## Precalculus (6th Edition) Blitzer

Solution set = $\displaystyle \{\frac{\ln 3}{2}\}.$ $x\approx 0.55$
Recognize that $e^{4x}=(e^{2x})^{2}$. Introduce a variable t, $t=e^{2x}.$ $t \gt 0$, because $e^{2x}$ is always positive. We solve $t^{2}+5t-24=0$ Factor by finding factors of -24 whose sum is 5: $(t+8)(t-3)=0$ $t+8=0$ or $t-3=0$ $t=-8$ is discarded, as $t \gt 0$, leaving $t=3\qquad$... bring back x $e^{2x}=3\qquad$... apply ln( ) $\ln e^{2x}=\ln 3\qquad$... apply $\log_{b}b^{x}=x$ $2x=\ln 3\qquad$... divide with 2 $x=\displaystyle \frac{\ln 3}{2}\qquad$... round to 2 decimal places $x\approx 0.55$