## College Algebra 7th Edition

$x=\frac{1}{2}\ln 3\approx 0.5493$
We need to solve: $e^{4x}+4e^{2x}-21=0$ We factor: $(e^{2x}+7)(e^{2x}-3)=0$ $(e^{2x}+7)=0$ or $(e^{2x}-3)=0$ $e^{2x}=-7$ or $e^{2x}=3$ $2x=\ln(-7)$ or $2x=\ln 3$ $x=\frac{1}{2}\ln(-7)$=no solution or $x=\frac{1}{2}\ln 3\approx 0.5493$