Class-8 Exponents and Powers

Introduction to Exponents and Powers

Negative Integers as Exponents

Laws of Exponents and Powers

Exponents and Powers Test

Exponents and Powers Worksheets

Answer Sheet

Introduction to Exponents and Powers

Any number that is written in the form of powers is known as power form or exponential form.

So, pⁿ = p × p × p × p × ... n times
If rational numbers multiply itself many times, then it is also written as exponential form.

(^p⁄_q)ⁿ = ^pⁿ⁄_qⁿ for any positive integer.

Example 1. ¹⁄₂ × ¹⁄₂ × ¹⁄₂ = (¹⁄₂)³

Example 2. ^-2⁄₃ × ^-2⁄₃ × ^-2⁄₃ × ^-2⁄₃ = (^-2⁄₃)⁴

Negative Integers as Exponents

Here, we will learn about powers with negative integers as exponents. For example, 3^-4 can be written as ¹⁄_3⁴. Let's figure out how we arrive at this.

3⁴ = 3 × 3 × 3 × 3 = 81

3³ = 3 × 3 × 3 = 27 = ⁸¹⁄₃

3² = 3 × 3 = 9 = ²⁷⁄₃

3¹ = 3 = ⁹⁄₃

3⁰ = 1 = ³⁄₃

3^-1 = 1 ÷ 3 = ¹⁄₃

3^-2 = ¹⁄₃ ÷ 3 = ¹⁄_(3×3) = ¹⁄_3²

3^-3 = ¹⁄_3² ÷ 3 = ¹⁄_(3×3×3) = ¹⁄_3³

3^-4 = ¹⁄_3³ ÷ 3 = ¹⁄_(3×3×3×3) = ¹⁄_3⁴

So from above pattern we can conclude that any non-zero integer 'p', p^-n = ¹⁄_pⁿ here, 'n' is positive integer.

p^-n is known as multiplicative inverse of pⁿ.

pⁿ is multiplicative inverse of p^-n.

If ^a⁄_b is non-zero rational number and 'n' is any positive integer, then

(^a⁄_b)^-n = (^b⁄_a)ⁿ = ^bⁿ⁄_aⁿ

Example 1. Find the value of 4^-3.

Solution. 4^-3 = ¹⁄_4³

= ¹⁄_(4×4×4)

= ¹⁄₆₄

Example 2. Find the value of (²⁄₅)^-5.

Solution. (²⁄₅)^-5 = ¹⁄_{(²⁄₅)⁵}

         = (⁵⁄₂)⁵ = (^5⁵⁄_2⁵)

         = ^{5×5×5×5×5}⁄_{2×2×2×2×2}

         = ³¹²⁵⁄₃₂

Example 3. Find the multiplicative inverse of 5^-2.

Solution. As we know, multiplicative inverse of p^-n= pⁿ

So, multiplicative inverse of 5^-2 = 5²

= 5 × 5 = 25

Example 4. Find the multiplicative inverse of ¹⁄₃³.

Solution. As we know, multiplicative inverse of p^-n= pⁿ

Multiplicative inverse of (¹⁄₃)³ = (¹⁄₃)^-3

        = ¹⁄_(¹⁄₃)³

        = (³⁄₁)³

        = 3×3×3 = 27

Laws of Exponents and Powers

If 'p' is a non-zero non-zero rational numbers and 'm', 'n' are two integers, then

p^m × pⁿ= p^m+n

Example 1. (-2)^-2 × (-2)^-3 = (-2)^{-2+(-3)} = (-2)^-5

Example 2. (²⁄₃)² × (²⁄₃)³ = (²⁄₃)⁽²⁺³⁾ = (²⁄₃)⁵

If 'p' is a non-zero rational numbers and 'm', 'n' are two integers, then

p^m ÷ p^m = p^m-n

Example 1. (3)² ÷ (3)^-3 = (3)^{2-(-3)} = (3)⁽²⁺³⁾ = (3)⁵

Example 2. (³⁄₅)⁴ ÷ (³⁄₅)³ = (³⁄₅)^(4-3) = (³⁄₅)¹ = ³⁄₅

If 'p' is a non-zero rational numbers and 'm', 'n' are two integers, then

(p^m)ⁿ = p^mn

Example 1. (2²)^-3 = 2^2×(-3) = 2^-6

If 'p' and 'q' are two non-zero rational numbers and 'm' is an integer, then

(pq)^m = p^m × q^m

Example 1. (3 × 5)² = 3² × 5².

Example 2. (²⁄₃ × ²⁄₅)² = (²⁄₃)² × (²⁄₅)².

If 'p' and 'q' are two non-zero rational numbers and 'm' is an integer, then

(^p⁄_q)^m = ^{p^m}⁄_q^m

Example 1. (²⁄₃)³ = ^2³⁄_3³

If 'p' is a non-zero rational numbers and m = 0, then

p⁰ = 1

Example 1. 5⁰ = 1.

