Solution 3.3:5e
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 \ln a^{b} = b\ln a, we obtain
| \displaystyle \ln \frac{1}{e^{2}} = \ln e^{-2} = (-2)\cdot\ln e = (-2)\cdot 1 = -2\,\textrm{.} |
