Let A = (81)^{3⁄4} - (1⁄32)^{-2⁄5} + (8)^{1⁄3}.(1⁄2)^{-1}.2^{0}
We know 81 = 3 x 3 x 3 x 3 = 3^{4}
32 = 2 x 2 x 2 x 2 x 2 = 2^{5} ⇒ 1⁄32 = 1⁄2^{5} = 2^{-5};
8 = 2 x 2 x 2 = 2^{3}; 1⁄2 = 2^{-1}; 2^{0} = 1;
∴ A = (3^{4})^{3⁄4} - (2^{-5})^{-2⁄5} + (2^{3})^{1⁄3}.(2^{-1})^{-1}.1
⇒ A = (3^{4 x 3⁄4}) - (2^{-5 x -2⁄5}) + (2^{3 x 1⁄3}).(2^{-1 x -1}).1 [since (a^{m})^{n} = a^{mn}]
= (3^{3}) - (2^{2}) + (2^{1}).(2^{1}).1
= (3 x 3 x 3) - (2 x 2) + (2).(2).1 = 27 - 4 + 4 = 27. Ans.

Evaluate (16⁄81)^{-3⁄4} x (49⁄9)^{3⁄2}⁄(343⁄216)^{2⁄3}

Solution to Example 2 of Negative Exponent:

Let A = (16⁄81)^{-3⁄4} x (49⁄9)^{3⁄2}⁄(343⁄216)^{2⁄3}
We know
16⁄81 = (2 x 2 x 2 x 2)⁄(3 x 3 x 3 x 3) = 2^{4}⁄3^{4} = (2⁄3)^{4}
49⁄9 = (7 x 7)⁄(3 x 3) = 7^{2}⁄3^{2} = (7⁄3)^{2}
343⁄216 = (7 x 7 x 7)⁄(6 x 6 x 6) = 7^{3}⁄6^{3} = (7⁄6)^{3}
∴ A = {(2⁄3)^{4}}^{-3⁄4} x {(7⁄3)^{2}}^{3⁄2}⁄{(7⁄6)^{3}}^{2⁄3}
= {(2⁄3)^{4 x -3⁄4}} x {(7⁄3)^{2 x 3⁄2}}⁄{(7⁄6)^{3 x 2⁄3}} [since (a^{m})^{n} = a^{mn}]
= {(2⁄3)^{-3}} x {(7⁄3)^{3}}⁄{(7⁄6)^{2}}
= {(3⁄2)^{3}} x {(7⁄3)^{3}}⁄{(7⁄6)^{2}} [since (2⁄3)^{-3} = (3⁄2)^{3}]
= {(3^{3}⁄2^{3})} x {(7^{3}⁄3^{3})}⁄{(7^{2}⁄6^{2})}
= {(3^{3}⁄2^{3})} x {(7^{3}⁄3^{3})} x {(6^{2}⁄7^{2})} [ Since division with (7^{2}⁄6^{2}) is same as multiplication with (6^{2}⁄7^{2})]
= (3^{3} x 7^{3} x 6^{2})⁄(2^{3} x 3^{3} x 7^{2})
= (3^{3 - 3} x 7^{3 - 2} x 6 x 6)⁄(2 x 2 x 2)
= (3^{0} x 7^{1} x 6 x 6)⁄(2 x 2 x 2)
= (1 x 7 x 3 x 3)⁄(2) = 63⁄2. Ans.

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Evaluate:
(64⁄125)^{-2⁄3} + 4^{0} x 9^{5⁄2} x 3^{-4} - √(25)⁄(64)^{1⁄3} x (1⁄3)^{-1}

For Answers, see at the bottom of the page.

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