Find:
(i) $ 9^{\frac{3}{2}} $
(ii) $ 32^{\frac{2}{5}} $
(iii) $ 16^{\frac{3}{4}} $
(iv) $ 125^{\frac{-1}{3}} $
To do:
We have to find the values of
(i) \( 9^{\frac{3}{2}} \)
(ii) \( 32^{\frac{2}{5}} \)
(iii) \( 16^{\frac{3}{4}} \)
(iv) \( 125^{\frac{-1}{3}} \)
Solution:
We know that,
$(a^m)^n=(a)^{mn}$
Therefore,
(i) $9^{\frac{3}{2}}=(3\times3)^{\frac{3}{2}}$
$=(3^2)^{\frac{3}{2}}$
$=(3)^{2\times\frac{3}{2}}$
$=3^3$
$=27$
Hence $9^{\frac{3}{2}}=27$
(ii) $32^{\frac{2}{5}}=(2\times2\times2\times2\times2)^{\frac{2}{5}}$
$=(2^5)^{\frac{2}{5}}$
$=(2)^{5\times\frac{2}{5}}$
$=2^2$
$=4$
Hence $32^{\frac{2}{5}}=4$
(iii) $16^{\frac{3}{4}}=(2\times2\times2\times2)^{\frac{3}{4}}$
$=(2^4)^{\frac{3}{4}}$
$=(2)^{4\times\frac{3}{4}}$
$=2^3$
$=8$
Hence $16^{\frac{3}{4}}=8$
(iv) $125^{\frac{-1}{3}}=(5\times5\times5)^{\frac{-1}{3}}$
$=(5^3)^{\frac{-1}{3}}$
$=(5)^{3\times\frac{-1}{3}}$
$=5^{-1}$
$=\frac{1}{5}$
Hence $125^{\frac{-1}{3}}=\frac{1}{5}$
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