smc

Let $A(6+m,2+n)$ and $B(6-m,2-n)$, since $(6,2)$ is their midpoint. Thus, we must find $2m$. We find two equations due to $A,B$ both lying on the function $y={\log }_{2}x$. The two equations are then ${\log }_2(6+m)=2+n$ and ${\log }_2(6-m)=2-n$. Now add these two equations to obtain ${\log }_2(6+m)+{\log }_2(6-m)=4$. By logarithm rules, we get ${\log }_2((6+m)(6-m))=4$. By raising 2 to the power of both sides, we obtain $(6+m)(6-m)=16$. We then get $$36-m^2=16\to m^2=20\to m=2\sqrt{5}$$. Since we're looking for $2m$, we obtain $(2)(2\sqrt{5})=\boxed{4\sqrt{5}}$