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stich3 [128]
3 years ago
8

Precalculus

Mathematics
1 answer:
luda_lava [24]3 years ago
6 0
\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}}
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