7.258. $2^{\frac{a+3}{a+2}} \cdot 32^{\frac{1}{x(a+2)}}=4^{\frac{1}{x}}$ (consider for all real values of $a$ ).
## Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ a \neq-2 .\end{array}\right.$

From the condition we have

$$
2^{\frac{a+3}{a+2}} \cdot 2^{\frac{5}{x(a+2)}}=2^{\frac{2}{x}} \Leftrightarrow 2^{\frac{a+3}{a+2}+\frac{5}{x(a+2)}}=2^{\frac{2}{x}} \Leftrightarrow \frac{a+3}{a+2}+\frac{5}{x(a+2)}=\frac{2}{x} \Rightarrow
$$

$\Rightarrow x=\frac{2 a-1}{a+3}$,

where $a \neq-3$, taking into account the domain of definition $x \neq 0, a \neq \frac{1}{2}$.

Answer: $\frac{2 a-1}{a+3}$ for $a \neq-2, a \neq-3$ and $a \neq \frac{1}{2}$; no solutions for $a=-2$, $a=-3$ and $a=\frac{1}{2}$.

Solve the systems of equations (7.259 - 7.294):