Prove the following
Question:

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

(a) $2^{-\frac{1}{6}}$

(b) $2^{-6}$

(c) $2^{\frac{1}{6}}$

(d) $2^{6}$

Solution:

(c)

$\sqrt[4]{\sqrt[3]{2^{2}}}=\left[\sqrt[3]{2^{2}}\right]^{1 / 4}=\left[\left(2^{2}\right)^{1 / 3}\right]^{1 / 4}$    $\left[\because \sqrt[n]{a}=a^{1 / n}\right]$

$=\left[2^{2 / 3}\right]^{1 / 4}=2^{\frac{2}{3} \cdot \frac{1}{4}}=2^{\frac{1}{6}}$    $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.