1. It is known that the inequality $a^{x} \geqslant x \geqslant \log _{a} x$ holds for all $x$ in the interval $(0,+\infty)$. Then the range of positive real number $a$ is $\qquad$
II. $1 . a \geqslant \mathrm{e}^{\frac{1}{6}}$.
Obviously, a necessary condition for the inequality to hold is $a>1$.
Let $f(x)=a^{x}-x$, then
$$
\begin{array}{l}
f^{\prime}(x)=a^{x} \ln a-1 \geqslant 0 \Leftrightarrow x \geqslant-\frac{\ln (\ln a)}{\ln a}, \\
f^{\prime}(x) \leqslant 0 \Leftrightarrow x \leqslant-\frac{\ln (\ln a)}{\ln a} .
\end{array}
$$

Therefore, $x=-\frac{\ln (\ln a)}{\ln a}$ is the unique minimum point of $f(x)$.
From $x=-\frac{\ln (\ln a)}{\ln a}$, we get
$\ln a^{x}=\ln \frac{1}{\ln a} \Rightarrow a^{x}=\frac{1}{\ln a}$.
Substituting into $f(x)=a^{x}-x$ gives
$f(x)_{\min }=\frac{1+\ln (\ln a)}{\ln a}=\frac{\ln (\mathrm{e} \ln a)}{\ln a}$.
Thus, $a^{x} \geqslant x$ holds for all $x \in (0,+\infty)$
$\Leftrightarrow f(x) \geqslant 0 \Leftrightarrow f(x)_{\min }=\frac{\ln (\mathrm{e} \ln a)}{\ln a} \geqslant 0$
$\Leftrightarrow \ln (\ln a) \geqslant-1 \Leftrightarrow a \geqslant \mathrm{e}^{\frac{1}{e}}$.
Since $x \geqslant \log _{a} x \Leftrightarrow a^{x} \geqslant x$, we have
$x \geqslant \log _{a} x$ holds for all $x \in (0,+\infty)$
$\Leftrightarrow a \geqslant \mathrm{e}^{\frac{1}{e}}$.
Therefore, $a^{x} \geqslant x \geqslant \log _{a} x$ holds for all $x \in (0,+\infty)$ $\Leftrightarrow a \geqslant \mathrm{e}^{\frac{1}{e}}$.