Question

The solution set of ${\log }_{1/2}{\log }_3\left(\frac{x+1}{x-1}\right)\ge 0$ is

• A. $x\in (-\infty ,-1)\cup (2,\infty )$
• B. $x\in (-\infty ,1)\cup [2,\infty )$
• C. $x\in (-\infty ,-1)$
• D. $x\in [2,\infty )$

Answer 1

The correct option is D $x\in [2,\infty )$

${\log }_{1/2}{\log }_3\left(\frac{x+1}{x-1}\right)\ge 0$
$\log$ is defined when
$\left(i\right)\left(\frac{x+1}{x-1}\right)>0$
$\Rightarrow x\in (-\infty ,-1)\cup (1,\infty )$
$\left(ii\right){\log }_3\left(\frac{x+1}{x-1}\right)>0$
$\Rightarrow \frac{x+1}{x-1}>1$
$\Rightarrow \frac{2}{x-1}>0$
$\Rightarrow x\in (1,\infty )$
From $\left(i\right)\cap \left(ii\right),$ we have
$\Rightarrow x\in (1,\infty )\dots \left(1\right)$

Now, ${\log }_{1/2}{\log }_3\left(\frac{x+1}{x-1}\right)\ge 0$
$\Rightarrow {\log }_3\left(\frac{x+1}{x-1}\right)\le 1$
$\Rightarrow \left(\frac{x+1}{x-1}\right)\le 3$
$\Rightarrow \frac{x+1}{x-1}-3\le 0$
$\Rightarrow \frac{x-2}{x-1}\ge 0$
$\Rightarrow x\in (-\infty ,1]\cup [2,\infty )\dots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we get
$x\in [2,\infty )$