# EXPONENTS AND RADICALS - FINDING Nth ROOTS OF NUMBERS BY APPLYING LAWS OF EXPONENTS

RATIONAL EXPONENTS BEFORE EXPONENTS AND RADICALS
if you have not already done so.

It is a prerequisite here.

There, we provided the explanation
for Rational Exponents.

We discuss how we can apply the same
7 Laws and the 2 Rules given for
whole number Exponents can be applied
for Fractional Exponents.

Here, we apply the Laws to find nth Root of Numbers.

### Solved Example 1 of Exponents and Radicals

Evaluate (125)1⁄3

Solution to Example 1 :

Let A = (125)1⁄3
As the denominator of the exponent fraction is 3, we check, if the base
can be expressed as a cube.
We know, 125 is divisible by 5 and 125⁄5 = 25.
∴ 125 = 5 x 25 = 5 x 5 x 5 = 53
∴ A = (125)1⁄3 = (53)1⁄3.
We know (am)n = amn (See Law 2)
Applying this Law, we get
A = (53)1⁄3 = 53 x 1⁄3 = 51 = 5. Ans.

### Solved Example 2 of Exponents and Radicals

Evaluate 82⁄3

Solution to Example 2 :

This is similar to Example 1 above.
82⁄3 = (23)2⁄3 = 23 x 2⁄3 = 22 = 4. Ans.

### Solved Example 3 of Exponents and Radicals

Evaluate (1⁄5)-2

Solution to Example 3 :

We know (1⁄a)-n = an
∴ (1⁄5)-2 = 52 = 25. Ans.

### Solved Example 4 of Exponents and Radicals

Evaluate (16)-3⁄4

Solution to Example 4 :

As the denominator of the exponent fraction is 4, we check, if the base
can be expressed as power 4. Then we arrive at 16 = 24
∴ (16)-3⁄4 = (24)-3⁄4 = 24 x -3⁄4 {since (am)n = amn}
= 2-3 = 1⁄23 ( since a-n = 1⁄an )
= 1⁄8. Ans.

### Solved Example 5 of Exponents and Radicals

Evaluate (32)-4⁄5

Solution to Example 5 :

As the denominator of the exponent fraction is 5,
we express 32 as 25. ( since 2 x 2 x 2 x 2 x 2 = 32)
(32)-4⁄5 = (25)-4⁄5 = 25 x -4⁄5 {since (am)n = amn}
= 2-4 = 1⁄24 ( since a-n = 1⁄an )
= 1⁄16. Ans.

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### Solved Example 6 of Exponents and Radicals

Evaluate 8⁄125)-1⁄3

Solution to Example 6 :

As the denominator of the exponent fraction is 3,
we express 8⁄125 as (2⁄5)3. ( since 2⁄5 x 2⁄5 x 2⁄5 = 8⁄125)
(8⁄125)-1⁄3 = {(2⁄5)3}-1⁄3 = (2⁄5)3 x -1⁄3 {since (am)n = amn}
= (2⁄5)-1 = (5⁄2)1 { since (pq)-n = (qp)n }
= 5⁄2. Ans.

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### Solved Example 7 of Exponents and Radicals

Evaluate (-27)2⁄3

Solution to Example 7 :

As the denominator of the exponent fraction is 3,
we express -27 as (-3)3. ( since -3 x -3 x -3 = -27)
(-27)2⁄3 = {(-3)3}2⁄3 = (-3)3 x 2⁄3 {since (am)n = amn}
= (-3)2 = 9. Ans.

### Solved Example 8 of Exponents and Radicals

Evaluate (0.001)-1⁄3

Solution to Example 8 :

As the denominator of the exponent fraction is 3,
we express 0.001 as (0.1)3. ( since 0.1 x 0.1 x 0.1 = 0.001)
(0.001)-1⁄3 = {(0.1)3}-1⁄3 = (0.1)3 x -1⁄3 {since (am)n = amn}
= (0.1)-1 = 1⁄(0.1) (since a-1 = 1⁄a)
= 10. Ans.

### Solved Example 9 of Exponents and Radicals

Evaluate (0.027)-2⁄3

Solution to Example 9 of Exponents and Radicals:

As the denominator of the exponent fraction is 3,
we express 0.027 as (0.3)3. ( since 0.3 x 0.3 x 0.3 = 0.027)
(0.027)-2⁄3 = {(0.3)3}-2⁄3 = (0.3)3 x -2⁄3 {since (am)n = amn}
= (0.3)-2 = (3⁄10)-2 = (10⁄3)2 { since (pq)-n = (qp)n }
= 102⁄32 { since (ab)m = ambm }
= 100⁄9. Ans.

### Exercise on Exponents and Radicals :

Evaluate

1. (64)1⁄3
2. 85⁄3
3. 93⁄2
4. If √2 = 1.414 and √3 = 1.732, find the value of
1. √(72)
2. √(147) + √(27)
3. 5√(128) + 2√(75) - 5√(108)
5. √(36)
6. (125)1⁄3
7. (81)1⁄4
8. (32)1⁄5
9. √(50)
10. (40)1⁄3

For Answers, see at the bottom of the page.

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