3. If $x_{1}$ satisfies $2 x+2^{x}=5, x_{2}$ satisfies $2 x+2 \log _{2}(x-1)=5$, then $x_{1}+x_{2}=$
3. $\frac{7}{2}$

Analysis: From the given, we have: $2^{x-1}=\frac{5}{2}-x, \log _{2}(x-1)=\frac{5}{2}-x$. Plot $y=2^{x-1}, y=\log _{2}(x-1), y=\frac{5}{2}-x$ on the same coordinate system. It can be seen that the entire graph is symmetric about $y=x-1$. Since $x-1=\frac{5}{2}-x \Leftrightarrow 2 x=\frac{7}{2}$, thus $x_{1}+x_{2}=\frac{7}{2}$;