7.077. $17 \cdot 2^{\sqrt{x^{2}-8 x}}-8=2 \cdot 4^{\sqrt{x^{2}-8 x}}$.
Solution.

Domain of definition: $x^{2}-8 x \geq 0, x \in(-\infty ; 0] \cup[8 ;+\infty)$

We have $2 \cdot 2^{2 \sqrt{x^{2}-8 x}}-17 \cdot 2^{\sqrt{x^{2}-8 x}}+8=0$. Solving this equation as a quadratic equation in terms of $2^{\sqrt{x^{2}-8 x}}$, we get $2^{\sqrt{x^{2}-8 x}}=2^{-1}$, from which $\sqrt{x^{2}-8}= -1, \varnothing$; or $2^{\sqrt{x^{2}-8 x}}=8$, from which $\sqrt{x^{2}-8 x}=3, x^{2}-8 x=9$, $x^{2}-8 x-9=0, x_{1}=-1, x_{2}=9$.

Answer: $-1 ; 9$