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

$-\ln(e^x+1)$
Use the fact that $\frac{1}{a}=a^{-1}$ to rewrite the expression as $\ln(e^x+1)^{-1}$. Since $\ln a^b=b\ln a$, this expression can be simplified to $-\ln(e^x+1)$. There is no way to simplify the log of a sum, so this is fully simplified.