4. Given that point $P$ is on the curve $y=2 e^{x}$, and point $Q$ is on the curve $y=\ln \frac{x}{2}$, then the minimum value of $|P Q|$ is
Solution: $\sqrt{2}(1+\ln 2)$.
The graphs of the functions $y=2 e^{x}$ and $y=\ln \frac{x}{2}$ are symmetric with respect to the line $y=x$, so we only need to find the minimum distance from a point on the curve $y=2 e^{x}$ to the line $y=x$.
Since the point on the curve $y=2 e^{x}$ that is closest to the line $y=x$ has a tangent line parallel to $y=x$, we have
$$
y^{\prime}=2\left(e^{x}\right)^{\prime}=2 e^{x}=1
$$

Therefore, $x=-\ln 2, y=1$.
Thus, when $P, Q$ are located at $(-\ln 2,1),(1,-\ln 2)$ respectively, the distance is the smallest, i.e.,
$$
|P Q|_{\text {min }}=\sqrt{2}(1+\ln 2)
$$