5. The solution set of the inequality $|\lg x|>|\lg 4 x|>|\lg 2 x|$ is $\qquad$
5. $\left(\frac{\sqrt{2}}{4}, \frac{1}{2}\right)$.
$0|\lg 4 x|$ leads to $-\lg x>\lg 4 x$, then $4 x^{2}1$, $x>\frac{\sqrt{2}}{4}$. Therefore, $\frac{\sqrt{2}}{4}<x<\frac{1}{2}$.