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

Please study
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 (a^m)ⁿ = a^mn (See Law 2)
Applying this Law, we get
A = (5³)^1⁄3 = 5^{3 x 1⁄3} = 5¹ = 5. Ans.

Solved Example 2 of Exponents and Radicals

Evaluate 8^2⁄3

Solution to Example 2 :

This is similar to Example 1 above.
8^2⁄3 = (2³)^2⁄3 = 2^{3 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 = 2^{4 x -3⁄4} {since (a^m)ⁿ = a^mn}
= 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 = 2^{5 x -4⁄5} {since (a^m)ⁿ = a^mn}
= 2^-4 = 1⁄2⁴ ( since a^-n = 1⁄aⁿ )
= 1⁄16. Ans.

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 (a^m)ⁿ = a^mn}
= (2⁄5)^-1 = (5⁄2)¹ { since (p⁄q)^-n = (q⁄p)ⁿ }
= 5⁄2. Ans.

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 (a^m)ⁿ = a^mn}
= (-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 (a^m)ⁿ = a^mn}
= (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 (a^m)ⁿ = a^mn}
= (0.3)^-2 = (3⁄10)^-2 = (10⁄3)² { since (p⁄q)^-n = (q⁄p)ⁿ }
= 10²⁄3² { since (a⁄b)^m = a^m⁄b^m }
= 100⁄9. Ans.

Exercise on Exponents and Radicals :

Evaluate

1. (64)^1⁄3
2. 8^5⁄3
3. 9^3⁄2
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.

Answers to Exercise :

1. 4.
2. 32.
3. 27.
4. 1. 8.484
   2. 17.32
   3. 21.92
5. 6.
6. 5.
7. 3.
8. 2.
9. 5√2.
10. 2 x 5^1⁄3.