Example 2. (⁵⁄₇)⁰ = 1

If 'p' is any rational number and 'n' is natural number ,then

(-p)ⁿ =(-1 × p)ⁿ = (-1)ⁿ × pⁿ = pⁿ (if n is even)

= (-1)ⁿ × pⁿ = -pⁿ (if n is odd)

Some different solved examples are provided below to understand how laws of exponents and powers are used to simplify sums.

Example 1. Express 8^-3 as a power with base 2.

Solution. We know that 8 can be written as 2³

Then we can write 8^-3 = {(2)³}^-3

Let's use (p^m)ⁿ =p^m×n law.

= 2^-9

Example 2. Simplify 3⁵ ÷ 3^-6.

Solution. 3⁵ ÷ 3^-6 = ^3⁵⁄_3^-6

Let's use ^{p^m}⁄_pⁿ = p^m-n

= 3^5-(-6) = 3¹¹

Example 3. Simplify 2² × 2⁵ × 2^-6.

Solution. 2² × 2⁵ × 2^-6

(Here we have to use p^m × pⁿ = p^m+n)

= 2²⁺⁵ × 2^-6 = 2⁷ × 2^-6

= 2^7-6 = 2¹ = 2

Example 4. Simplify and write in the exponential form.
2³ × 3² + (-13)² + 2^-3 ÷ 2^-8 − (-2/5)⁰

Solution. 2³ × 3² + (-13)² + 2^-3 ÷ 2^-8 − (-2/5)⁰

First simplify all exponent by using exponent laws.

         = 8 × 9 + 169 + 2^{-3-(-8)} − 1

         = 72 + 169 + 2⁵ − 1

         = 72 + 169 + 32 − 1 = 272

         = 2 × 2 × 2 × 2 × 17

         = 2⁴ × 17¹

Example 5. Simplify {(¹⁄₅)^-2 − (¹⁄₃)^-3} ÷ (¹⁄₆)^-2.

Solution. {(¹⁄₅)^-2 − (¹⁄₃)^-3} ÷ (¹⁄₆)^-2

Use law of reciprocal of exponent (^a⁄_b)^-m = (^b⁄_a)^m

         = {(⁵⁄₁)² − (³⁄₁)³} ÷ (⁶⁄₁)²

         = (5² − 3³) ÷ 6²

         = (25 − 9) ÷ 36 = ¹⁶⁄₃₆ = ⁴⁄₉

Example 6. What is the value of ^{(6ⁿ⁺² − 6ⁿ⁺¹)}⁄_6ⁿ⁺³

Solution. ^{(6ⁿ⁺² − 6ⁿ⁺¹)}⁄_6ⁿ⁺³

Here, use law of exponent p^m+n = (p^m × pⁿ)

         = ^{{(6ⁿ × 6²) − (6ⁿ × 6¹)}}⁄_{(6ⁿ × 6³)}

         = ^{6ⁿ(6²-6¹)}⁄_{(6ⁿ × 6³)}

Here 6ⁿ is common in both numerator and denominator, so cancel them out.

         = ^36-6⁄₂₁₆

         = ³⁰⁄₂₁₆ = ⁵⁄₃₆

Example 7. Find the value of x from the below given equation
         2¹⁰ ÷ 2⁴ = 2^-2 × 2^2x

Solution. 2¹⁰ ÷ 2⁴ = 2^-2 × 2^2x

Simplify by using law of exponent p^m × pⁿ = p^m+n

         => 2^10-4 = 2^-2+2x

         => 2⁶ = 2^-2+2x

Let's consider exponent law p^m = pⁿ, m = n

         => 6 = -2 + 2x

         => 2x = 6 + 2

         => 2x = 8

         => x = ⁸⁄₂

         => x = 4

Example 8. If 3^x+2 = 216 + 3^x, then find the value of x.

Solution. As given 3^x+2 = 216 + 3^x

         => 3^x × 3² = 216 + 3^x

         => 3^x × 9 − 3^x = 216

         => 3^x(9 − 1) = 216

         => 3^x × 8 = 216

         => 3^x = ²¹⁶⁄₈

         => 3^x = 27

         => 3^x = 3³

By using exponent law p^m = pⁿ => m = n

         => x = 3

Example 9. By what number should (^-3⁄₄)^-2 be divided to get (¹⁄₅)^-1?

Solution. Let's assume the number required number is 'x'.

(^-3⁄₄)^-2 ÷ x = (¹⁄₅)^-1

         => (^-4⁄₃)² × ¹⁄_x = 5

         => ¹⁶⁄₉ × ¹⁄_x = 5

         => ¹⁄_x = 5 ÷ ¹⁶⁄₉

         => ¹⁄_x = 5 × ⁹⁄₁₆

         => ¹⁄_x = ^5×9⁄₁₆

         => ¹⁄_x = ⁴⁵⁄₁₆

         => x = ¹⁶⁄₄₅

Hence, required number is ¹⁶⁄₄₅.