3. The solution set of the inequality $\sqrt{\log _{2} x-1}+\frac{1}{2} \log _{\frac{1}{2}} x^{3}+2>0$ is ( ).
(A) $[2,3)$
(B) $(2,3]$
(C) $[2,4)$
(D) $(2,4]$
3.C.

The original inequality is equivalent to
$$
\left\{\begin{array}{l}
\sqrt{\log _{2} x-1}-\frac{3}{2} \log _{2} x+\frac{3}{2}+\frac{1}{2}>0 \\
\log _{2} x-1 \geqslant 0
\end{array}\right.
$$

Let $\sqrt{\log _{2} x-1}=t$, then we have
$$
\left\{\begin{array}{l}
t-\frac{3}{2} t^{2}+\frac{1}{2}>0, \\
t \geqslant 0
\end{array} \Rightarrow 0 \leqslant t<1 .\right.
$$

Therefore, $0 \leqslant \log _{2} x-1<1$, so, $2 \leqslant x<4$.