9 If for any real number $x$, there is $f(x)=\log _{a}\left(2+\mathrm{e}^{x-1}\right) \leqslant-1$, then the range of real number $a$ is $\qquad$ . (where $e$ is an irrational number, $e=2.71828 \cdots$ )
(9) $\left[\frac{1}{2}, 1\right)$ Hint: When $01$, from the condition we get $2+\mathrm{e}^{x-1} \leqslant \frac{1}{a}$, which means $a \leqslant \frac{1}{2+\mathrm{e}^{x-1}}$ always holds, but this is impossible. Therefore, the range of real number $a$ is $\left[\frac{1}{2}, 1\right)$.