| Classwise Concept with Examples | ||||||
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| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
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| Powers with Negative Exponents | Laws of Exponents | Use of Exponents to Express Small Numbers in Stardard Form |
| Comparing Very Large and Very Small Numbers | ||
Chapter 12 Exponents and Powers (Concepts)
Welcome back to the powerful world of Exponents and Powers! This chapter revisits the fundamental concepts and laws you encountered in Class 7, but significantly extends their application and utility. We will solidify your understanding of how exponents provide a concise way to represent repeated multiplication and explore the rules that govern operations with powers more deeply. The key advancements in this chapter involve a thorough exploration of negative exponents and the expansion of Standard Form (Scientific Notation) to efficiently handle not only extremely large numbers but also incredibly small ones. Mastering these concepts is crucial for simplifying complex calculations and for effectively representing numbers encountered in various scientific and mathematical contexts.
Let's begin by reinforcing the basics. Recall that an expression like $a^n$ represents the base 'a' multiplied by itself 'n' times, where 'n' is the exponent or power. This compact notation is governed by a set of fundamental Laws of Exponents, which streamline calculations involving powers:
- Product Rule: $a^m \times a^n = a^{m+n}$ (Same base, add exponents)
- Quotient Rule: $\frac{a^m}{a^n} = a^{m-n}$ (Same base, subtract exponents, $a \neq 0$)
- Power of a Power Rule: $(a^m)^n = a^{m \times n}$ (Multiply exponents)
- Product to a Power Rule: $(ab)^m = a^m b^m$ (Distribute exponent to bases in product)
- Quotient to a Power Rule: $\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}$ (Distribute exponent to bases in quotient, $b \neq 0$)
- Zero Exponent Rule: $a^0 = 1$ (Any non-zero base to the power zero equals 1, $a \neq 0$)
A major focus of this chapter is the comprehensive understanding and application of negative exponents. The rule $a^{-n} = \frac{1}{a^n}$ (where $a \neq 0$) is central. This definition establishes that a negative exponent does not make the number negative (unless the base itself is negative); rather, it signifies the reciprocal of the base raised to the corresponding positive exponent. For instance, $3^{-2} = \frac{1}{3^2} = \frac{1}{9}$, and $5^{-1} = \frac{1}{5^1} = \frac{1}{5}$. Understanding this reciprocal relationship is key to simplifying expressions.
We will dedicate significant practice to simplifying and evaluating expressions that involve both positive and negative integer exponents, applying the established laws consistently. This requires careful attention to the rules and the order of operations. For example, to simplify $(2^{-3}) \times (2^5)$, we apply the product rule: $2^{-3+5} = 2^2 = 4$. Similarly, simplifying $\frac{x^{-4}}{x^{-6}}$ using the quotient rule gives $x^{-4 - (-6)} = x^{-4+6} = x^2$. Evaluating $(4^{-1} + 8^{-1}) \div (\frac{2}{3})^{-1}$ would involve applying the negative exponent rule first: $(\frac{1}{4} + \frac{1}{8}) \div (\frac{3}{2}) = (\frac{2+1}{8}) \times \frac{2}{3} = \frac{3}{8} \times \frac{2}{3} = \frac{1}{4}$.
A highly significant application of exponents, revisited and extended here, is expressing numbers in Standard Form, also known as Scientific Notation. This format represents any number as $k \times 10^n$, where $k$ is a number such that $1 \le k < 10$, and $n$ is an integer exponent. While Class 7 primarily used this for very large numbers (where $n$ is positive, e.g., $345,000,000 = 3.45 \times 10^8$), Class 8 emphasizes its use for very small numbers as well, utilizing negative exponents for $n$. For example, the very small number $0.000056$ can be concisely written in standard form as $5.6 \times 10^{-5}$. We will practice converting numbers between their usual decimal form and standard form (and vice versa) for both large and small values. This notation dramatically simplifies the comparison and calculation of numbers with vastly different magnitudes, proving indispensable in scientific disciplines dealing with astronomical distances or microscopic dimensions. It greatly enhances our computational toolkit for handling a wider range of numerical values efficiently.
