Solve the following equation:

$$
16^{x}+16^{1-x}=10
$$
Our equation can also be written as

$$
16^{x}+\frac{16}{16^{x}}=10
$$

or

$$
16^{2 x}-10 \cdot 16^{x}+16=0
$$

or, if $16^{x}=y$,

$$
y^{2}-10 y+16=0
$$

from which

$$
y_{1}=2, \quad y_{2}=8
$$

thus

$$
\begin{aligned}
& 16^{x_{1}}=2, \text { so } x_{1}=\frac{1}{4} \\
& 16^{x_{2}}=8, \text { so } x_{2}=\frac{3}{4}
\end{aligned}
$$