Answer
-2
Work Step by Step
We know that $y=log_{b}x$ is equivalent to $b^{y}=x$ (where $x\gt0$, $b\gt0$, and $b\ne1$).
Also, based on the definition of the common logarithm, we know that $ln(x)=log_{e}x$.
Therefore, $ln(\frac{1}{e^{2}})=log_{e}\frac{1}{e^{2}}=-2$, because $e^{-2}=\frac{1}{e^{2}}$.
