$7.80 \quad\left\{\begin{array}{l}x^{x-2 y}=36, \\ 4(x-2 y)+\log _{6} x=9\end{array}\right.$

(find only integer solutions).
7.80 Here $x>0$. By taking the logarithm of the first equation to the base 6, we arrive at the system

$$
\left\{\begin{array} { l } 
{ ( x - 2 y ) \log _ { 6 } x = 2 , } \\
{ 4 ( x - 2 y ) + \log _ { 6 } x = 9 , }
\end{array} \text { or } \left\{\begin{array}{l}
4(x-2 y) \log _{6} x=8 \\
4(x-2 y)+\log _{6} x=9
\end{array}\right.\right.
$$

Thus, the values of the expressions $4(x-2 y)$ and $\log _{6} x$ can be considered as the roots of the quadratic equation $z^{2}-9 z+8=0$, from which $z_{1}=8, z_{2}=1$. As a result, we obtain two systems of equations

$$
\left\{\begin{array} { l } 
{ 4 ( x - 2 y ) = 8 , } \\
{ \log _ { 6 } x = 1 }
\end{array} \text { and } \left\{\begin{array}{l}
4(x-2 y)=1 \\
\log _{6} x=8
\end{array}\right.\right.
$$

The first of these has the solution $x=6, y=2$, while the solution to the second system does not satisfy the required condition, as the value of $y$ is not an integer.

Answer: $(6 ; 2)$.