Powers with Negative Exponents
In previous classes, you learned about exponents and powers. An exponent indicates how many times a base number is multiplied by itself. For example, $5^4$ means 5 is multiplied by itself 4 times ($5 \times 5 \times 5 \times 5$). Here, 5 is the base and 4 is the exponent or power. $a^n$ is read as "$a$ raised to the power of $n$" or "$a$ to the $n$-th power". If $a$ is a non-zero rational number and $n$ is a natural number, $a^n = a \times a \times \dots \times a$ (n times).
Extending the Concept of Exponents
Until now, we have primarily dealt with exponents that are positive integers (natural numbers) or zero. Let's explore how to define powers when the exponent is a negative integer.
Consider the powers of a non-zero base, say 2, and observe the pattern as the exponent decreases:
- $2^4 = 2 \times 2 \times 2 \times 2 = 16$
- $2^3 = 2 \times 2 \times 2 = 8$ (Notice that $8 = 16 \div 2$, or $2^3 = 2^4 \div 2^1$)
- $2^2 = 2 \times 2 = 4$ (Notice that $4 = 8 \div 2$, or $2^2 = 2^3 \div 2^1$)
- $2^1 = 2$ (Notice that $2 = 4 \div 2$, or $2^1 = 2^2 \div 2^1$)
Following this pattern, to get the next value as the exponent decreases by 1, we divide the previous value by the base (2 in this case). What should $2^0$ be?
$2^0 = 2^1 \div 2 = 2 \div 2 = 1$
This aligns with the general rule you may have learned: for any non-zero base $a$, $a^0 = 1$.
$a^0 = 1$
[For any non-zero rational number $a$]
Powers with Negative Exponents
Let's continue the pattern observed above into negative exponents:
- $2^0 = 1$ (Following the pattern, the next step is to divide by the base, 2)
- $2^{-1} = 2^0 \div 2 = 1 \div 2 = \frac{1}{2}$
Let's continue again:
- $2^{-2} = 2^{-1} \div 2 = \frac{1}{2} \div 2 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. Note that $\frac{1}{4} = \frac{1}{2^2}$.
- $2^{-3} = 2^{-2} \div 2 = \frac{1}{4} \div 2 = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$. Note that $\frac{1}{8} = \frac{1}{2^3}$.
- $2^{-4} = 2^{-3} \div 2 = \frac{1}{8} \div 2 = \frac{1}{8} \times \frac{1}{2} = \frac{1}{16}$. Note that $\frac{1}{16} = \frac{1}{2^4}$.
From this pattern, we can see a clear relationship between a power with a negative exponent and a power with a positive exponent:
For any non-zero rational number $a$ and any positive integer $n$, the power $a^{-n}$ is defined as the reciprocal of $a^n$.
$a^{-n} = \frac{1}{a^n}$
[For any non-zero rational number $a$, and positive integer $n$] ... (i)
Conversely, the reciprocal of a power with a positive exponent is a power with a negative exponent: $\frac{1}{a^n} = a^{-n}$.
Also, if we have a power with a negative exponent in the denominator, its reciprocal is a power with a positive exponent in the numerator:
$\frac{1}{a^{-n}} = \frac{1}{\frac{1}{a^n}} = 1 \times \frac{a^n}{1} = a^n$
[For any non-zero rational number $a$, and positive integer $n$] ... (ii)
For a fraction raised to a negative exponent:
$(\frac{a}{b})^{-n} = \frac{1}{(\frac{a}{b})^n} = \frac{1}{\frac{a^n}{b^n}} = 1 \times \frac{b^n}{a^n} = \frac{b^n}{a^n} = (\frac{b}{a})^n$
[For non-zero rational numbers $a, b$, and positive integer $n$] ... (iii)
So, a fraction raised to a negative exponent is equal to the reciprocal of the fraction raised to the corresponding positive exponent.
