## Precalculus (6th Edition) Blitzer

$\approx 0.97$
We are given that $e^{5x}-32=96$ Use rule: $\log_a{a^x}=x$ and simplify to obtain: $e^{5x} =128$ Take $\ln$ of each side. $\ln e^{5x} =\ln 128$ So, $x=\dfrac{\ln 128}{5} \approx 0.97$