Question

Solve the following inequality:
${\log }_{1/2}{\log }_3\frac{x+1}{x-1}⩾0$

Answer 1

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

Here, $\frac{(x+1)}{(x-1)}>0(x+1)(x-1)>0$

$xϵ(-\infty ,-1)\bigcup (1,\infty )$

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

$-{\log }_2{\log }_3\left(\frac{x+1}{x-1}\right)\ge 0={\log }_2{\log }_3\left(\frac{x+1}{x-1}\right)\le 0$

${\log }_3\left(\frac{x+1}{x-1}\right)\le 1$

$\frac{(x+1)}{(x-1)}\le 3$

$(x-1)(x+1)\le 3{(x-1)}^2$

$x^2-1\le 3x^2-6x+3$

$2x^2-6x+4\ge 0$

$(x^2-3x+2)\ge 0$

$(x-2)(x-1)\ge 0$

$xϵ(-\infty ,1)\bigcup (2,\infty )$

Taking intersection we get $xϵ(-\infty ,1)\bigcup (2,\infty )$