7.170. $\log _{2}(2-x)-\log _{2}(2-\sqrt{x})=\log _{2} \sqrt{2-x}-0.5$.
Solution.

Domain of Definition (DOD): $\left\{\begin{array}{l}2-x>0, \\ 2-\sqrt{x}>0,\end{array}, 0 \leq x<2\right.$.

From the condition, we have $\log _{2} \frac{2-x}{2-\sqrt{x}}=\log _{2} \frac{\sqrt{2-x}}{\sqrt{2}} \Leftrightarrow \frac{2-x}{2-\sqrt{x}}=\frac{\sqrt{2-x}}{\sqrt{2}} \Leftrightarrow$ $\Leftrightarrow \frac{2-x}{2-\sqrt{x}}-\frac{\sqrt{2-x}}{\sqrt{2}}=0 \Leftrightarrow \sqrt{2-x}\left(\frac{\sqrt{2-x}}{2-\sqrt{x}}-\frac{1}{\sqrt{2}}\right)=0$, from which $\frac{\sqrt{2-x}}{2-\sqrt{x}}-\frac{1}{\sqrt{2}}=0$, $\sqrt{4-2 x}=2-\sqrt{x}, 4-2 x=4-4 \sqrt{x}+x, 3 x-4 \sqrt{x}=0, \sqrt{x}(3 \sqrt{x}-4)=0$. Thus, $x_{1}=0, x_{2}=\frac{16}{9}$.

Answer: $0 ; \frac{16}{9}$.