7.292. $\left\{\begin{array}{l}\log _{x y} \frac{y}{x}-\log _{y}^{2} x=1, \\ \log _{2}(y-x)=1 .\end{array}\right.$
## Solution.

OD3: $\left\{\begin{array}{l}0x .\end{array}\right.$

In the first equation of the system, we transition to the base $y$:

$$
\frac{\log _{y} \frac{y}{x}}{\log _{y} x y}-\log _{y}^{2} x=1 \Leftrightarrow \frac{1-\log _{y} x}{1+\log _{y} x}-\log _{y}^{2} x-1=0 \Leftrightarrow
$$

$\Leftrightarrow \log _{y}^{3} x+\log _{y}^{2} x+2 \log _{y} x=0 \Leftrightarrow \log _{y} x\left(\log _{y}^{2} x+\log _{y} x+2\right)=0$,

from which $\log _{y} x=0, x=y^{0}=1, \log _{y}^{2} x+\log _{y} x+2 \neq 0$. From the second equation, we get $y-x=2$. Hence, $y=3$.

Answer: $(1 ; 3)$.