Example 1 (2004 National High School Competition Question) The solution set of the inequality $\sqrt{\log _{2} x-1}+\frac{1}{2} \log _{\frac{1}{2}} x^{3}+2>0$ is $(\quad$.
A. $[2,3)$
B. $(2,3]$
C. $[2,4)$
D. $(2,4]$
Solution: Choose C. Reason: The original inequality can be reduced to the equivalent system of inequalities $\left\{\begin{array}{l}\sqrt{\log _{2} x-1}-\frac{3}{2} \log _{2} x+\frac{3}{2}+\frac{1}{2}>0, \\ \log _{2} x-1 \geqslant 0 .\end{array}\right.$ Let $\sqrt{\log _{2} x-1}=t$, then we have $\left\{\begin{array}{l}t-\frac{3}{2} t^{2}+\frac{1}{2}>0, \\ t \geqslant 0 .\end{array}\right.$ Solving this, we get $0 \leqslant t<1$, which means $0 \leqslant \log _{2} x-1<1$. Therefore, $2 \leqslant x<4$.