# How do you solve 4^(2-5x)=1/64?

Notice $\frac{1}{64} = {4}^{-} 3$ hence
${4}^{2 - 5 x} = {4}^{-} 3 \implies \log \left({4}^{2 - 5 x}\right) = \log {4}^{-} 3 \implies \left(2 - 5 x\right) = - 3 \implies 5 = 5 x \implies x = 1$