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Proportion

√√22 equals t...

Question

$\sqrt[4]{\sqrt[3]{2^{2}}}$ equals to

A.) $2^{\frac{- 1}{6}}$
B.) $2^{- 6}$
C.) $2^{\frac{1}{6}}$
D.) $2^{6}$

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Solution

The answer is C.
$\sqrt[4]{{\sqrt[3]{2}}^{2}} = [{\sqrt[3]{2}}^{2}]^{\frac{1}{4}} = [∵ \sqrt[n]{a} = a^{1 / n}]$
$= [2^{2 / 3}]^{1 / 4} = 2^{\frac{2}{3} . \frac{1}{4}} = 2^{\frac{1}{6}} [(a^{m})^{n} = a^{m n}]$

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$\sqrt[4]{\sqrt[3]{2^{2}}}$ equals to

A.) $2^{\frac{- 1}{6}}$
B.) $2^{- 6}$
C.) $2^{\frac{1}{6}}$
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