Answer
$x\approx0.90$
Work Step by Step
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, $2x=\ln(6)$
$x=\frac{\ln(6)}{2}$
$x\approx0.90$
