# Class-8 Exponents and Powers

Introduction to Exponents and Powers

Negative Integers as Exponents

Exponents and Powers Worksheets

## Introduction to Exponents and Powers

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

So, p^{n} = 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.

3^{4} = 3 × 3 × 3 × 3 = 81

3^{3} = 3 × 3 × 3 = 27 = ^{81}⁄_{3}

3^{2} = 3 × 3 = 9 = ^{27}⁄_{3}

3^{1} = 3 = ^{9}⁄_{3}

3^{0} = 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 p^{n}.*

*p*^{n}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}= p^{n}

So, multiplicative inverse of 5^{-2} = 5^{2}

= 5 × 5 = 25

**Example 4.** Find the multiplicative inverse of ^{1}⁄_{3}^{3}.

**Solution.** As we know, multiplicative inverse of p^{-n}= p^{n}

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} × p^{n}= 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.** (2^{2})^{-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} = 3^{2} × 5^{2}.

**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} = ^{pm}⁄_{qm}*

**Example 1.** (^{2}⁄_{3})^{3} = ^{23}⁄_{33}

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

*p ^{0} = 1*

**Example 1.** 5^{0} = 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} × p^{n} = p^{n} (if n is even)

= (-1)^{n} × p^{n} = -p^{n} (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 2^{3}

Then we can write 8^{-3} = {(2)^{3}}^{-3}

Let's use (p^{m})^{n} =p^{m×n} law.

= 2^{-9}

**Example 2.** Simplify 3^{5} ÷ 3^{-6}.

**Solution.** 3^{5} ÷ 3^{-6} = ^{35}⁄_{3-6}

Let's use ^{pm}⁄_{pn} = p^{m-n}

= 3^{5-(-6)} = 3^{11}

**Example 3.** Simplify 2^{2} × 2^{5} × 2^{-6}.

**Solution.** 2^{2} × 2^{5} × 2^{-6}

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

= 2^{2+5} × 2^{-6} = 2^{7} × 2^{-6}

= 2^{7-6} = 2^{1} = 2

**Example 4.** Simplify and write in the exponential form.

2^{3} × 3^{2} + (-13)^{2} + 2^{-3} ÷ 2^{-8} − (-2/5)^{0}

**Solution.** 2^{3} × 3^{2} + (-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 + 2^{5} − 1

= 72 + 169 + 32 − 1 = 272

= 2 × 2 × 2 × 2 × 17

= 2^{4} × 17^{1}

**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}

= (5^{2} − 3^{3}) ÷ 6^{2}

= (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} × p^{n})

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

= ^{6n(62-61)}⁄_{(6n × 63)}

Here 6^{n} 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

2^{10} ÷ 2^{4} = 2^{-2} × 2^{2x}

**Solution.** 2^{10} ÷ 2^{4} = 2^{-2} × 2^{2x}

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

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

=> 2^{6} = 2^{-2+2x}

Let's consider exponent law p^{m} = p^{n}, 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} × 3^{2} = 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} = 3^{3}

By using exponent law p^{m} = p^{n} => 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}.

## Class-8 Exponents and Powers Test

## Class-8 Exponents and Powers Worksheet

Exponents & Powers Worksheet - 1

Exponents & Powers Worksheet - 2

Exponents & Powers Worksheet - 3

Exponents & Powers Worksheet - 4

## Answer Sheet

**Exponents-And-Powers-Answer**Download the pdf

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