13. Find the range of positive real number $a$ such that the inequality $\log _{a} x \leqslant x \leqslant a^{x}$ holds for all $x$ in the interval $(0,+\infty)$; and determine the value of $a$ for which the graphs of the functions $y=a^{x}$ and $y=\log _{a} x$ intersect.
Let $f(x)=\log _{a} x-x, f^{\prime}(x)=\frac{1}{x \ln a}-1=\frac{1}{x}\left(\frac{1}{\ln a}-x\right)$,
When $0<a<1$, $f(x)$ is monotonically decreasing on $\left(0, \frac{1}{\ln a}\right)$ and monotonically increasing on $\left(\frac{1}{\ln a},+\infty\right)$,
When $a>1$, $f(x)$ is monotonically increasing on $\left(0, \frac{1}{\ln a}\right)$ and monotonically decreasing on $\left(\frac{1}{\ln a},+\infty\right)$,
$$
f(x)_{\max }=f\left(\frac{1}{\ln a}\right)=\log _{a} \frac{1}{\ln a}-\frac{1}{\ln a}=-\frac{1}{\ln a}[\ln (\ln a)+1] \leqslant 0 \Rightarrow a \geqslant \mathrm{e}^{\mathrm{e}^{-1}} ;
$$

Let $g(x)=a^{x}-x, g^{\prime}(x)=a^{x} \ln a-1=\ln a\left(a^{x}-\frac{1}{\ln a}\right)$,
When $0<a<1$, $g(x)$ is monotonically increasing on $\left(0, \log _{a} \frac{1}{\ln a}\right)$ and monotonically decreasing on $\left(\log _{a} \frac{1}{\ln a},+\infty\right)$,
When $a>1$, $g(x)$ is monotonically decreasing on $\left(0, \log _{a} \frac{1}{\ln a}\right)$ and monotonically increasing on $\left(\log _{a} \frac{1}{\ln a},+\infty\right)$, $g(x)_{\min }=g\left(\log _{a} \frac{1}{\ln a}\right)=\frac{1}{\ln a}-\log _{a} \frac{1}{\ln a}=\frac{1}{\ln a}[\ln (\ln a)+1] \geqslant 0 \Rightarrow a \geqslant \mathrm{e}^{\mathrm{e}^{-1}}$. In summary, the range of positive real number $a$ is $\left[\mathrm{e}^{\mathrm{e}^{-1}},+\infty\right)$. When $a \in(0,1) \cup\left(1, \mathrm{e}^{\mathrm{e}^{-1}}\right]$, the graphs of the functions $y=a^{x}$ and $y=\log _{a} x$ have intersection points.