## Functions Modeling Change: A Preparation for Calculus, 5th Edition

$-5$
Since $a^{-b}=\frac{1}{a^b}$, the expression can be rewritten as $\ln(e^{-5})$. Also, the identity that $\ln e^a=a$ can be used to simplify the expression to $\ln(1/e^5)=\ln e^{-5}=-5$