7.293. $\left\{\begin{array}{l}(x+y)^{x}=(x-y)^{y}, \\ \log _{2} x-\log _{2} y=1\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
(x+y)^{x}=(x-y)^{y}, \\
\log _{2} x-\log _{2} y=1
\end{array}\right.
\]
## Solution.

Domain of definition: $\left\{\begin{array}{l}x>0, \\ y>0, \\ x \neq \pm y .\end{array}\right.$

From the second equation of the system, we have $\log _{2} \frac{x}{y}=1$, from which $\frac{x}{y}=2$, $x=2 y$. Then from the first equation of the system, we get $(3 y)^{2 y}=y^{y}$, $\left(9 y^{2}\right)^{y}=y^{y}$, from which $9 y^{2}=y, y=\frac{1}{9}, x=2 \cdot \frac{1}{9}=\frac{2}{9}$.

Answer: $\left(\frac{2}{9} ; \frac{1}{9}\right)$.