7.177. $\frac{\log _{2}\left(9-2^{x}\right)}{3-x}=1$.

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7.177. $\frac{\log _{2}\left(9-2^{x}\right)}{3-x}=1$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}9-2^{x}>0, \\ 3-x \neq 0,\end{array} \Leftrightarrow 3 \neq x<\log _{2} 9\right.$.

## From the condition

$$
\log _{2}\left(9-2^{x}\right)=3-x \Leftrightarrow 9-2^{x}=2^{3-x} \Leftrightarrow 2^{2 x}-9 \cdot 2^{x}+8=0
$$

Solving it as a quadratic equation in terms of $2^{x}$, we find $\left(2^{x}\right)=1$, hence $x_{1}=0$ or $\left(2^{x}\right)_{2}=8$, hence $x_{2}=3 ; x_{2}=3$ does not satisfy the domain of definition.

Answer: 0.