Example 1. Write the following with positive exponents:
(i) $3^{-2}$
(ii) $p^{-5}$
(iii) $(\frac{2}{5})^{-3}$
Answer:
(i) Using the definition $a^{-n} = \frac{1}{a^n}$ with $a=3$ and $n=2$:
$3^{-2} = \frac{1}{3^2}$
The expression with a positive exponent is $\frac{1}{3^2}$.
(ii) Using the definition $a^{-n} = \frac{1}{a^n}$ with $a=p$ and $n=5$ (assuming $p \neq 0$):
$p^{-5} = \frac{1}{p^5}$
The expression with a positive exponent is $\frac{1}{p^5}$.
(iii) Using the property $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$ with $a=2$, $b=5$, and $n=3$ (assuming $2, 5 \neq 0$):
$(\frac{2}{5})^{-3} = (\frac{5}{2})^3$
The expression with a positive exponent is $(\frac{5}{2})^3$.
Example 2. Evaluate:
(i) $4^{-3}$
(ii) $(-3)^{-2}$
(iii) $(\frac{1}{2})^{-4}$
Answer:
(i) Evaluate $4^{-3}$:
$4^{-3} = \frac{1}{4^3}$
[Using Formula (i)]
Calculate the power in the denominator:
$= \frac{1}{4 \times 4 \times 4} = \frac{1}{64}$
The value is $\frac{1}{64}$.
(ii) Evaluate $(-3)^{-2}$:
$(-3)^{-2} = \frac{1}{(-3)^2}$
[Using Formula (i)]
Calculate the power in the denominator. $(-3)^2 = (-3) \times (-3) = 9$.
$= \frac{1}{9}$
The value is $\frac{1}{9}$.
(iii) Evaluate $(\frac{1}{2})^{-4}$:
Using the property $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$ with $a=1$, $b=2$, and $n=4$:
$(\frac{1}{2})^{-4} = (\frac{2}{1})^4$
[Using Formula (iii)]
Simplify and calculate the power:
$= 2^4 = 2 \times 2 \times 2 \times 2 = 16$
The value is 16.
Laws of Exponents
The power of exponents lies in the fact that they follow certain rules or laws that simplify calculations involving multiplication, division, and raising powers to other powers. These laws, which you learned for positive integer exponents, extend and remain valid for negative integer exponents as well, provided the base is not zero.
Laws of Exponents for Integer Exponents
Let $a$ and $b$ be non-zero rational numbers (bases), and $m$ and $n$ be any integers (positive, negative, or zero, exponents). The following laws hold true:
Law 1: Product of Powers with the Same Base
This law states that when you multiply two exponential expressions that have the same base, you can combine them into a single expression by keeping the base and adding the exponents.
$a^m \times a^n = a^{m+n}$
... (i)
Example (Positive exponents): Simplify $2^3 \times 2^5$.
Using the law: $2^3 \times 2^5 = 2^{3+5} = 2^8$.
Example (Negative exponents): Simplify $3^4 \times 3^{-2}$.
Using the law: $3^4 \times 3^{-2} = 3^{4+(-2)} = 3^{4-2} = 3^2 = 9$.
Verification: $3^4 \times 3^{-2} = 3^4 \times \frac{1}{3^2} = \frac{3 \times 3 \times 3 \times 3}{3 \times 3} = 3 \times 3 = 9$. The law holds.
Law 2: Quotient of Powers with the Same Base
When dividing two powers with the same base, you can combine them into a single expression by keeping the base and subtracting the exponent of the denominator from the exponent of the numerator.
$\frac{a^m}{a^n} = a^{m-n}$
... (ii)
Example (Positive exponents): Simplify $\frac{5^7}{5^3}$.
Using the law: $\frac{5^7}{5^3} = 5^{7-3} = 5^4$.
Example (Negative exponents): Simplify $\frac{a^{-4}}{a^{-6}}$.
Using the law: $\frac{a^{-4}}{a^{-6}} = a^{-4 - (-6)} = a^{-4 + 6} = a^2$.
