Answer
$x\approx1.10$
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.
$e^{2x}-2e^x-3=0$
$(e^x-3)(e^x+1)=0$
$e^x-3=0, e^x+1=0$
$e^x=3, e^x=-1$
$x\approx1.10$, not possible
$x\approx1.10$
