Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.7 AMC 10/12 Find the answer China

The solution set of the inequality $\sqrt{\log_2 x - 1} + \frac{1}{2} \log_{\frac{1}{2}} x^3 + 2 > 0$ is:

Pick one

Solution

$\begin{cases} \sqrt{\log_2 x - 1} - \frac{3}{2} \log_2 x + \frac{3}{2} + \frac{1}{2} > 0, \\ \log_2 x - 1 \ge 0. \end{cases}$
Let $t = \sqrt{\log_2 x - 1}$ , we have
$\begin{cases} t - \frac{3}{2}t^2 + \frac{1}{2} > 0, \\ t \ge 0. \end{cases}$

The solution of the above inequalities is $0 \le t < 1$ , or $0 \le \log_2 x - 1 < 1$ , which implies that $2 \le x < 4$ . Answer: C.

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