Verification: $\frac{a^{-4}}{a^{-6}} = \frac{1/a^4}{1/a^6} = \frac{1}{a^4} \times \frac{a^6}{1} = \frac{a^6}{a^4} = a^{6-4} = a^2$. The law holds.
Law 3: Power of a Power
When raising a power to another exponent, you can simplify the expression by keeping the base and multiplying the exponents.
$(a^m)^n = a^{mn}$
... (iii)
Example (Positive exponents): Simplify $(3^2)^4$.
Using the law: $(3^2)^4 = 3^{2 \times 4} = 3^8$.
Example (Negative exponents): Simplify $ ((-2)^{-3})^2 $.
Using the law: $( (-2)^{-3} )^2 = (-2)^{(-3) \times 2} = (-2)^{-6} = \frac{1}{(-2)^6} = \frac{1}{64}$.
Verification: $( (-2)^{-3} )^2 = (\frac{1}{(-2)^3})^2 = (\frac{1}{-8})^2 = \frac{1^2}{(-8)^2} = \frac{1}{64}$. The law holds.
Law 4: Power of a Product
When a product of two or more factors is raised to an exponent, the exponent is distributed to each factor in the product.
$(ab)^m = a^m b^m$
... (iv)
Example (Positive exponents): Simplify $(2 \times 3)^5$.
Using the law: $(2 \times 3)^5 = 2^5 \times 3^5$.
Example (Negative exponents): Simplify $(4p)^{-2}$.
Using the law: $(4p)^{-2} = 4^{-2} p^{-2} = \frac{1}{4^2} \times \frac{1}{p^2} = \frac{1}{16p^2}$.
Law 5: Power of a Quotient
When a quotient (fraction) is raised to an exponent, the exponent is distributed to both the numerator and the denominator.
$(\frac{a}{b})^m = \frac{a^m}{b^m}$
... (v)
Example (Positive exponents): Simplify $(\frac{2}{5})^3$.
Using the law: $(\frac{2}{5})^3 = \frac{2^3}{5^3} = \frac{8}{125}$.
Example (Negative exponents): Simplify $(\frac{x}{y})^{-4}$.
Using the law: $(\frac{x}{y})^{-4} = \frac{x^{-4}}{y^{-4}} = \frac{1/x^4}{1/y^4} = \frac{y^4}{x^4} = (\frac{y}{x})^4$. An important consequence is that a negative exponent flips the fraction: $(\frac{a}{b})^{-m} = (\frac{b}{a})^m$.
Law 6: Zero Exponent
Any non-zero number raised to the power of 0 is always equal to 1.
$a^0 = 1$, for $a \neq 0$.
... (vi)
Derivation from Law 2: We know that any non-zero number divided by itself is 1. For example, $\frac{a^m}{a^m} = 1$. Using Law 2, we can also write $\frac{a^m}{a^m} = a^{m-m} = a^0$. Therefore, $a^0 = 1$.
Example: $7^0 = 1$; $(-5)^0 = 1$; $(\frac{3}{4})^0 = 1$.
Example 1. Simplify and write the result in power notation with a positive exponent:
$(\frac{1}{3})^{-2} \times 3^{-5}$
Answer:
Given expression: $(\frac{1}{3})^{-2} \times 3^{-5}$.
First, simplify the term $(\frac{1}{3})^{-2}$. Using the property $(\frac{a}{b})^{-m} = (\frac{b}{a})^m$:
$(\frac{1}{3})^{-2} = (\frac{3}{1})^2 = 3^2$
Now, substitute this back into the original expression:
$= 3^2 \times 3^{-5}$
Now, the bases are the same. We can use Law 1 ($a^m \times a^n = a^{m+n}$):
$= 3^{2+(-5)} = 3^{2-5} = 3^{-3}$
The result is $3^{-3}$. The question requires the answer with a positive exponent. Using the definition $a^{-n} = \frac{1}{a^n}$:
$3^{-3} = \frac{1}{3^3}$
The simplified result with a positive exponent is $\frac{1}{3^3}$.
