[tex]\bf 4^{-x}=\sqrt[8]{\cfrac{1}{32}}\qquad
\begin{cases}
4=2^2\\
32=2^5
\end{cases}\implies (2^2)^{-x}=\sqrt[8]{\cfrac{1}{2^5}}
\\\\\\
2^{-2x}=\sqrt[8]{2^{-5}}\implies 2^{-2x}=2^{\frac{-5}{8}}\impliedby
\begin{array}{llll}
\textit{same base, thus the}\\
\textit{exponents must the same}
\end{array}
\\\\\\
-2x=-\cfrac{5}{8}\implies \boxed{x=\cfrac{5}{16}}[/tex]
