Question

The value of ${\left[\frac{256x^{16}}{81y^4}\right]}^{-\frac{1}{4}}$ is
(a) $\frac{3y}{8x^4}$

(b) $\frac{3y}{4x^4}$

(c) $\frac{4y}{5x^4}$

(d) $\frac{4x^4}{3y}$

Answer 1

$$\begin{gathered} {\left[\frac{256x^{16}}{81y^4}\right]}^{-\frac{1}{4}}={\left[\frac{2^8{x}^{16}}{3^4{y}^4}\right]}^{-\frac{1}{4}} \\ ={\left[\frac{3^4{y}^4}{2^8{x}^{16}}\right]}^{\frac{1}{4}} \\ =\left[\frac{{\left(3^4\right)}^{{\displaystyle \frac{1}{4}}}{\left(y^4\right)}^{{\displaystyle \frac{1}{4}}}}{{\left(2^8\right)}^{{\displaystyle \frac{1}{4}}}{\left(x^{16}\right)}^{{\displaystyle \frac{1}{4}}}}\right] \\ =\left[\frac{3y}{2^2{x}^4}\right] \\ =\frac{3y}{4x^4} \\ \therefore \mathrm{The}\mathrm{value}\mathrm{of}{\left[\frac{256x^{16}}{81y^4}\right]}^{-\frac{1}{4}}\mathrm{is}\frac{3y}{4x^4}. \end{gathered}$$

Hence, the correct option is (b).