Answer
The solution set is $\left\{0, \frac{1}{2}\right\}$.
Work Step by Step
To solve the given equation, make the two sides have the same base.
Note that $4 = 2^2$ so the given equation is equivalent to:
$(2^2)^{x^2}=2^x$
Use the rule $(a^m)^n = a^{mn}$ to obtain:
$2^{2x^2} = 2^{x}$
Use the rule $a^m=a^n \longrightarrow m=n$ to obtain:
$2x^2 = x$
Subtract $x$ to both sides of the equation to obtain:
$2x^2-x=0$
Factor out the GCF $x$ to obtain:
$x(2x-1)=0$
Equate each factor to $0$, then solve each equation to obtain:
$\begin{array}{ccc}
&x=0 &\text{ or } &2x-1=0
\\&x=0 &\text{ or } &2x=1
\\&x=0 &\text{ or } &x = \frac{1}{2}
\end{array}$
Thus, the solution set is $\left\{0, \frac{1}{2}\right\}$.
