2. Given $\log _{\sqrt{7}}(5 a-3)=\log _{\sqrt{a^{2}+1}} 5$. Then the real number
$$
a=
$$
. $\qquad$
2. 2 .

Simplify the original equation to
$$
\log _{7}(5 a-3)=\log _{a^{2}+1} 5 \text {. }
$$

Since $f(x)=\log _{7}(5 x-3)$ is an increasing function for $x>\frac{3}{5}$, and $g(x)=\log _{5}\left(x^{2}+1\right)$ is also an increasing function, and $f(2)=$ $g(2)=1$, therefore, $a=2$.