7.281. $\left\{\begin{array}{l}y \cdot x^{\log _{y} x}=x^{2.5} \\ \log _{3} y \cdot \log _{y}(y-2 x)=1 .\end{array}\right.$
Solution.

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

Logarithmize the first equation of the system with base 3, we have

$$
\begin{aligned}
& \log _{3}\left(y \cdot x^{\log _{y} x}\right)=\log _{3} x^{2.5} \Leftrightarrow \log _{3} y+\log _{3} x^{\log _{y} x}=2.5 \log _{3} x \Leftrightarrow \\
& \Leftrightarrow \log _{3} y+\log _{y} x \log _{3} x=2.5 \log _{3} x
\end{aligned}
$$

Switch to base 3 in the first and second equations of the system. We have

$$
\begin{aligned}
& \left\{\begin{array}{l}
\log _{3} y+\frac{\log _{3} x}{\log _{3} y} \cdot \log _{3} x=2.5 \log _{3} x, \\
\frac{\log _{3} y \cdot \log _{3}(y-2 x)}{\log _{3} y}=1
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
\log _{3}^{2} y+\log _{3}^{2} x=2.5 \log _{3} x \log _{3} y, \\
\log _{3}(y-2 x)=1
\end{array} \Rightarrow\right. \\
& \Rightarrow \log _{3}^{2} y-2.5 \log _{3} x \cdot \log _{3} y+\log _{3}^{2} x=0 .
\end{aligned}
$$

Solve it as a quadratic equation in $\log _{3} y$, and find $\log _{3} y=\frac{1}{2} \log _{3} x$, hence $y_{1}=\sqrt{x}$, or $\log _{3} y=2 \log _{3} x$, hence $y_{2}=x^{2}$. From the second equation of this system, we get $y-2 x=3, x=\frac{y-3}{2}$. Then the given system of equations is equivalent to the following:

1) $\left\{\begin{array}{l}y=\sqrt{x}, \\ x=\frac{y-3}{2} ;\end{array}\right.$ has no solutions;
2) $\left\{\begin{array}{l}y=x^{2}, \\ x=\frac{y-3}{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}x=3, \\ y=9,\end{array}\right.\right.$ considering the domain of definition.

Answer: $(3 ; 9)$.