5. The solution set of the inequality $|\lg x|>|\lg 4 x|>|\lg 2 x|$ is . $\qquad$
5. $\left\{x \left\lvert\, \frac{\sqrt{2}}{4}\lg x>\lg 2 x$
$>\lg 4 x$.
$$
\begin{aligned}
\therefore & |\lg x||\lg 4 x|=\lg 4 x \\
\text { pos } 0 & >\lg x+\lg 4 x=\lg 4 x^{2},
\end{aligned}
$$

Thus, $4 x^{2}<1$.
Combining (1) and (2), we get
$$
\frac{1}{4}<x<\frac{1}{2} \text {. }
$$

Therefore, $\frac{1}{2}<2 x<1<4 x$.
Also, $-\lg 2 x=|\lg 2 x|<|\lg 4 x|=\lg 4 x$,
Thus, $0<\lg 2 x+\lg 4 x=\lg 8 x^{2}$.
${ }^{\prime}$ (1) Combining and solving, we get $\frac{\sqrt{2}}{4}<x<\frac{1}{2}$.