7.083. $x^{\log _{4} x-2}=2^{3\left(\log _{4} x-1\right)}$.

7.083. $x^{\log _{4} x-2}=2^{3\left(\log _{4} x-1\right)}$.
Solution.

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

Taking the logarithm of both sides of the equation with base 4, we have

$$
\begin{aligned}
& \log _{4} x^{\log _{4} x-2}=\log _{4} 2^{3\left(\log _{4} x-1\right)},\left(\log _{4} x-2\right) \log _{4} x=3\left(\log _{4} x-1\right) \log _{4} 2, \\
& \log _{4}^{2} x-2 \log _{4} x=\frac{3}{2}\left(\log _{4} x-1\right), 2 \log _{4}^{2} x-7 \log _{4} x+3=0
\end{aligned}
$$

Solving this equation as a quadratic in $\log _{4} x$, we find $\left(\log _{4} x\right)_{1}=\frac{1}{2},\left(\log _{4} x\right)_{2}=3$. Therefore, $x_{1}=4^{\frac{1}{2}}=2, x_{2}=4^{3}=64$.

Answer: $2 ; 64$.