From Förberedande kurs i matematik 1

The argument of ln can be written as

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

and with the logarithm law, \displaystyle \lg a^{b}=b\lg a, we obtain

\displaystyle \ln \frac{1}{e^{2}}=\ln e^{-2}=\left( -2 \right)\centerdot \ln e=\left( -2 \right)\centerdot 1=-2