7.110. $\log _{2}\left(9-2^{x}\right)=10^{\lg (3-x)}$.

7.110. $\log _{2}\left(9-2^{x}\right)=10^{\lg (3-x)}$.
## Solution.

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

We have $\log _{2}\left(9-2^{x}\right)=3-x, 9-2^{x}=2^{3-x}, 2^{2 x}-9 \cdot 2^{x}+8=0$. Solving this equation as a quadratic in terms of $2^{x}$, we get $\left(2^{x}\right)=1$ or $\left(2^{x}\right)_{2}=8$, hence $x_{1}=0, x_{2}=3 ; x_{2}=3$ does not satisfy the domain of definition.

Answer: 0.