7.218. $\left(\frac{3}{5}\right)^{2 \log _{9}(x+1)} \cdot\left(\frac{125}{27}\right)^{\log _{27}(x-1)}=\frac{\log _{5} 27}{\log _{5} 243}$.
Solution.

Domain of definition: $x>1$.

From the condition we have

$\left(\frac{3}{5}\right)^{\log _{3}(x+1)} \cdot\left(\frac{3}{5}\right)^{\log _{3}(x-1)}=\frac{3}{5} \Leftrightarrow\left(\frac{3}{5}\right)^{\log _{3}(x+1)+\log _{3}(x-1)}=\frac{3}{5} \Rightarrow$

$\Rightarrow \log _{3}(x+1)+\log _{3}(x-1)=1 \Rightarrow \log _{3}\left(x^{2}-1\right)=1, x^{2}-1=3, x^{2}=4$.

From this, $x_{1}=-2, x_{2}=2 ; x_{1}=-2$ does not satisfy the domain of definition.

Answer: 2.