What is the distance between the closest points on the graphs of the functions $x \mapsto a^{x}$ and $x \mapsto \log _{a} x$ (where $a>1$)?
The two functions are inverses of each other, so their graphs are mirror images with respect to the line $y=x$. If the function $x \mapsto a^{x}$ has a point on the line $y=x$, then this point is also a point on the graph of the function $x \mapsto \log _{a} x$, and in this case, the sought distance is zero. Therefore, let's examine the function $f(x)=a^{x}-x$. If the graph of the function $x \mapsto a^{x}$ has a common point with the line $y=x$, then the set of values of $f(x)$ will include non-positive numbers. The function $f(x)$ is continuous, and $f^{\prime}(x)=\left(a^{x}-x\right)^{\prime}=a^{x} \cdot \ln a-1$ is strictly increasing. Since $a^{x} \cdot \ln a-1$ is exactly 0 when $x=\log _{a} \log _{a} e$, the function $f(x)$ has an absolute minimum at this point. The function value here is:

$$
a^{\log _{a} \log _{a} e}-\log _{a} \log _{a} e=\log _{a} e-\log _{a} \log _{a} e=\log _{a} \frac{e}{\log _{a} e}
$$

This function value is non-positive exactly when $\frac{e}{\log _{a} e} \leqslant 1$, i.e., $e \cdot \ln a \leqslant 1$, or $a \leqslant e^{\frac{1}{e}}$. From this, we can see that the two function graphs have no common point exactly when $a>e^{\frac{1}{e}}$.

The points on the two graphs that are closest to each other will be those points where the tangents are parallel to the line $y=x$. Therefore, let's form the derivative of the function $x \mapsto a^{x}$: $\left(a^{x}\right)^{\prime}=a^{x} \cdot \ln a$. If the point of tangency is the point $\left(x_{0} ; y_{0}\right)$, then $a^{x_{0}} \cdot \ln a=1$, from which $x_{0}=\frac{\ln \ln a}{\ln a}$ and $y_{0}=\frac{1}{\ln a}$. Since the point of tangency on the graph of the function $x \mapsto \log _{a} x$ is the reflection of the point $\left(x_{0} ; y_{0}\right)$ over the line $y=x$, which is $\left(y_{0} ; x_{0}\right)$, the sought distance is

$$
d=\sqrt{\left(\frac{1}{\ln a}+\frac{\ln \ln a}{\ln a}\right)^{2}+\left(\frac{1}{\ln a}+\frac{\ln a \ln a}{\ln a}\right)^{2}}=\sqrt{2} \cdot \frac{1+\ln \ln a}{\ln a}
$$

Summarizing the above, the distance between the closest points on the graphs of the two functions is:

$$
d= \begin{cases}0, & \text { if } 1<a \leqslant e^{\frac{1}{e}} \\ \sqrt{2} \cdot \frac{1+\ln \ln a}{\ln a}, & \text { if } a>e^{\frac{1}{e}}\end{cases}
$$