Answer: D

Solution:

Taking $\log_2,$ the equation becomes $2^{\sqrt2}\log_2 x=2^{x-1}.$ Substituting $x=\sqrt2$ gives $2^{\sqrt2}\cdot\tfrac12=2^{\sqrt2-1},$ which holds, so $x=\sqrt2$ is a solution.

Let $f(x)=2^{x-1}-2^{\sqrt2}\log_2 x.$ Then $f(1)\gt 0,$ $f(\sqrt2)=0,$ $f(2)\lt 0,$ and $f(4)\gt 0,$ so there is a second root $x_0$ between $2$ and $4.$

To prove there are no others, divide the equation by $2^x$ and consider $h(x)=\ln x/2^x$ for $x\gt1.$ Its derivative has the sign of $1/x-(\ln2)(\ln x),$ a strictly decreasing expression. Thus $h$ increases once and then decreases, so a horizontal line meets its graph at most twice. The two roots already found are all the solutions. Since $2\lt x_0\lt4,$ we have $2\lt S=\sqrt2+x_0\lt6.$

Thus, the correct answer is D.