Question

Fill in the blanks to make the statement true:${\left[{\left(\frac{-1}{4}\right)}^{16}\right]}^2={\left(\frac{-1}{4}\right)}^{\_\_\_}$

Answer 1

Find the missing exponent.

Given equation: ${\left[{\left(\frac{-1}{4}\right)}^{16}\right]}^2={\left(\frac{-1}{4}\right)}^{\_\_\_}$

$$\begin{gathered} {\left[{\left(\frac{-1}{4}\right)}^{16}\right]}^2={\left(\frac{-1}{4}\right)}^{16\times 2}\left[\because {\left(a^m\right)}^n=a^{mn}\right] \\ ={\left(\frac{-1}{4}\right)}^{32} \end{gathered}$$

So the missing exponent is $32$.

Hence, ${\left[{\left(\frac{-1}{4}\right)}^{16}\right]}^2={\left(\frac{-1}{4}\right)}^{\underline{32}}$