Determine all values of $x$ for which $\left(2 \cdot 4^{x^{2}-3 x}\right)^{2}=2^{x-1}$.
Using exponent laws, we obtain the following equivalent equations:

$$
\begin{aligned}
\left(2 \cdot 4^{x^{2}-3 x}\right)^{2} & =2^{x-1} \\
2^{2} \cdot 4^{2\left(x^{2}-3 x\right)} & =2^{x-1} \\
2^{2} \cdot 4^{2 x^{2}-6 x} & =2^{x-1} \\
2^{2} \cdot\left(2^{2}\right)^{2 x^{2}-6 x} & =2^{x-1} \\
2^{2} \cdot 2^{2\left(2 x^{2}-6 x\right)} & =2^{x-1} \\
\frac{2^{2} \cdot 2^{4 x^{2}-12 x}}{2^{x-1}} & =1 \\
2^{2+4 x^{2}-12 x-(x-1)} & =1 \\
2^{4 x^{2}-13 x+3} & =1
\end{aligned}
$$

Since $2^{0}=1$, this last equation is true exactly when $4 x^{2}-13 x+3=0$ or $(4 x-1)(x-3)=0$. Therefore, $x=\frac{1}{4}$ or $x=3$.

We can check by substitution that each of these values of $x$ satisfies the original equation.

ANSWER: $\frac{1}{4}, 3$