Example 2. Simplify: $5^{-1} \times 2^{-1} \times ( \frac{1}{10} )^{-1}$
Answer:
Given expression: $5^{-1} \times 2^{-1} \times ( \frac{1}{10} )^{-1}$.
Method 1: Using the definition of negative exponents first.
$5^{-1} = \frac{1}{5}$
[Using $a^{-n} = \frac{1}{a^n}$]
$2^{-1} = \frac{1}{2}$
[Using $a^{-n} = \frac{1}{a^n}$]
$( \frac{1}{10} )^{-1} = (\frac{10}{1})^1 = 10$
[Using $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$]
Substitute these values back into the expression:
$= \frac{1}{5} \times \frac{1}{2} \times 10$
$= \frac{10}{10} = 1$
Method 2: Using the laws of exponents first.
For the first two terms, $5^{-1} \times 2^{-1}$, we can use Law 4 ($a^m b^m = (ab)^m$):
$5^{-1} \times 2^{-1} = (5 \times 2)^{-1} = 10^{-1}$
Substitute this into the expression:
$= 10^{-1} \times ( \frac{1}{10} )^{-1}$
Now, rewrite $\frac{1}{10}$ as $10^{-1}$:
$= 10^{-1} \times (10^{-1})^{-1}$
Using Law 3 ($(a^m)^n = a^{mn}$) on the second term:
$(10^{-1})^{-1} = 10^{(-1) \times (-1)} = 10^1 = 10$
Substitute this back:
$= 10^{-1} \times 10^1$
Using Law 1 ($a^m \times a^n = a^{m+n}$):
$= 10^{-1+1} = 10^0$
Using Law 6 ($a^0 = 1$):
$= 1$
Both methods yield the same result, 1.
Use of Exponents to Express Small Numbers in Standard Form
In science, mathematics, and everyday life, we often encounter numbers that are either extremely large (like the distance from the Earth to the Sun, or the number of stars in a galaxy) or extremely small (like the size of an atom, or the mass of a subatomic particle). Writing these numbers in their standard decimal form can be very lengthy, cumbersome, and prone to errors in counting the zeros.
To make it easier to work with such numbers, we use a special form called standard form or scientific notation, which utilises powers of 10.
Standard Form (Scientific Notation)
A number is expressed in standard form if it is written in the form $m \times 10^n$, where:
- $m$ is a decimal number such that $1 \leq m < 10$. This means $m$ has only one non-zero digit to the left of the decimal point.
- $n$ is an integer (it can be positive, negative, or zero).
This form separates the significant digits of the number ($m$) from its magnitude (determined by the power of 10, $10^n$).
Expressing Very Large Numbers in Standard Form
For very large numbers (greater than or equal to 10), the exponent $n$ in $m \times 10^n$ will be a positive integer. To convert a very large number into standard form, we shift the decimal point to the left until there is only one non-zero digit remaining to the left of the decimal point. The number of places the decimal point is shifted is the positive exponent of 10.
Example 1. Express 345000000 in standard form.
Answer:
Given number: 345000000.
This is a very large number. The decimal point is implicitly at the end of the number (345000000.0).
We need to shift the decimal point to the left so that only one non-zero digit (which is 3) is before the decimal point. The new position of the decimal point should be between 3 and 4.
Original: 3 4 5 0 0 0 0 0 0 .
Target: 3 . 4 5 0 0 0 0 0 0
Count the number of places the decimal point was moved to the left: 8 places.
The number becomes 3.45, and the exponent of 10 is the number of places shifted to the left, which is +8.
So, $345000000 = 3.45 \times 10^8$.
Here, $m = 3.45$ (which satisfies $1 \leq 3.45 < 10$) and $n=8$ (a positive integer).
Example 2. Write the speed of light, approximately $299792458$ metres per second, in standard form.
Answer:
Given number: 299792458.
The decimal point is at the end. Shift it to the left so that only one non-zero digit (2) is before the decimal point. The new position is between 2 and 9.
