7.147. $\left\{\begin{array}{l}\log _{x y}(x-y)=1, \\ \log _{x y}(x+y)=0 .\end{array}\right.$
## Solution.

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

## We have

$\left\{\begin{array}{l}x-y=x y, \\ x+y=1\end{array} \Rightarrow y=1-x, \quad x-(1-x)-x(1-x)=0, x^{2}+x-1=0\right.$, from which $x_{1}=\frac{-1-\sqrt{5}}{2}, x_{2}=\frac{-1+\sqrt{5}}{2}, y_{1}=\frac{3+\sqrt{5}}{2}, y_{2}=\frac{3-\sqrt{5}}{2}$.

Then, taking into account the domain of definition, we have $x=\frac{-1+\sqrt{5}}{2}, y=\frac{3-\sqrt{5}}{2}$.

Answer: $\left(\frac{-1+\sqrt{5}}{2} ; \frac{3-\sqrt{5}}{2}\right)$