by itself 4 times.

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

Since the exponent is even, the result will be positive.

$(-3) \times (-3) = 9$

$(-3) \times (-3) = 9$

So, $9 \times 9 = 81$.

$(-3)^4 = 81$

To compare $3^2$ and $2^3$, we need to evaluate each expression.

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

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

Comparing the values, we have $9 > 8$.

Therefore, $3^2$ is greater than $2^3$.

To simplify $3^2 \times 3^4$, we use the law of exponents for multiplying powers with the same base, which states that $a^m \times a^n = a^{m+n}$.

Here, the base is 3, $m=2$, and $n=4$.

So, we add the exponents:

$3^2 \times 3^4 = 3^{2+4}$

$= 3^6$

The simplified expression is $3^6$.

To simplify $5^7 \div 5^3$, we use the law of exponents for dividing powers with the same base, which states that $a^m \div a^n = a^{m-n}$.

Here, the base is 5, $m=7$, and $n=3$.

So, we subtract the exponents:

$5^7 \div 5^3 = 5^{7-3}$

$= 5^4$

Alternatively, we can evaluate $5^4$:

$5^4 = 5 \times 5 \times 5 \times 5 = 25 \times 25 = 625$

So the simplified form is $5^4$ or $625$.

To simplify $(2^3)^4$, we use the law of exponents for power of a power, which states that $(a^m)^n = a^{m \times n}$.

Here, the base is 2, the inner exponent is $m=3$, and the outer exponent is $n=4$.

So, we multiply the exponents:

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

$= 2^{12}$

The simplified expression is $2^{12}$.

To simplify $(a \times b)^3$, we use the law of exponents for the power of a product, which states that $(ab)^m = a^m b^m$.

Here, the bases are $a$ and $b$, and the exponent is $m=3$.

So, we apply the exponent to each factor inside the parenthesis:

$(a \times b)^3 = a^3 \times b^3$

The simplified expression is $a^3 b^3$.

To simplify $(\frac{x}{y})^5$, we use the law of exponents for the power of a quotient, which states that $(\frac{a}{b})^m = \frac{a^m}{b^m}$.

Here, the numerator is $x$, the denominator is $y$, and the exponent is $m=5$.

So, we apply the exponent to both the numerator and the denominator:

$(\frac{x}{y})^5 = \frac{x^5}{y^5}$

The simplified expression is $\frac{x^5}{y^5}$.

To evaluate $(-1)^{100}$, we consider the base which is $-1$ and the exponent which is 100.

When the base is $-1$ and the exponent is an even number, the result is always 1.

The exponent 100 is an even number.

Therefore, $(-1)^{100} = 1$.

To evaluate $(-1)^{101}$, we consider the base which is $-1$ and the exponent which is 101.

When the base is $-1$ and the exponent is an odd number, the result is always $-1$.

The exponent 101 is an odd number.

Therefore, $(-1)^{101} = -1$.

For any non-zero number $a$, the value of $a^0$ is always 1.

This is a fundamental property of exponents.

Example:

Let's consider the number 3.

According to the rule, $3^0 = 1$.

Let's consider the number $-5$.

According to the rule, $(-5)^0 = 1$.

We need to simplify the expression $2^0 + 3^0 + 4^0$.

We know that any non-zero number raised to the power of 0 is equal to 1. That is, for any $a \neq 0$, $a^0 = 1$.

Applying this rule to each term in the expression:

$2^0 = 1$

$3^0 = 1$

$4^0 = 1$

Now, substitute these values back into the expression:

$2^0 + 3^0 + 4^0 = 1 + 1 + 1$

$= 3$

The simplified value of the expression is 3.

We need to simplify the expression $(2^2 \times 3^3) \div (2 \times 3)$.

We can rewrite the expression as a fraction:

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

We can treat $2$ as $2^1$ and $3$ as $3^1$. The expression becomes:

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

Using the law of exponents for division with the same base, $a^m \div a^n = a^{m-n}$, we can simplify the terms with base 2 and base 3 separately.

For the base 2: $2^2 \div 2^1 = 2^{2-1} = 2^1$

For the base 3: $3^3 \div 3^1 = 3^{3-1} = 3^2$

Now, multiply the simplified terms:

$2^1 \times 3^2$

Evaluate the powers:

$2^1 = 2$

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

Multiply the results:

$2 \times 9 = 18$

The simplified value of the expression is 18.

