7.145. $\left\{\begin{array}{l}y=1+\log _{4} x, \\ x^{y}=4^{6}\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
y=1+\log _{4} x, \\
x^{y}=4^{6}
\end{array}\right.
\]
## Solution.

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

Taking the logarithm of the second equation of the system with base 4, we have $\log _{4} x^{y}=\log _{4} 4^{6}, y \log _{4} x=6$. Therefore,

$$
\left\{\begin{array}{l}
y=1+\log _{4} x, \\
y \log _{4} x=6
\end{array} \Rightarrow\left(1+\log _{4} x\right) \log _{4} x=6, \log _{4}^{2} x+\log _{4} x-6=0\right.
$$

Solving this equation as a quadratic equation in terms of $\log _{4} x$, we find $\left(\log _{4} x\right)_{1}=-3,\left(\log _{4} x\right)_{2}=2, x_{1}=\frac{1}{64}, x_{2}=16$. Then $y_{1}=-2, y_{2}=3$.

Answer: $\left(\frac{1}{64} ;-2\right),(16 ; 3)$.