Question

Precalculus

Answer 1

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