7.272. $\left\{\begin{array}{l}3^{1+2 \log _{3}(y-x)}=48, \\ 2 \log _{5}(2 y-x-12)-\log _{5}(y-x)=\log _{5}(y+x) .\end{array}\right.$
Solution.

Domain of definition: $\left\{\begin{array}{l}y-x>0 \\ y+x>0 \\ 2 y-x-12>0 .\end{array}\right.$

Rewrite the system of equations as

$$
\begin{aligned}
& \left\{\begin{array}{l}
3 \cdot 3^{\log _{3}(y-x)^{2}}=48, \\
\log _{5}(2 y-x-12)^{2}-\log _{5}(y-x)=\log _{5}(y+x)
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
(y-x)^{2}=16 \\
\frac{(2 y-x-12)^{2}}{y-x}=y+x .
\end{array}\right.
\end{aligned}
$$

From the first equation of the system $y-x=4, y=x+4$. From the second equation of the system we get $(2(x+4)-x-12)^{2}=(x+4)^{2}-x^{2}, x^{2}-16 x=0$, from which $x_{1}=0, x_{2}=16$. Then $y_{1}=4, y_{2}=20$.

$$
\left\{\begin{array}{l}
x_{1}=0, \\
y_{1}=4
\end{array}\right. \text { does not satisfy the domain of definition. }
$$

Answer: $(16 ; 20)$.