## Precalculus (10th Edition)

$x=3$.
$\left(\frac{1}{4}\right)^x=4^{-x}$, $\left(\frac{1}{64}\right)=\left(\frac{1}{4}\right)^3=4^{-3}$, hence the equation becomes: $4^{-x}=4^{-3}.$ The base is same on the 2 sides on the equation (and it is not $1$ or $-1$), hence they will be equal if the exponents are equal. Hence $-x=-3\\x=3$.