Answer
$\log _{a}(x+\sqrt{x^{2}-1})+\log _{a}(x-\sqrt{x^{2}-1})=\log_a{(x^2-}(x^2-1))=\log_a{1}=0$
Work Step by Step
Given: $\quad \log _{a}(x+\sqrt{x^{2}-1})+\log _{a}(x-\sqrt{x^{2}-1})=0$
Use the rules $\quad \log_aA+\log_aB=\log_a AB\quad$ and $\quad(a+b)(a-b)=a^2-b^2$ to simplify the LHS::
\begin{align*}
\log _{a}(x+\sqrt{x^{2}-1})+\log _{a}(x-\sqrt{x^{2}-1})&=\log _{a}(x+\sqrt{x^{2}-1})(x-\sqrt{x^{2}-1})\\
&=\log _{a}\left(x^{2}-\left(\sqrt{\left(x^{2}-1\right)}\right)^{2}\right)\\
&=\log _{a}\left(x^{2}-(x^2-1)\right)\\
&=\log _{a}\left(x^{2}-x^2+1\right)\\
&=\log _{a}\left(1\right)\\
&=0 \quad\text{(since } \log_a{1}=0)\\
&=\text{RHS}
\end{align*}