To express 729 as a power of 3, we need to find how many times 3 is multiplied by itself to get 729. This can be done by finding the prime factorisation of 729 using the base 3.

$\begin{array}{c|cc} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

From the prime factorisation, we see that 729 is obtained by multiplying 3 by itself 6 times.

$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$

In exponential form, this is written as $3^6$.

Therefore, $729 = 3^6$.

To express 243 as a power of 3, we need to find how many times 3 is multiplied by itself to get 243. This can be done by finding the prime factorisation of 243 using the base 3.

$\begin{array}{c|cc} 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

From the prime factorisation, we see that 243 is obtained by multiplying 3 by itself 5 times.

$243 = 3 \times 3 \times 3 \times 3 \times 3$

In exponential form, this is written as $3^5$.

Therefore, $243 = 3^5$.

We need to simplify the expression $\frac{2^5 \times 3^4}{2^2 \times 3^2}$.

We can apply the division law of exponents, which states that for the same base, $a^m \div a^n = a^{m-n}$. We apply this separately to the terms with base 2 and the terms with base 3.

For the base 2:

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

For the base 3:

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

Now, we combine the results by multiplication:

$\frac{2^5 \times 3^4}{2^2 \times 3^2} = 2^3 \times 3^2$

Finally, we evaluate the powers:

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

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

Multiply the values:

$8 \times 9 = 72$

The simplified value of the expression is 72.

We need to simplify the expression $(5^2 \times 5^4) \div 5^3$.

First, we simplify the multiplication inside the parenthesis using the law $a^m \times a^n = a^{m+n}$:

$5^2 \times 5^4 = 5^{2+4} = 5^6$

Now, the expression becomes $5^6 \div 5^3$.

We use the law of exponents for division with the same base, $a^m \div a^n = a^{m-n}$:

$5^6 \div 5^3 = 5^{6-3} = 5^3$

The simplified expression is $5^3$. We can also evaluate this value:

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

The simplified value is 125.

To compare $10^2$ and $2^{10}$, we need to evaluate each expression.

Evaluate $10^2$:

$10^2 = 10 \times 10 = 100$

Evaluate $2^{10}$:

$2^{10} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$2^{10} = (2 \times 2 \times 2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2 \times 2)$

$2^{10} = 32 \times 32$

$2^{10} = 1024$

Now, we compare the values: $100$ and $1024$.

$100 < 1024$

Therefore, $2^{10}$ is greater than $10^2$.

$2^{10} > 10^2$

To write 64 in exponential form with base 2, we find how many times 2 must be multiplied by itself to get 64. We can do this by prime factorization.

$\begin{array}{c|cc} 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

From the factorization, we see that $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2$.

So, in exponential form with base 2, $64 = 2^6$.

To write 64 in exponential form with base 8, we find how many times 8 must be multiplied by itself to get 64.

We know that $8 \times 8 = 64$.

So, in exponential form with base 8, $64 = 8^2$.

To express 125 as a power of 5, we need to find how many times 5 is multiplied by itself to get 125. This can be done by finding the prime factorisation of 125 using the base 5.

$\begin{array}{c|cc} 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

From the prime factorisation, we see that 125 is obtained by multiplying 5 by itself 3 times.

$125 = 5 \times 5 \times 5$

In exponential form, this is written as $5^3$.

Therefore, $125 = 5^3$.

We need to simplify the expression $(7^2)^3 \div 7^4$ using the laws of exponents.

First, we apply the power of a power rule, $(a^m)^n = a^{m \times n}$, to the term $(7^2)^3$.

$(7^2)^3 = 7^{2 \times 3} = 7^6$

Now the expression becomes $7^6 \div 7^4$.

Next, we apply the division law of exponents, $a^m \div a^n = a^{m-n}$, to $7^6 \div 7^4$.

$7^6 \div 7^4 = 7^{6-4} = 7^2$

The simplified expression is $7^2$.

We can also evaluate the final value:

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

The simplified value of the expression is 49.

We are given the equation $2^x = 32$.

To find the value of $x$, we need to express the number 32 as a power of the same base as the left side, which is 2.

We find the prime factorization of 32:

$\begin{array}{c|cc} 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

From the prime factorization, we see that $32 = 2 \times 2 \times 2 \times 2 \times 2$.

So, 32 can be written in exponential form with base 2 as $2^5$.

