## College Algebra 7th Edition

$x=1$
Letting $u=3^x$, the given equation becomes: $u^2-u-6=0$ Factor the trinomial to obtain: $(u-3)(u+2)=0$ Equate each factor to zero then solve each equation to obtain: $\begin{array}{ccc} &u-3=0 &\text{ or } &u+2=0 \\&u=3 &\text{ or } &u=-2 \end{array}$ Replace $u$ with $3^x$ to obtain: $3^x=3 \text{ or } 3^x=-2 \\3^x=3^1 \text{ or } 3^x=-2$ Use the rule $a^m=a^n\longrightarrow m=n$ to solve the first equation and have: $x=1$ The second equation has no solution since there is no real number that will make $3^x=-2$. Thus, the only solution is $x=1$.