2. The function $f(x)=\log _{a} \frac{x+1}{x-1}$ has a range of $(1,+\infty)$ on the interval $(1, a-2)$. Then $a=$ $\qquad$ .
2. $2+\sqrt{3}$.

From $13$.
Since $\frac{x+1}{x-1}=1+\frac{2}{x-1}$ is decreasing on the interval $(1,+\infty)$, therefore, $f(x)$ is decreasing on the interval $(1,+\infty)$.
Hence $f(a-2)=1$
$\Rightarrow a^{2}-4 a+1=0$
$\Rightarrow a=2 \pm \sqrt{3}$ (discard the smaller root).