Question

The solution set of ${\log }_3(x^2-2)<{\log }_3(\frac{3}{2}|x|-1)$ contains

• A. $(-2,-\sqrt{2})$
• B. $(-2,\sqrt{2})$
• C. $(\sqrt{2},2)$
• D. $(-\sqrt{2},2)$

Answer 1

Answer: (A), (C)

$(\sqrt{2},2)$

We must have
$x^2-2>0,\frac{3}{2}|x|-1>0\text{and}{x}^2-2<\frac{3}{2}|x|-1\Rightarrow x^2>2,|x|>\frac{2}{3}\text{and}|x{|}^2-\frac{3}{2}|x|+\frac{9}{16}<\frac{25}{16}\Rightarrow |x|>\sqrt{2}\text{and}{(|x|-\frac{3}{4})}^2<{\left(\frac{5}{4}\right)}^2\Rightarrow |x|>\sqrt{2}\text{and}-\frac{5}{4}<(|x|-\frac{3}{4})<\frac{5}{4}\Rightarrow |x|>\sqrt{2}\text{and}|x|<2\Rightarrow x\in (-2,-\sqrt{2})\cup (\sqrt{2},2)$