smc

Note that $$\begin{array}{rl}{x}^{2^{\sqrt{2}}}& ={\sqrt{2}}^{2^x}\\ 2^{\sqrt{2}}{\log }_{2}x& =2^x{\log }_2\sqrt{2}.\end{array}$$ (At this point we see by inspection that $x=\sqrt{2}$ is a solution.) We simplify the RHS, then take the base-$2$ logarithm for both sides: $$\begin{array}{rl}{2}^{\sqrt{2}}{\log }_{2}x& =2^{x-1}\\ {\log }_2\left(2^{\sqrt{2}}{\log }_{2}x\right)& =x-1\\ \sqrt{2}+{\log }_2{\log }_{2}x& =x-1\\ {\log }_2{\log }_{2}x& =x-1-\sqrt{2}.\end{array}$$ The RHS is a line; the LHS is a concave curve that looks like a logarithm and has $x$ intercept at $(2,0).$ There are at most two solutions, one of which is $\sqrt{2}.$ But note that at $x=2,$ we have ${\log }_2{\log }_{2}2=0>2-1-\sqrt{2},$ meaning that the log log curve is above the line, so it must intersect the line again at a point $x>2.$ Now we check $x=4$ and see that ${\log }_2{\log }_{2}4=1<4-1-\sqrt{2},$ which means at $x=4$ the line is already above the log log curve. Thus, the second solution lies in the interval $(2,4).$ The answer is $\boxed{2\le S<6}.$