Answer
solution set is $\left\{-6, 2\right\}$
Work Step by Step
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 $\left\{-6, 2\right\}$.