Now, substitute this into the given equation:

$2^x = 2^5$

When two exponential expressions with the same non-zero and non-one base are equal, their exponents must be equal.

Since the bases are both 2, which is not 0, 1, or -1, we can equate the exponents.

Therefore, $x = 5$.

The value of $x$ is 5.

To write the number $1,40,000$ in standard form (scientific notation), we need to express it in the form $a \times 10^n$, where $a$ is a number such that $1 \leq a < 10$, and $n$ is an integer.

The number is $140,000$. The decimal point is currently at the end, i.e., $140000.$

We need to move the decimal point to the left until the number is between 1 and 10. The new position of the decimal point should be after the first non-zero digit.

Moving the decimal point from the right end to after the digit 1:

$1.40000$

We moved the decimal point 5 places to the left.

The number of places the decimal point is moved gives the exponent $n$. Since we moved the decimal point to the left, the exponent is positive.

So, the exponent is $n = 5$.

The number $a$ is the number with the decimal point moved to the correct position, which is $1.4$ (we can drop the trailing zeros after the decimal point).

Therefore, the standard form of $1,40,000$ is $1.4 \times 10^5$.

The concept of exponents and powers is a way to represent repeated multiplication of the same number by itself. Instead of writing out the multiplication many times, we use a shorter notation.

The general expression for an exponential form is $a^n$.

In this expression:

• The letter 'a' represents the base. The base is the number that is being multiplied repeatedly.
• The letter 'n' represents the exponent or power. The exponent tells us how many times the base is used as a factor in the multiplication.

For example, consider the expression $2^3$.

• Here, the base is 2.
• The exponent is 3.

This means we multiply the base (2) by itself 3 times:

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

So, $2^3$ is read as "2 raised to the power of 3" or "2 cubed", and its value is 8.

Now, let's write the requested numbers in exponential form:

1000 in exponential form with base 10:

We need to find how many times 10 is multiplied by itself to get 1000.

$10 \times 10 = 100$

$10 \times 10 \times 10 = 1000$

So, 1000 is obtained by multiplying 10 by itself 3 times.

In exponential form with base 10, $1000 = 10^3$.

128 in exponential form with base 2:

We need to find how many times 2 is multiplied by itself to get 128. We can do this by finding the prime factorization of 128 using the base 2.

$\begin{array}{c|cc} 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

From the prime factorization, we see that 128 is obtained by multiplying 2 by itself 7 times.

$128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

In exponential form with base 2, $128 = 2^7$.

There are several laws of exponents that simplify calculations involving powers. Here, we explain two of the fundamental laws.

1. Law of Multiplication of Powers with the Same Base

Statement: For any non-zero number $a$ and positive integers $m$ and $n$, the law is stated as: $a^m \times a^n = a^{m+n}$.

Explanation: This law states that when you multiply powers with the same base, you can keep the base the same and add the exponents.

This makes sense because $a^m$ means $a$ multiplied by itself $m$ times, and $a^n$ means $a$ multiplied by itself $n$ times. When you multiply $a^m$ by $a^n$, you are multiplying $a$ by itself a total of $m+n$ times.

Example 1: Simplify $2^3 \times 2^2$.

By expansion: $2^3 \times 2^2 = (2 \times 2 \times 2) \times (2 \times 2) = 2 \times 2 \times 2 \times 2 \times 2 = 2^5$.

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

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

Both methods give the same result, $2^5$, which is $32$.

Example 2: Simplify $5^4 \times 5^1$.

Using the law $a^m \times a^n = a^{m+n}$ with $a=5$, $m=4$, $n=1$:

$5^4 \times 5^1 = 5^{4+1} = 5^5$.

The simplified expression is $5^5$, which is $3125$.

2. Law of Division of Powers with the Same Base

Statement: For any non-zero number $a$ and positive integers $m$ and $n$ with $m > n$, the law is stated as: $a^m \div a^n = a^{m-n}$.

Explanation: This law states that when you divide powers with the same base, you can keep the base the same and subtract the exponent of the denominator from the exponent of the numerator.

This makes sense because $a^m \div a^n$ can be written as a fraction $\frac{a^m}{a^n} = \frac{a \times a \times ... \times a \text{ ($m$ times)}}{a \times a \times ... \times a \text{ ($n$ times)}}$. When you cancel out $n$ factors of $a$ from both the numerator and the denominator, you are left with $m-n$ factors of $a$ in the numerator.

Example 1: Simplify $3^5 \div 3^2$.

