Value of sqrt[4]{(81)^{-2}} is | vedclass

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(D) Given expression: $\sqrt[4]{(81)^{-2}}$We know that $a^{-n} = \frac{1}{a^n}$,so $(81)^{-2} = \frac{1}{81^2}$.Thus,$\sqrt[4]{(81)^{-2}} = \sqrt[4]{

Vedclass · Accepted Answer

$\frac{1}{9}$

Vedclass · Answer

(D) Given expression: $\sqrt[4]{(81)^{-2}}$We know that $a^{-n} = \frac{1}{a^n}$,so $(81)^{-2} = \frac{1}{81^2}$.Thus,$\sqrt[4]{(81)^{-2}} = \sqrt[4]{\frac{1}{81^2}} = \left(\frac{1}{81^2}ight)^{1/4}$.Since $81 = 9^2$,we have $81^2 = (9^2)^2 = 9^4$.Substituting this back,we get $\left(\frac{1}{9^4}ight)^{1/4} = \left(\frac{1}{9}ight)^{4 	imes \frac{1}{4}} = \frac{1}{9}$.Hence,$(d)$ is the correct answer.

Value of $\sqrt[4]{(81)^{-2}}$ is

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