7.160. Simplify the expression $\log _{a+b} m+\log _{a-b} m-2 \log _{a+b} m \cdot \log _{a-b} m$, given that $m^{2}=a^{2}-b^{2}$.
Solution.

$\log _{a+b} m+\log _{a-b} m-2 \log _{a+b} m \cdot \log _{a-b} m=\log _{a+b} m+\frac{\log _{a+b} m}{\log _{a+b}(a-b)}-$

$-2 \frac{\log _{a+b} m \cdot \log _{a+b} m}{\log _{a+b}(a-b)}=\log _{a+b} m \cdot\left(1+\frac{1}{\log _{a+b}(a-b)}-\frac{2 \log _{a+b} m}{\log _{a+b}(a-b)}\right)=$

$=\frac{\log _{a+b} m\left(\log _{a+b}(a-b)+1-2 \log _{a+b} m\right)}{\log _{a+b}(a-b)}$.

Since $m=\sqrt{a^{2}-b^{2}}$, we have

$$
\begin{aligned}
& \frac{\log _{a+b} \sqrt{a^{2}-b^{2}}\left(\log _{a+b}(a-b)+1-2 \log _{a+b} \sqrt{a^{2}-b^{2}}\right)}{\log _{a+b}(a-b)}= \\
& =\frac{\log _{a+b} \sqrt{a^{2}-b^{2}}\left(\log _{a+b}(a-b)+1-\log _{a+b}(a-b)-1\right)}{\log _{a+b}(a-b)}= \\
& =\frac{\log _{a+b} \sqrt{a^{2}-b^{2}} \cdot 0}{\log _{a+b}(a-b)}=0
\end{aligned}
$$

Answer: 0.