By expansion and cancellation: $3^5 \div 3^2 = \frac{3^5}{3^2} = \frac{\cancel{3} \times \cancel{3} \times 3 \times 3 \times 3}{\cancel{3} \times \cancel{3}} = 3 \times 3 \times 3 = 3^3$.

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

$3^5 \div 3^2 = 3^{5-2} = 3^3$.

Both methods give the same result, $3^3$, which is $27$.

Example 2: Simplify $10^7 \div 10^3$.

Using the law $a^m \div a^n = a^{m-n}$ with $a=10$, $m=7$, $n=3$:

$10^7 \div 10^3 = 10^{7-3} = 10^4$.

The simplified expression is $10^4$, which is $10,000$.

Here, we explain two more important laws of exponents that help simplify expressions involving powers.

1. Law of Power of a Power

Statement: For any non-zero number $a$ and integers $m$ and $n$, the law is stated as: $(a^m)^n = a^{m \times n}$.

Explanation: This law states that when you raise a power ($a^m$) to another power ($n$), you can keep the base ($a$) the same and multiply the exponents ($m$ and $n$).

This is because $(a^m)^n$ means $a^m$ multiplied by itself $n$ times: $(a^m)^n = a^m \times a^m \times ... \times a^m$ ($n$ times). Since each $a^m$ is $a$ multiplied by itself $m$ times, you end up with $a$ multiplied by itself a total of $m \times n$ times.

Example 1: Simplify $(2^3)^2$.

By expansion: $(2^3)^2 = 2^3 \times 2^3 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$.

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

$(2^3)^2 = 2^{3 \times 2} = 2^6$.

Both methods give the same result, $2^6$, which is $64$.

Example 2: Simplify $(x^4)^5$.

Using the law $(a^m)^n = a^{m \times n}$ with $a=x$, $m=4$, $n=5$:

$(x^4)^5 = x^{4 \times 5} = x^{20}$.

The simplified expression is $x^{20}$.

2. Law of Multiplication of Powers with the Same Exponents

Statement: For any non-zero numbers $a$ and $b$ and integer $m$, the law is stated as: $a^m \times b^m = (ab)^m$.

Explanation: This law states that when you multiply powers with different bases but the same exponent, you can multiply the bases first and then raise the product to the common exponent.

This works because $a^m \times b^m = (a \times a \times ... \times a \text{ ($m$ times)}) \times (b \times b \times ... \times b \text{ ($m$ times)})$. By rearranging the terms (due to commutativity and associativity of multiplication), we can group pairs of $a$ and $b$: $(a \times b) \times (a \times b) \times ... \times (a \times b)$ ($m$ times). This is equivalent to $(ab)^m$.

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

By expansion: $2^3 \times 5^3 = (2 \times 2 \times 2) \times (5 \times 5 \times 5) = (2 \times 5) \times (2 \times 5) \times (2 \times 5) = 10 \times 10 \times 10 = 10^3$.

Using the law $a^m \times b^m = (ab)^m$ with $a=2$, $b=5$, $m=3$:

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

Both methods give the same result, $10^3$, which is $1000$.

Example 2: Simplify $3^4 \times y^4$.

Using the law $a^m \times b^m = (ab)^m$ with $a=3$, $b=y$, $m=4$:

$3^4 \times y^4 = (3 \times y)^4 = (3y)^4$.

The simplified expression is $(3y)^4$.

Here, we explain two additional important laws of exponents that help simplify expressions involving powers.

1. Law of Division of Powers with the Same Exponents

Statement: For any non-zero numbers $a$ and $b$ and integer $m$, the law is stated as: $\frac{a^m}{b^m} = (\frac{a}{b})^m$.

Explanation: This law states that when you divide powers with different bases but the same exponent, you can divide the bases first and then raise the quotient to the common exponent.

This is because $\frac{a^m}{b^m} = \frac{a \times a \times ... \times a \text{ ($m$ times)}}{b \times b \times ... \times b \text{ ($m$ times)}}$. This can be rewritten as $\frac{a}{b} \times \frac{a}{b} \times ... \times \frac{a}{b}$ ($m$ times), which is equivalent to $(\frac{a}{b})^m$.

Example 1: Simplify $\frac{6^3}{2^3}$.

By expansion: $\frac{6^3}{2^3} = \frac{6 \times 6 \times 6}{2 \times 2 \times 2} = \frac{216}{8} = 27$.

