Answer
Solution set = $\displaystyle \{\frac{\ln 3}{2}\}.$
$ x\approx 0.90$
Work Step by Step
Recognize that $ e^{4x}=(e^{2x})^{2}$. Introduce a new variable t, $ t=e^{2x}.$
$ t \gt 0 $, because $ e^{2x}$ is always positive.
We solve
$ t^{2}-3t-18=0$
Factor by finding factors of -18 whose sum is -3:
$(t-6)(t+3)=0$
$ t-6=0$ or $ t+3=0$
$ t=6$ or $ t=-3$
$ t=-3$ is discarded, as $ t \gt 0 $, leaving
$ t=6\qquad $... bring back x
$ e^{2x}=6\qquad $... apply ln( )
$\ln e^{2x}=\ln 6\qquad $... apply $\log_{b}b^{x}=x $
$ 2x=\ln 6\qquad $... divide with 2
$ x=\displaystyle \frac{\ln 6}{2}\qquad $... round to 2 decimal places
$ x\approx 0.90$
