Determine all real values of $x$ for which $16^{x}-\frac{5}{2}\left(2^{2 x+1}\right)+4=0$.
Starting with the given equation, we obtain the following equivalent equations:

$$
\begin{aligned}
16^{x}-\frac{5}{2}\left(2^{2 x+1}\right)+4 & =0 \\
\left(4^{2}\right)^{x}-\frac{5}{2}\left(2 \cdot\left(2^{2}\right)^{x}\right)+4 & =0 \\
\left(4^{x}\right)^{2}-5\left(4^{x}\right)+4 & =0 \\
\left(4^{x}-4\right)\left(4^{x}-1\right) & =0
\end{aligned}
$$

Therefore, $4^{x}=4$ (which gives $x=1$ ) or $4^{x}=1$ (which gives $x=0$ ).

Thus, the solutions are $x=0,1$.