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) $(2,4]$
(C)
3. 【Analysis and Solution】The original inequality is equivalent to $\left\{\begin{array}{l}\sqrt{\log _{2} x-1}-\frac{3}{2} \log _{2} x+\frac{3}{2}+\frac{1}{2}>0 . \\ \log _{2} x-1,0\end{array}\right.$.
Let $\sqrt{\log _{2} x-1}=t$, then we have $\left\{\begin{array}{l}t-\frac{3}{2} t^{2}+\frac{1}{2}>0 \\ t \geqslant 0\end{array} \quad\right.$ Solving this gives $0 \leqslant t<1$,
which means $0 \leqslant \log _{2} x-1<1, \therefore 2 \leqslant x<4$. Therefore, the correct choice is C.