## Precalculus (6th Edition) Blitzer

$x= \ln 1, \ln 5$ or, $x= 0 ; x=1.609438$
We are given that $e^{2x}-6e^x +5=0$ or, $(e^x -5)(e^x-1) =0$ Solve by equating both factors to $0$: $e^5 -5 =0 \implies x =\ln 5$ and $e^x -1 =0 \implies x= \ln 1$ So, $x= \ln 1, \ln 5$ or, $x= 0 ; x=1.609438$