Using the law $\frac{a^m}{b^m} = (\frac{a}{b})^m$ with $a=6$, $b=2$, $m=3$:

$\frac{6^3}{2^3} = (\frac{6}{2})^3 = 3^3 = 3 \times 3 \times 3 = 27$.

Both methods give the same result, $27$.

Example 2: Simplify $\frac{15^2}{3^2}$.

Using the law $\frac{a^m}{b^m} = (\frac{a}{b})^m$ with $a=15$, $b=3$, $m=2$:

$\frac{15^2}{3^2} = (\frac{15}{3})^2 = 5^2 = 5 \times 5 = 25$.

The simplified value is $25$.

2. Zero Exponent Law

Statement: For any non-zero number $a$, the law is stated as: $a^0 = 1$.

Explanation: This law states that any non-zero base raised to the power of 0 is equal to 1.

We can understand this using the division law of exponents. Consider $\frac{a^m}{a^m}$ where $a \neq 0$. Using the division law, $\frac{a^m}{a^m} = a^{m-m} = a^0$. However, any non-zero number divided by itself is always 1. So, $\frac{a^m}{a^m} = 1$. By equating the two results, we get $a^0 = 1$. Note that the base cannot be zero ($0^0$) as it is undefined.

Example 1: Evaluate $5^0$.

Using the zero exponent law $a^0 = 1$ with $a=5$ (which is non-zero):

$5^0 = 1$.

The value is $1$.

Example 2: Evaluate $(-7)^0$.

Using the zero exponent law $a^0 = 1$ with $a=-7$ (which is non-zero):

$(-7)^0 = 1$.

The value is $1$.

(a) Simplify $(2^3 \times 2^5)^2 \div (2^4)^3$

Start with the expression: $(2^3 \times 2^5)^2 \div (2^4)^3$

First, simplify the multiplication inside the first parenthesis using the law $a^m \times a^n = a^{m+n}$:

$2^3 \times 2^5 = 2^{3+5} = 2^8$

The expression becomes: $(2^8)^2 \div (2^4)^3$

Next, apply the power of a power rule $(a^m)^n = a^{m \times n}$ to both terms:

$(2^8)^2 = 2^{8 \times 2} = 2^{16}$

$(2^4)^3 = 2^{4 \times 3} = 2^{12}$

The expression becomes: $2^{16} \div 2^{12}$

Now, apply the division law of exponents $a^m \div a^n = a^{m-n}$:

$2^{16} \div 2^{12} = 2^{16-12} = 2^4$

Finally, evaluate the result:

$2^4 = 2 \times 2 \times 2 \times 2 = 16$

The simplified value is 16.

(b) Simplify $\frac{3^2 \times 10^5 \times 25}{5^3 \times 6^2}$

Start with the expression: $\frac{3^2 \times 10^5 \times 25}{5^3 \times 6^2}$

Express all bases as prime factors. $10 = 2 \times 5$, $25 = 5^2$, $6 = 2 \times 3$.

Substitute these prime factors into the expression:

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

Apply the power of a product rule $(ab)^m = a^m b^m$ to $(2 \times 5)^5$ and $(2 \times 3)^2$:

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

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

Substitute these back into the expression:

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

Combine terms with the same base in the numerator using $a^m \times a^n = a^{m+n}$. For base 5: $5^5 \times 5^2 = 5^{5+2} = 5^7$.

The numerator is now $3^2 \times 2^5 \times 5^7$.

The denominator is $5^3 \times 2^2 \times 3^2$ (rearranged for clarity).

The expression becomes: $\frac{2^5 \times 3^2 \times 5^7}{2^2 \times 3^2 \times 5^3}$

Now, simplify the division for each base using the law $a^m \div a^n = a^{m-n}$:

For base 2: $\frac{2^5}{2^2} = 2^{5-2} = 2^3$

For base 3: $\frac{3^2}{3^2} = 3^{2-2} = 3^0 = 1$

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

Multiply the simplified terms:

$2^3 \times 1 \times 5^4 = 2^3 \times 5^4$

Evaluate the powers and multiply:

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

$5^4 = 5 \times 5 \times 5 \times 5 = 625$

$8 \times 625$

$\begin{array}{cc}& & 6 & 2 & 5 \\ \times & & & & 8 \\ \hline & 5 & 0 & 0 & 0 \\ \hline \end{array}$

$8 \times 625 = 5000$

The simplified value is 5000.

