## Precalculus (10th Edition)

$x=-1+ln{\frac{5}{4}}$
We know that by definition if $y=a^x$, then $log_a {y}=x$ , also $log_e x=ln x$, hence if $y=e^x$, then $ln{e^x}=log_e {e^x}=x$ and vice versa and that $ln{\frac{1}{x}}=-ln{x}$. Hence if $4e^{x+1}=5$, then $e^{x+1}=\frac{5}{4}$. Solve the equation above to obtain (after taking $ln$ of both sides): \begin{align*}x+1=ln{\frac{5}{4}}\end{align*}\begin{align*}x=-1+ln{\frac{5}{4}}\end{align*}