## Precalculus (6th Edition) Blitzer

Solution set = $\displaystyle \{\frac{\ln 3}{\ln 2}\}.$ $x\approx 1.58$
Recognize that $2^{2x}=(2^{x})^{2}$. Introduce a new variable t: $t=2^{x}.$ $t \gt 0$, because $2^{x}$ is always positive. We solve $t^{2}+t-12=0$ Factor by finding factors of $-12$ whose sum is $+1$: $(t+4)(t-3)=0$ $t+4=0$ or $t-3=0$ $t=-4$ or $t=3$ $t=-4$ is discarded, as $t \gt 0$, leaving $t=3\qquad$... bring back x $2^{x}=3\qquad$... apply ln( ) $\ln 2^{x}=\ln 3\qquad$... apply $\log_{b}a^{p}=p\log_{b}a$ $x\ln 2=\ln 3\qquad$... divide with $\ln 2$ $x=\displaystyle \frac{\ln 3}{\ln 2}\qquad$... round to 2 decimal places $x\approx 1.58$