Original: 2 9 9 7 9 2 4 5 8 .
Target: 2 . 9 9 7 9 2 4 5 8
Count the number of places the decimal point was moved to the left: 8 places.
The number becomes 2.99792458, and the exponent of 10 is +8.
So, $299792458 = 2.99792458 \times 10^8$.
Often, for very large numbers with many trailing zeros or for approximations, we might round the value of $m$. For example, $345000000 \approx 3.45 \times 10^8$ or even $3.5 \times 10^8$ depending on the required precision.
Expressing Very Small Numbers in Standard Form
For very small numbers (numbers between 0 and 1), the exponent $n$ in $m \times 10^n$ will be a negative integer. To convert a very small decimal number into standard form, we shift the decimal point to the right until the first non-zero digit is just before the decimal point. The number of places the decimal point is shifted to the right gives the absolute value of the negative exponent of 10.
Example 3. Express 0.000007 in standard form.
Answer:
Given number: 0.000007.
This is a very small number (between 0 and 1). The first non-zero digit is 7. We need to shift the decimal point to the right so that it is after the digit 7.
Original: 0 . 0 0 0 0 0 7
Target: 0 0 0 0 0 7 .
Count the number of places the decimal point was moved to the right: 6 places.
The number becomes 7. The exponent of 10 is negative, and its absolute value is the number of places shifted to the right, which is 6. So, the exponent is -6.
So, $0.000007 = 7 \times 10^{-6}$.
Here, $m=7$ (which satisfies $1 \leq 7 < 10$) and $n=-6$ (a negative integer).
Example 4. Express 0.0000000325 in standard form.
Answer:
Given number: 0.0000000325.
This is a very small number. The first non-zero digit is 3. We need to shift the decimal point to the right so that it is after the digit 3.
Original: 0 . 0 0 0 0 0 0 0 3 2 5
Target: 0 0 0 0 0 0 0 3 . 2 5
Count the number of places the decimal point was moved to the right: 8 places.
The number becomes 3.25. The exponent of 10 is negative, and its value is -8.
So, $0.0000000325 = 3.25 \times 10^{-8}$.
Here, $m=3.25$ (which satisfies $1 \leq 3.25 < 10$) and $n=-8$ (a negative integer).
Expressing Numbers from Standard Form to Usual Form
To convert a number from standard form ($m \times 10^n$) back to its usual decimal form, we look at the exponent $n$.
- If $n$ is positive, move the decimal point in $m$ to the right by $n$ places. Add zeros if necessary to fill the places.
- If $n$ is negative, move the decimal point in $m$ to the left by $|n|$ places. Add zeros if necessary to fill the places between the decimal point and $m$.
- If $n$ is zero, the number is simply $m$, as $10^0=1$.
Example 5. Write $2.85 \times 10^7$ in usual form.
Answer:
Given number in standard form: $2.85 \times 10^7$.
The exponent is $n=7$, which is positive. This means the original number is large.
Move the decimal point in $m=2.85$ to the right by 7 places.
Start with 2.85
Move 1 place: 28.5
Move 2 places: 285.
We need to move 5 more places. Add 5 zeros to the right of 285.
28500000.
So, $2.85 \times 10^7 = 28,500,000$ (2 crore 85 lakh in Indian numbering system). This is a large number.
Example 6. Write $1.004 \times 10^{-5}$ in usual form.
Answer:
Given number in standard form: $1.004 \times 10^{-5}$.
The exponent is $n=-5$, which is negative. This means the original number is small (between 0 and 1).
Move the decimal point in $m=1.004$ to the left by $|-5|=5$ places.
Start with 1.004
Move 1 place: 0.1004
Move 2 places: 0.01004
Move 3 places: 0.001004
Move 4 places: 0.0001004
Move 5 places: 0.00001004
We needed to add 4 zeros between the decimal point and the digit 1 to move the decimal 5 places to the left.
So, $1.004 \times 10^{-5} = 0.00001004$. This is a very small number.
