3B. Solve the inequality

$$
\left(\frac{1}{2}\right)^{-|x-3|-\frac{1}{|x-3|}}=4
$$
Solution. Using the properties of exponents, we sequentially obtain

$$
\begin{aligned}
& 2^{|x-3|+\frac{1}{|x-3|}}=2^{2} \\
& |x-3|+\frac{1}{|x-3|}=2
\end{aligned}
$$

$$
\begin{aligned}
& |x-3|^{2}-2|x-3|+1=0 \\
& (|x-3|-1)^{2}=0
\end{aligned}
$$

from which $|x-3|=1$, that is, $x=4$ or $x=2$.