Question:
The value of $\sqrt[4]{(81)^{-2}}$ is
(a) $\frac{1}{9}$
(b) $\frac{1}{3}$
(c) 9
(d) $\frac{1}{81}$
Solution:
(a)
$\sqrt[4]{(81)^{-2}}=\sqrt[4]{\frac{1}{(81)^{2}}}=\frac{1}{(81)^{2 / 4}}=\frac{1}{(81)^{1 / 2}}$ $\left[\because \sqrt[m]{a^{n}}=a^{n / m}\right]$
$=\frac{1}{\left(9^{2}\right)^{1 / 2}}=\frac{1}{9^{2 / 2}}=\frac{1}{9}$ $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$
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