Comparing Very Large and Very Small Numbers
One of the primary benefits of expressing very large and very small numbers in standard form is that it makes comparing them significantly easier than counting the zeros in their usual decimal form. Comparing numbers in standard form $m \times 10^n$ involves comparing the exponents of 10 and then the decimal parts.
Comparing Numbers in Standard Form
To compare two numbers written in standard form, say $m_1 \times 10^{n_1}$ and $m_2 \times 10^{n_2}$ (where $1 \leq m_1 < 10$ and $1 \leq m_2 < 10$, and $n_1, n_2$ are integers), follow these steps:
- Compare the Exponents of 10: Look at the exponents $n_1$ and $n_2$. The number with the larger exponent of 10 is the greater number.
If $n_1 > n_2$, then $m_1 \times 10^{n_1} > m_2 \times 10^{n_2}$.
If $n_1 < n_2$, then $m_1 \times 10^{n_1} < m_2 \times 10^{n_2}$.
- Compare the Decimal Parts (if exponents are equal): If the exponents are the same ($n_1 = n_2$), then compare the decimal numbers $m_1$ and $m_2$. The number with the larger decimal part is the greater number.
If $n_1 = n_2$ and $m_1 > m_2$, then $m_1 \times 10^{n_1} > m_2 \times 10^{n_2}$.
If $n_1 = n_2$ and $m_1 < m_2$, then $m_1 \times 10^{n_1} < m_2 \times 10^{n_2}$.
Comparing exponents is the primary step because a higher power of 10 indicates a much larger magnitude. Only if the magnitudes (powers of 10) are the same, we compare the decimal parts.
Example 1. Which is greater: $1.5 \times 10^9$ or $8.2 \times 10^8$?
Answer:
We are comparing $1.5 \times 10^9$ and $8.2 \times 10^8$. Both numbers are in standard form.
Step 1: Compare the exponents of 10.
For $1.5 \times 10^9$, the exponent is $n_1 = 9$.
For $8.2 \times 10^8$, the exponent is $n_2 = 8$.
Compare 9 and 8. Since $9 > 8$, the number with the exponent 9 is greater.
$1.5 \times 10^9 > 8.2 \times 10^8$
This makes sense because $10^9$ is 10 times larger than $10^8$.
Verification (optional):
$1.5 \times 10^9 = 1.5 \times 1,000,000,000 = 1,500,000,000$
$8.2 \times 10^8 = 8.2 \times 100,000,000 = 820,000,000$
Clearly, 1,500,000,000 > 820,000,000.
Example 2. Which is smaller: $3.1 \times 10^{-5}$ or $5.2 \times 10^{-6}$?
Answer:
We are comparing $3.1 \times 10^{-5}$ and $5.2 \times 10^{-6}$. Both numbers are in standard form.
Step 1: Compare the exponents of 10.
For $3.1 \times 10^{-5}$, the exponent is $n_1 = -5$.
For $5.2 \times 10^{-6}$, the exponent is $n_2 = -6$.
Compare -5 and -6. Recall that on the number line, numbers to the left are smaller. -6 is to the left of -5. So, $-6 < -5$.
The number with the smaller exponent (-6) is the smaller number.
$5.2 \times 10^{-6} < 3.1 \times 10^{-5}$
This makes sense because $10^{-6} = \frac{1}{10^6}$ is smaller than $10^{-5} = \frac{1}{10^5}$.
Verification (optional):
$3.1 \times 10^{-5} = 0.000031$
$5.2 \times 10^{-6} = 0.0000052$
Clearly, 0.0000052 < 0.000031.
Example 3. Compare $4.5 \times 10^6$ and $7.8 \times 10^{-2}$.
Answer:
We are comparing $4.5 \times 10^6$ and $7.8 \times 10^{-2}$. Both numbers are in standard form.
Step 1: Compare the exponents of 10.
For $4.5 \times 10^6$, the exponent is $n_1 = 6$.
For $7.8 \times 10^{-2}$, the exponent is $n_2 = -2$.
