7.253. $\log _{a} \sqrt{4+x}+3 \log _{a^{2}}(4-x)-\log _{a^{4}}\left(16-x^{2}\right)^{2}=2$. For which values of $a$ does the equation have a solution?
Solution.

Domain of definition: $\left\{\begin{array}{l}-4<x<4, \\ 0<a \neq 1 .\end{array}\right.$

Transition to base $a$. We have

$$
\begin{aligned}
& \frac{1}{2} \log _{a}(4+x)+\frac{3}{2} \log _{a}(4-x)-\frac{1}{2} \log _{a}(4-x)-\frac{1}{2} \log _{a}(4+x)=2 \Leftrightarrow \\
& \Leftrightarrow \log _{a}(4-x)=2 \Leftrightarrow 4-x=a^{2}, x=4-a^{2}
\end{aligned}
$$

From this, we have

$$
\left\{\begin{array} { l } 
{ - 4 < 4 - a ^ { 2 } < 4 , } \\
{ 0 < a \neq 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a<2 \sqrt{2} \\
0<a \neq 1
\end{array}\right.\right.
$$

Answer: $\quad x=4-a^{2}$, where $a \in(0 ; 1) \cup(1 ; 2 \sqrt{2})$.