Example 1 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. $(2,4]$
Solution: Choose C. Reason: The original inequality is equivalent to $\sqrt{\log _{2} x-1}-\frac{3}{2} \log _{2} x+2>0$.
Let $\sqrt{\log _{2} x-1}=t$. Then we have $\left\{\begin{array}{l}t-\frac{3}{2}\left(t^{2}+1\right)+2>0 \\ t \geqslant 0\end{array}\right.$, solving this yields $0 \leqslant t<1$, hence $0 \leqslant \log _{2} x-1<1$, so $2 \leqslant x<4$, therefore choose C.