Find:
(i) $ 9^{\frac{3}{2}} $
(ii) $ 32^{\frac{2}{5}} $
(iii) $ 16^{\frac{3}{4}} $
(iv) $ 125^{\frac{-1}{3}} $

AcademicMathematicsNCERTClass 9

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 $125^{\frac{-1}{3}}=\frac{1}{5}$

raja
Updated on 10-Oct-2022 13:38:51

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