## College Algebra (10th Edition)

The solution set is $\left\{-6, 2\right\}$.
Simplify the left side using the rules $(a^m)^n=a^{mn}$ and $a^m \cdot a^n=a^{m+n}$ to obtain: $e^{4x} \cdot e^{x^2} = e^{12} \\e^{4x+x^2}=e^{12} \\e^{x^2+4x} = e^{12}$ Use the rule $a^m=a^n \longrightarrow m=n$ to obtain: $x^2+4x=12$ Subtract $12$ on both sides of the equation to obtain: $\begin{array}{ccc} &x^2+4x-12&= &12-12 \\&x^2+4x-12 &= &0\end{array}$ Factor the trinomial to obtain: $(x+6)(x-2)=0$ Equate each factor to zero, and then solve each equation to obtain: $\begin{array}{ccc} &x+6=0 &\text{ or } &x-2=0 \\&x=-6 &\text{ or } &x=2\end{array}$ Thus, the solution set is $\left\{-6, 2\right\}$.