Answer
$log_{4}\frac{1}{16}=-2$
Work Step by Step
We know that if $b\gt0$ and $b\ne1$, then $y=log_{b}x$ is equivalent to $x=b^{y}$ (where $x\gt0$ and $y$ is a real number).
Therefore, $4^{-2}=\frac{1}{16}$ can be written as the logarithmic equation $log_{4}\frac{1}{16}=-2$.