Question

The complete solution set of the inequality ${\log }_5(x^2-2)<{\log }_5(\frac{3}{2}|x|-1)$ is

• A. $(-\infty ,-\frac{2}{3})\cup (\frac{2}{3},\infty )$
• B. $(-2,-\frac{2}{3})\cup (\frac{2}{3},2)$
• C. $(-2,-\sqrt{2})\cup (\sqrt{2},2)$
• D. $(-2,-\sqrt{2})\cup (\frac{2}{3},2)$

Answer 1

The correct option is C $(-2,-\sqrt{2})\cup (\sqrt{2},2)$

${\log }_5(x^2-2)<{\log }_5(\frac{3}{2}|x|-1)$ is defined if
$x^2-2>0$ and $\frac{3}{2}|x|-1>0$
$\Rightarrow x\in (-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )\cdots \left(1\right)$
and $x\in (-\infty ,-\frac{2}{3})\cup (\frac{2}{3},\infty )\cdots \left(2\right)$

Now, ${\log }_5(x^2-2)<{\log }_5(\frac{3}{2}|x|-1)$
$\Rightarrow x^2-2<\frac{3}{2}|x|-1\Rightarrow x^2-\frac{3}{2}|x|-1<0\Rightarrow 2x^2-3|x|-2<0$
$\Rightarrow 2|x{|}^2-3|x|-2<0(\because |x{|}^2=x^2,\forall x\in R)$
$\Rightarrow (2|x|+1)(|x|-2)<0$
$\Rightarrow \frac{-1}{2}<|x|<2$
But $|x|\ge 0$
So, $0\le |x|<2$
$\Rightarrow x\in (-2,2)\cdots \left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right),$ we have
$x\in (-2,-\sqrt{2})\cup (\sqrt{2},2)$