Answer
$x=\ln5\approx1.609438$ and $x=\ln3\approx1.098612$
Work Step by Step
$e^{x}+15e^{-x}-8=0$
Rewrite the second term of this equation as $\dfrac{15}{e^{x}}$:
$e^{x}+\dfrac{15}{e^{x}}-8=0$
Multiply the whole equation by $e^{x}$:
$e^{x}\Big(e^{x}+\dfrac{15}{e^{x}}-8=0\Big)$
$e^{2x}+15-8e^{x}=0$
$e^{2x}-8e^{x}+15=0$
Factor this equation:
$(e^{x}-5)(e^{x}-3)=0$
Set both factors equal to $0$ and solve each individual equation:
$e^{x}-5=0$
Take the $-5$ to the right side:
$e^{x}=5$
Apply $\ln$ to both sides:
$\ln e^{x}=\ln5$
Take the exponent $x$ to multiply in front of the $\ln$:
$x\ln e=\ln5$
Since $\ln e=1$, the solution is:
$x=\ln5\approx1.609438$
Let's move on to the other equation:
$e^{x}-3=0$
Take the $-3$ to the right side of the equation:
$e^{x}=3$
Apply $\ln$ to both sides:
$\ln e^{x}=\ln3$
Take the exponent $x$ down to multiply in front of its respective $\ln$:
$x\ln e=\ln3$
Since $\ln e=1$, the solution is:
$x=\ln3\approx1.098612$
The two solutions for this equations are:
$x=\ln5\approx1.609438$ and $x=\ln3\approx1.098612$