#### Answer

$e^{-2}=\frac{1}{e^{2}}$

#### 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, $log_{e}\frac{1}{e^{2}}=-2$ can be written as the exponential equation $e^{-2}=\frac{1}{e^{2}}$.