## College Algebra 7th Edition

$x=\ln 5\approx 1.6094$ or $x=\ln 3\approx 1.0986$
We need to solve: $e^{x}+15e^{-x}-8=0$ We multiply through by $e^x$ (always positive): $e^x(e^{x}+15e^{-x}-8)=e^x(0)$ $e^{2x}+15-8e^{x}=0$ $(e^{x}-5)(e^{x}-3)=0$ $(e^{x}-5)=0$ or $(e^{x}-3)=0$ $e^x=5$ or $e^x=3$ $x=\ln 5\approx 1.6094$ or $x=\ln 3\approx 1.0986$