Compare 6 and -2. A positive exponent is always greater than a negative exponent. So, $6 > -2$.
The number with the larger exponent (6) is greater.
$4.5 \times 10^6 > 7.8 \times 10^{-2}$
This confirms that a very large number is greater than a very small number.
Example 4. Which is greater: $2.5 \times 10^{-4}$ or $1.9 \times 10^{-4}$?
Answer:
We are comparing $2.5 \times 10^{-4}$ and $1.9 \times 10^{-4}$. Both are in standard form.
Step 1: Compare the exponents of 10. For both numbers, the exponent is -4. The exponents are equal.
Step 2: Since the exponents are equal, compare the decimal parts $m_1$ and $m_2$.
For $2.5 \times 10^{-4}$, $m_1 = 2.5$.
For $1.9 \times 10^{-4}$, $m_2 = 1.9$.
Compare 2.5 and 1.9. Since $2.5 > 1.9$, the number with the larger decimal part is greater.
$2.5 \times 10^{-4} > 1.9 \times 10^{-4}$
Verification (optional):
$2.5 \times 10^{-4} = 0.00025$
$1.9 \times 10^{-4} = 0.00019$
Clearly, 0.00025 > 0.00019.
Operations with Numbers in Standard Form
Standard form is also useful for performing multiplication and division of very large or very small numbers. We use the properties of exponents to simplify these operations.
To multiply $(m_1 \times 10^{n_1}) \times (m_2 \times 10^{n_2})$:
Multiply the decimal parts ($m_1 \times m_2$) and add the exponents of 10 ($n_1 + n_2$).
$(m_1 \times 10^{n_1}) \times (m_2 \times 10^{n_2}) = (m_1 \times m_2) \times 10^{n_1 + n_2}$
To divide $(m_1 \times 10^{n_1}) \div (m_2 \times 10^{n_2})$:
Divide the decimal parts ($m_1 \div m_2$) and subtract the exponents of 10 ($n_1 - n_2$).
$(m_1 \times 10^{n_1}) \div (m_2 \times 10^{n_2}) = (m_1 \div m_2) \times 10^{n_1 - n_2}$
After performing the operation, the result might need to be adjusted to be in standard form (ensure the decimal part is between 1 and 10 by shifting the decimal and adjusting the exponent accordingly).
Example 5. Multiply $(3 \times 10^5) \times (2 \times 10^3)$.
Answer:
Given: $(3 \times 10^5) \times (2 \times 10^3)$. Both numbers are in standard form.
Multiply the decimal parts and the powers of 10 separately:
$(3 \times 10^5) \times (2 \times 10^3) = (3 \times 2) \times (10^5 \times 10^3)$
Calculate the product of the decimal parts and use the product law of exponents ($a^m \times a^n = a^{m+n}$) for the powers of 10:
$= 6 \times 10^{5+3}$
$= 6 \times 10^8$
The result is $6 \times 10^8$. The decimal part is 6, which is between 1 and 10 ($1 \leq 6 < 10$). The exponent is 8, which is an integer. So, the result is already in standard form.
Example 6. Divide $(8 \times 10^{-4})$ by $(2 \times 10^{-7})$.
Answer:
Given: $(8 \times 10^{-4}) \div (2 \times 10^{-7})$. Both numbers are in standard form.
Divide the decimal parts and the powers of 10 separately:
$\frac{8 \times 10^{-4}}{2 \times 10^{-7}} = (\frac{8}{2}) \times (\frac{10^{-4}}{10^{-7}})$
Calculate the division of the decimal parts and use the quotient law of exponents ($\frac{a^m}{a^n} = a^{m-n}$) for the powers of 10:
$= 4 \times 10^{-4 - (-7)}$
$= 4 \times 10^{-4 + 7} = 4 \times 10^3$
The result is $4 \times 10^3$. The decimal part is 4 ($1 \leq 4 < 10$), and the exponent is 3 (an integer). So, the result is already in standard form.
These examples show how the laws of exponents greatly simplify calculations with numbers in standard form.