Find the Exact Value natural log of 1/(e^2)

$\ln (\frac{1}{e^{2}})$

Step 1

Move $e^{2}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$\ln (e^{- 2})$

Step 2

Expand $\ln (e^{- 2})$ by moving $- 2$ outside the logarithm.

$- 2 \ln (e)$

Step 3

The natural logarithm of $e$ is $1$ .

$- 2 \cdot 1$

Step 4

Multiply $- 2$ by $1$ .

$- 2$