The value of ${[{[{(2401)}^{- 1 / 2}]}^{- 1 / 4}]}^{2}$ is:

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{7}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $7$
$[[(2401)^{- 1 / 2}]^{- 1 / 4}]^{2} = [(2401)^{- 1 / 2}]^{- 1 / 4 \times 2}$
$= [(2401)^{- 1 / 2}]^{- 1 / 2}$
$= (2401)^{- 1 / 2 \times - 1 / 2}$ $= (2401)^{1 / 4}$
$= (7^{4})^{1 / 4}$ $= 7^{4 \times 1 / 4}$ $= 7$
Option B is correct.

Suggest Corrections

Similar questions

128^{\frac{- 2}{7}} - (625^{- 3})^{\frac{- 1}{4}} + 14 (2401)^{\frac{- 1}{4}}

D_{k} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a & 2^{k} & 2^{16} - 1 b & 3 (4^{k}) & 2 (4^{16} - 1) c & 7 (8^{k}) & 4 (8^{16} - 1) \end{matrix} ∣ ∣ ∣ ∣ ∣

16 \sum k = 1 D_{k}

\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}}

\frac{1}{\sqrt{9} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{1}}

x

\sqrt{1 + \sqrt{1 - \frac{2176}{2401}}} = 1 + \frac{x}{7}

Join BYJU'S Learning Program