9. For any real number $x$, we have
$$
f(x)=\log _{a}\left(2+\mathrm{e}^{\mathrm{x}-1}\right) \leqslant-1 \text {, }
$$

then the range of real number $a$ is $\qquad$
9. $\left[\frac{1}{2}, 1\right)$.

When $0 < a < 1$, the condition is always satisfied.

When $a > 1$, from the condition we get
$$
2+\mathrm{e}^{x-1} \leqslant \frac{1}{a} \Rightarrow a \leqslant \frac{1}{2+\mathrm{e}^{x-1}}
$$

which is impossible to hold constantly.
Therefore, the range of real number $a$ is $\left[\frac{1}{2}, 1\right)$.