Evaluate natural log of 1/(e^9)

$\ln (\frac{1}{e^{9}})$

Move $e^{9}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$\ln (e^{- 9})$

Expand $\ln (e^{- 9})$ by moving $- 9$ outside the logarithm.

$- 9 \ln (e)$

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

$- 9 \cdot 1$

Multiply $- 9$ by $1$ .

$- 9$

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