## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x=\displaystyle \{1,\quad \frac{\log 5}{\log 3}\}$
Substituting $\quad t=3^{x}\qquad (t\gt 0)$ the equation becomes $t^{2}-8t+15=0$ Factor by finding two factors of 15 with sum $-8$... they are $-3$ and $-5$ $(t-3)(t-5)=0$ $t=3, t=5$ $t=3\Rightarrow\quad 3^{x}=3\Rightarrow\quad x=1.$ $t=5\Rightarrow\quad 3^{x}=5\quad$apply log(...) in both sides $x\log 3=\log 5$ $x=\displaystyle \frac{\log 5}{\log 3}$ $x=\displaystyle \{1,\quad \frac{\log 5}{\log 3}\}$