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From Förberedande kurs i matematik 1

The argument of ln can be written as

1e2=e−2

and with the logarithm law, lgab=blga, we obtain

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