To compare the values of $3^5$ and $5^3$, we need to evaluate each expression.

Calculate the value of $3^5$:

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

$3 \times 3 = 9$

$9 \times 3 = 27$

$27 \times 3 = 81$

$81 \times 3 = 243$

So, $3^5 = 243$.

Calculate the value of $5^3$:

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

$5 \times 5 = 25$

$25 \times 5 = 125$

So, $5^3 = 125$.

Now, we compare the calculated values: $243$ and $125$.

$243 > 125$

Therefore, $3^5$ is greater than $5^3$.

$3^5 > 5^3$

To express $1024 \times 81$ as a product of powers of prime factors, we first find the prime factorization of each number separately.

Find the prime factorization of 1024:

$\begin{array}{c|cc} 2 & 1024 \\ \hline 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, $1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{10}$.

Find the prime factorization of 81:

$\begin{array}{c|cc} 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $81 = 3 \times 3 \times 3 \times 3 = 3^4$.

Now, we express the product $1024 \times 81$ using their prime factor forms:

$1024 \times 81 = 2^{10} \times 3^4$

The number $1024 \times 81$ expressed as a product of powers of prime factors is $2^{10} \times 3^4$.

To write a number in standard form (scientific notation), we express it as a number between 1 and 10 (inclusive of 1, exclusive of 10) multiplied by a power of 10. The general form is $a \times 10^n$, where $1 \leq a < 10$ and $n$ is an integer.

(a) Write $5,00,00,000$ in standard form.

The number is $50,000,000$.

To get a number between 1 and 10, we place the decimal point after the first digit 5: $5.0$.

We count how many places we moved the decimal point from its original position (at the end of the number). The original number is $50,000,000.$, and the new number is $5.0$. We moved the decimal 7 places to the left.

Since the decimal was moved to the left, the exponent of 10 is positive, and the number of places moved is 7.

So, the standard form is $5 \times 10^7$.

(b) Write $3,25,000$ in standard form.

The number is $325,000$.

To get a number between 1 and 10, we place the decimal point after the first digit 3: $3.25$.

We count how many places we moved the decimal point from its original position (at the end of the number). The original number is $325,000.$, and the new number is $3.25$. We moved the decimal 5 places to the left.

Since the decimal was moved to the left, the exponent of 10 is positive, and the number of places moved is 5.

So, the standard form is $3.25 \times 10^5$.

(c) Write the distance $384,000,000$ meters in standard form.

The number is $384,000,000$.

To get a number between 1 and 10, we place the decimal point after the first digit 3: $3.84$.

We count how many places we moved the decimal point from its original position (at the end of the number). The original number is $384,000,000.$, and the new number is $3.84$. We moved the decimal 8 places to the left.

Since the decimal was moved to the left, the exponent of 10 is positive, and the number of places moved is 8.

So, the standard form is $3.84 \times 10^8$ meters.

To write a number from standard form ($a \times 10^n$) to usual form, we move the decimal point of the number $a$ by $n$ places.

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

(a) Write $3.5 \times 10^4$ in usual form.

The number is $3.5$. The exponent is $4$. We need to move the decimal point 4 places to the right.

$3.5 \times 10^4 = 3.5000 \times 10^4$

Moving the decimal point 4 places to the right gives:

$\text{35000}$

So, $3.5 \times 10^4 = 35,000$.

(b) Write $7.08 \times 10^6$ in usual form.

The number is $7.08$. The exponent is $6$. We need to move the decimal point 6 places to the right.

$7.08 \times 10^6 = 7.080000 \times 10^6$

Moving the decimal point 6 places to the right gives:

$\text{7080000}$

So, $7.08 \times 10^6 = 7,080,000$.

(c) Write $1.15 \times 10^9$ in usual form.

The number is $1.15$. The exponent is $9$. We need to move the decimal point 9 places to the right.

$1.15 \times 10^9 = 1.150000000 \times 10^9$

Moving the decimal point 9 places to the right gives:

$\text{1150000000}$

So, $1.15 \times 10^9 = 1,150,000,000$.

We need to simplify the expression $\frac{(3^2 \times 3^5)^3 \times 2^6}{ (2^3)^2 \times 3^7 }$ using the laws of exponents and express the result in exponential form.

