7.19

$$
\frac{\log _{a} \sqrt{a^{2}-1} \cdot \log _{\frac{1}{a}}^{2} \sqrt{a^{2}-1}}{\log _{a^{2}}\left(a^{2}-1\right) \cdot \log _{\sqrt[3]{a}} \sqrt[6]{a^{2}-1}}
$$
7.19 By the definition and property $1^{0}$ of the logarithmic function, we have

$a>0, a \neq 1$

and

$a^{2}-1>0 \Rightarrow|a|>1 \Rightarrow a>1$.

From (1) and (2), it follows that $a>1$. Further, since the denominator of the fraction must be non-zero, then $a^{2}-1 \neq 1$ and, therefore, $\quad a \neq \pm \sqrt{2}$. Now, notice that $\log _{\frac{1}{a}} \sqrt{a^{2}-1}=-\log _{a} \sqrt{a^{2}-1}$ (according to (7.7)), and using formula (7.9), we find

$A=\frac{\log _{a} \sqrt{a^{2}-1} \cdot \log _{a}^{2} \sqrt{a^{2}-1}}{\log _{a} \sqrt{a^{2}-1} \cdot \log _{a} \sqrt{a^{2}-1}}=\log _{a} \sqrt{a^{2}-1}$,

where $a>1$ and $a \neq \sqrt{2}$.