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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

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.

Class 8 Exponents and Powers

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

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-AnswerDownload the pdf










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