7.267. $\left\{\begin{array}{l}y-\log _{3} x=1 \\ x^{y}=3^{12}\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
y - \log_{3} x = 1 \\
x^{y} = 3^{12}
\end{array}\right.
\]
## Solution.

Domain of definition: $0<x \neq 1$.

By logarithmizing the second equation with base 3, we get

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ y - \log _ { 3 } x = 1 , } \\
{ \log _ { 3 } x ^ { y } = \log _ { 3 } 3 ^ { 12 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y=1+\log _{3} x, \\
y \log _{3} x=12
\end{array} \Rightarrow\left(1+\log _{3} x\right) \log _{3} x=12\right.\right. \\
& \log _{3}^{2} x+\log _{3} x-12=0 \Rightarrow \log _{3} x=-4, x_{1}=\frac{1}{81} \text { or } \log _{3} x=3, x_{2}=27
\end{aligned}
$$

Then $y_{1}=-3, y_{2}=4$.

Answer: $\left(\frac{1}{81} ;-3\right),(27 ; 4)$.