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 = 5^{3}
∴ A = (125)^{1⁄3} = (5^{3})^{1⁄3}.
We know (a^{m})^{n} = a^{mn} (See Law 2)
Applying this Law, we get
A = (5^{3})^{1⁄3} = 5^{3 x 1⁄3} = 5^{1} = 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} = (2^{3})^{2⁄3} = 2^{3 x 2⁄3} = 2^{2} = 4. Ans.

Solved Example 3 of Exponents and Radicals

Evaluate (1⁄5)^{-2}

Solution to Example 3 :

We know (1⁄a)^{-n} = a^{n}
∴ (1⁄5)^{-2} = 5^{2} = 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 = 2^{4}
∴ (16)^{-3⁄4} = (2^{4})^{-3⁄4} = 2^{4 x -3⁄4} {since (a^{m})^{n} = a^{mn}}
= 2^{-3} = 1⁄2^{3} ( since a^{-n} = 1⁄a^{n} )
= 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 2^{5}. ( since 2 x 2 x 2 x 2 x 2 = 32)
(32)^{-4⁄5} = (2^{5})^{-4⁄5} = 2^{5 x -4⁄5} {since (a^{m})^{n} = a^{mn}}
= 2^{-4} = 1⁄2^{4} ( since a^{-n} = 1⁄a^{n} )
= 1⁄16. Ans.

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.

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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} }
= 10^{2}⁄3^{2} { since (a⁄b)^{m} = a^{m}⁄b^{m} }
= 100⁄9. Ans.

Exercise on Exponents and Radicals :

Evaluate

(64)^{1⁄3}

8^{5⁄3}

9^{3⁄2}

If √2 = 1.414 and √3 = 1.732,
find the value of

√(72)

√(147) + √(27)

5√(128) + 2√(75) - 5√(108)

√(36)

(125)^{1⁄3}

(81)^{1⁄4}

(32)^{1⁄5}

√(50)

(40)^{1⁄3}

For Answers, see at the bottom of the page.

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