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# 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

## 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.

(pq)n = pnqn for any positive integer.

Example 1. 12 × 12 × 12 = (12)3

Example 2. -23 × -23 × -23 × -23 = (-23)4

## Negative Integers as Exponents

Here, we will learn about powers with negative integers as exponents. For example, 3-4 can be written as 134. Let's figure out how we arrive at this.

34 = 3 × 3 × 3 × 3 = 81

33 = 3 × 3 × 3 = 27 = 813

32 = 3 × 3 = 9 = 273

31 = 3 = 93

30 = 1 = 33

3-1 = 1 ÷ 3 = 13

3-2 = 13 ÷ 3 = 1(3×3) = 132

3-3 = 132 ÷ 3 = 1(3×3×3) = 133

3-4 = 133 ÷ 3 = 1(3×3×3×3) = 134

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

p-n is known as multiplicative inverse of pn.

pn is multiplicative inverse of p-n.

If ab is non-zero rational number and 'n' is any positive integer, then

(ab)-n = (ba)n = bnan

Example 1. Find the value of 4-3.

Solution. 4-3 = 143

= 1(4×4×4)

= 164

Example 2. Find the value of (25)-5.

Solution. (25)-5 = 1(25)5

= (52)5 = (5525)

= 5×5×5×5×52×2×2×2×2

= 312532

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 133.

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

Multiplicative inverse of (13)3 = (13)-3

= 1(13)3

= (31)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

pm × pn= pm+n

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

Example 2. (23)2 × (23)3 = (23)(2+3) = (23)5

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

pm ÷ pm = pm-n

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

Example 2. (35)4 ÷ (35)3 = (35)(4-3) = (35)1 = 35

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

(pm)n = pmn

Example 1. (22)-3 = 22×(-3) = 2-6

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

(pq)m = pm × qm

Example 1. (3 × 5)2 = 32 × 52.

Example 2. (23 × 25)2 = (23)2 × (25)2.

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

(pq)m = pmqm

Example 1. (23)3 = 2333

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

p0 = 1

Example 1. 50 = 1.

Example 2. (57)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 (pm)n =pm×n law.

= 2-9

Example 2. Simplify 35 ÷ 3-6.

Solution. 35 ÷ 3-6 = 353-6

Let's use pmpn = pm-n

= 35-(-6) = 311

Example 3. Simplify 22 × 25 × 2-6.

Solution. 22 × 25 × 2-6

(Here we have to use pm × pn = pm+n)

= 22+5 × 2-6 = 27 × 2-6

= 27-6 = 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 {(15)-2 − (13)-3} ÷ (16)-2.

Solution. {(15)-2 − (13)-3} ÷ (16)-2

Use law of reciprocal of exponent (ab)-m = (ba)m

= {(51)2 − (31)3} ÷ (61)2

= (52 − 33) ÷ 62

= (25 − 9) ÷ 36 = 1636 = 49

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 pm+n = (pm × 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.

= 36-6216

= 30216 = 536

Example 7. Find the value of x from the below given equation
210 ÷ 24 = 2-2 × 22x

Solution. 210 ÷ 24 = 2-2 × 22x

Simplify by using law of exponent pm × pn = pm+n

=> 210-4 = 2-2+2x

=> 26 = 2-2+2x

Let's consider exponent law pm = pn, m = n

=> 6 = -2 + 2x

=> 2x = 6 + 2

=> 2x = 8

=> x = 82

=> x = 4

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

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

=> 3x × 32 = 216 + 3x

=> 3x × 9 − 3x = 216

=> 3x(9 − 1) = 216

=> 3x × 8 = 216

=> 3x = 2168

=> 3x = 27

=> 3x = 33

By using exponent law pm = pn => m = n

=> x = 3

Example 9. By what number should (-34)-2 be divided to get (15)-1?

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

(-34)-2 ÷ x = (15)-1

=> (-43)2 × 1x = 5

=> 169 × 1x = 5

=> 1x = 5 ÷ 169

=> 1x = 5 × 916

=> 1x = 5×916

=> 1x = 4516

=> x = 1645

Hence, required number is 1645.

## Class-8 Exponents and Powers Test

Exponents & Powers - 1

Exponents & Powers - 2

Exponents & Powers - 3

## Class-8 Exponents and Powers Worksheet

Exponents & Powers Worksheet - 1

Exponents & Powers Worksheet - 2

Exponents & Powers Worksheet - 3

Exponents & Powers Worksheet - 4