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Q.

What is the value of $\sqrt[4]{\sqrt[3]{2^{2}}}$ ?

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a

2^{\frac{- 1}{6}}

b

2^{- 6}

c

2^{\frac{1}{6}}

d

2^{6}

answer is C.

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Detailed Solution

Given number is

\sqrt[4]{\sqrt[3]{2^{2}}}

.
We know that

\sqrt[b]{a} = a^{\frac{1}{b}}

where a and b are real numbers.

\Rightarrow \sqrt[4]{\sqrt[3]{2^{2}}} = {(\sqrt[3]{2^{2}})}^{\frac{1}{4}}

\Rightarrow \sqrt[4]{\sqrt[3]{2^{2}}} = (\sqrt[3]{4})^{\frac{1}{4}}

\Rightarrow \sqrt[4]{\sqrt[3]{2^{2}}} = (4^{\frac{1}{3}})^{\frac{1}{4}}

We know that

(a^{b})^{c} = a^{bc}

where a, b and c are real numbers.

\Rightarrow \sqrt[4]{\sqrt[3]{2^{2}}} = 4^{\frac{1}{12}}

\Rightarrow \sqrt[4]{\sqrt[3]{2^{2}}} = (2^{2})^{\frac{1}{12}}

\Rightarrow \sqrt[4]{\sqrt[3]{2^{2}}} = 2^{\frac{1}{6}}

Hence, option 3 is correct.

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