Answer
The solution set is $\left\{-4, 2\right\}$.
Work Step by Step
To solve the given equation, make the two sides have the same base.
Note that $4=2^2$ and $16=2^4$, so the given equation is equivalent to:
$(2^2)^x \cdot 2^{x^2} = (2^4)^2$
Use the rule $(a^m)^n = a^{mn}$ to obtain:
$2^{2x} \cdot 2^{x^2} = 2^{4(2)}
\\2^{2x} \cdot 2^{x^2}=2^{8}$
Use the rule $a^m \cdot a^n = a^{m+n}$ to obtain:
$2^{2x+x^2} = 2^8
\\2^{x^2+2x} = 2^8$
Use the rule $a^m=a^n \longrightarrow m=n$ to obtain:
$x^2+2x=8$
Subtract $8$ to both sides of the equation to obtain:
$\begin{array}{ccc}
&x^2+2x -8 &= &8-8
\\&x^2+2x-8 &= &0
\end{array}$
Factor the trinomial to obtain:
$(x+4)(x-2)=0$
Equate each factor to zero, and then solve each equation to obtain:
$\begin{array}{ccc}
&x+4=0 &\text{ or } &x-2-0
\\&x=-4 &\text{ or } &x=2
\end{array}$
Thus, the solution set is $\left\{-4, 2\right\}$.
