6 If point $P$ is on the curve $y=\frac{1}{2} e^{x}$, and point $Q$ is on the curve $y=\ln (2 x)$, then the minimum value of $|P Q|$ is $\qquad$
Solution: Note that the graphs of $y=\frac{1}{2} e^{x}$ and $y=\ln (2 x)$ are symmetric with respect to the line $y=x$. We only need to consider the distance from point $Q$ to the line $y=x$,
where $d=\frac{|\ln (2 x)-x|}{\sqrt{2}}=\frac{x-\ln (2 x)}{\sqrt{2}}$,
Let $f(x)=x-\ln (2 x)$, then $f^{\prime}(x)=1-\frac{1}{x}$,
If $x \in(0,1)$, then $f^{\prime}(x)0$;
Thus $\min _{x \in(0,+\infty)} f(x)=f(1)=1-\ln 2$,
Therefore, $\min |P Q|=2 \min d=\sqrt{2}(1-\ln 2)$.