First, simplify the expression inside the parenthesis in the numerator, $3^2 \times 3^5$, using the law $a^m \times a^n = a^{m+n}$:

$3^2 \times 3^5 = 3^{2+5} = 3^7$

The numerator term becomes $(3^7)^3$. Now apply the power of a power rule $(a^m)^n = a^{m \times n}$ to this term:

$(3^7)^3 = 3^{7 \times 3} = 3^{21}$

So the numerator is $3^{21} \times 2^6$.

Next, simplify the term in the denominator $(2^3)^2$ using the power of a power rule $(a^m)^n = a^{m \times n}$:

$(2^3)^2 = 2^{3 \times 2} = 2^6$

So the denominator is $2^6 \times 3^7$.

The expression now looks like this:

$\frac{3^{21} \times 2^6}{2^6 \times 3^7}$

Now, we can rearrange the terms and apply the division law of exponents $\frac{a^m}{a^n} = a^{m-n}$ separately for each base (base 2 and base 3):

$\frac{3^{21}}{3^7} \times \frac{2^6}{2^6}$

For the base 3:

$\frac{3^{21}}{3^7} = 3^{21-7} = 3^{14}$

For the base 2:

$\frac{2^6}{2^6} = 2^{6-6} = 2^0$

The expression simplifies to the product of these terms:

$3^{14} \times 2^0$

Using the zero exponent law, $a^0 = 1$ (for $a \neq 0$), we have $2^0 = 1$.

$3^{14} \times 1$

Multiplying by 1 does not change the value:

$3^{14}$

The simplified result in exponential form is $3^{14}$.

We need to evaluate the expression $\frac{(5^2)^3 \times 5^4}{5^7 \times 5^2}$ using the laws of exponents.

First, let's simplify the numerator: $(5^2)^3 \times 5^4$.

Apply the power of a power rule $(a^m)^n = a^{m \times n}$ to $(5^2)^3$:

$(5^2)^3 = 5^{2 \times 3} = 5^6$

The numerator becomes: $5^6 \times 5^4$.

Now, apply the product rule $a^m \times a^n = a^{m+n}$ to $5^6 \times 5^4$:

$5^6 \times 5^4 = 5^{6+4} = 5^{10}$

So, the simplified numerator is $5^{10}$.

Next, let's simplify the denominator: $5^7 \times 5^2$.

Apply the product rule $a^m \times a^n = a^{m+n}$ to $5^7 \times 5^2$:

$5^7 \times 5^2 = 5^{7+2} = 5^9$

So, the simplified denominator is $5^9$.

Now the expression is simplified to $\frac{5^{10}}{5^9}$.

Apply the division rule $a^m \div a^n = a^{m-n}$:

$\frac{5^{10}}{5^9} = 5^{10-9} = 5^1$

Finally, evaluate the result:

$5^1 = 5$

The value of the expression is 5.

We are given the mass of the Earth and the mass of the Moon in standard form and need to find their total mass and express the result in standard form.

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

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

To find the total mass, we need to add these two values:

Total Mass = Mass of Earth + Mass of Moon

Total Mass = $(5.97 \times 10^{24}) + (7.35 \times 10^{22})$ kg

To add numbers in scientific notation, the powers of 10 must be the same. We can rewrite $7.35 \times 10^{22}$ with the exponent $10^{24}$.

$10^{22} = 10^{24-2} = 10^{24} \times 10^{-2}$

So, $7.35 \times 10^{22} = 7.35 \times 10^{-2} \times 10^{24}$.

To calculate $7.35 \times 10^{-2}$, we move the decimal point 2 places to the left:

$7.35 \times 10^{-2} = 0.0735$

Thus, $7.35 \times 10^{22} = 0.0735 \times 10^{24}$ kg.

Now, we can add the masses:

Total Mass = $(5.97 \times 10^{24}) + (0.0735 \times 10^{24})$

Factor out the common term $10^{24}$:

Total Mass = $(5.97 + 0.0735) \times 10^{24}$

Perform the addition of the coefficients:

$\begin{array}{c} & 5.9700 \\ + & 0.0735 \\ \hline & 6.0435 \\ \hline \end{array}$

$5.97 + 0.0735 = 6.0435$

So, the total mass is $6.0435 \times 10^{24}$ kg.

This result is already in standard form because the coefficient $6.0435$ is between 1 and 10 ($1 \leq 6.0435 < 10$).

The total mass of the Earth and the Moon in standard form is $6.0435 \times 10^{24}$ kg.