## Algebra and Trigonometry 10th Edition

$x=1.609$
$e^{2x}-4e^x-5=0$ $(e^x)^2-4e^x-5=0~~$ ($a=1,b=-4,c=-5$): $e^x=\frac{-b±\sqrt {b^2-4ac}}{2a}=\frac{-(-4)±\sqrt {(-4)^2-4(1)(-5)}}{2(1)}=\frac{4±\sqrt {36}}{2}=\frac{4±6}{2}=2±3$ $e^x=5$ $\ln e^x=\ln5$ $x=1.609$ $e^x=-1$. But, there is no $x$ such that $e^x=-1$.