## College Algebra (6th Edition)

$x\approx0.90$
Begin by factoring the terms using trial and error. Then set the terms in each set of parentheses equal to $0$ and solve for $x$ by using the natural log on both sides of each equation. $e^{4x}-3e^{2x}-18=0$ $(e^{2x}+3)(e^{2x}-6)=0$ $e^{2x}=-3, e^{2x}=6$ not possible, $2x=\ln(6)$ $x=\frac{\ln(6)}{2}$ $x\approx0.90$