21. Let $S$ be the sum of all positive real roots satisfying the equation $x^{2 \sqrt{2}}=\sqrt{2}^{2^{x}}$. Then the correct option among the following is ( ).
(A) $S<\sqrt{2}$
(B) $S=\sqrt{2}$
(C) $\sqrt{2}<S<2$
(D) $2 \leqslant S<6$
(E) $S \geqslant 6$
21. D.
$$
\begin{array}{l}
\text { Given } x^{2 \sqrt{2}}=\sqrt{2}^{2 x} \\
\Rightarrow 2^{\sqrt{2}} \log _{2} x=2^{x} \log _{2} \sqrt{2}=2^{x-1} \\
\Rightarrow \sqrt{2}+\log _{2}\left(\log _{2} x\right)=x-1 \\
\Rightarrow \log _{2}\left(\log _{2} x\right)=x-1-\sqrt{2} \text {. } \\
\end{array}
$$

Notice that, $f(x)=\log _{2}\left(\log _{2} x\right)$ is a convex function on $x \in (0,+\infty)$, and as shown in Figure 14, it is easy to see that
$$
\begin{array}{l}
\log _{2}\left(\log _{2} 2\right)=0>2-1-\sqrt{2}, \\
\log _{2}\left(\log _{2} 3\right)4-1-\sqrt{2} .
\end{array}
$$

Thus, the positive real root of the equation $\log _{2}\left(\log _{2} x\right)=x-1-\sqrt{2}$ lies in the interval $(2,4)$.
Therefore, $S \in[2,6)$.