Question

Solve the following inequality:
${\log }_{\frac{x+3}{x-3}}4<2\left({\log }_{1/2}\right(x-3)-{\log }_{\frac{\sqrt{2}}{2}}\sqrt{x+3})$

Answer 1

${\log }_{\left(\frac{x+3}{x-3}\right)}4<2[{\log }_{1/2}(x-3)-{\log }_{\left(\frac{\sqrt{2}}{2}\right)}\sqrt{x+3}]$

${\log }_{\left(\frac{x+3}{x-3}\right)}4<2[-{\log }_2(x-3)+\frac{2}{2}{\log }_2(x+3)]$

${\log }_{\left(\frac{x+3}{x-3}\right)}4<2{\log }_2\left(\frac{x+3}{x-3}\right)$

$2{\log }_{\left(\frac{x+3}{x-3}\right)}2<2{\log }_2\left(\frac{x+3}{x-3}\right)$

$\frac{\log 2}{\log \left(\frac{x+3}{x-3}\right)}<\log \left(\frac{x+3}{x-3}\right)\frac{1}{\left(\log 2\right)}$

${\left(\log 2\right)}^2\log \left(\frac{x+3}{x-3}\right)<{\left[\log \left(\frac{x+3}{x-3}\right)\right]}^3$

$\log \left(\frac{x+3}{x-3}\right)[{\left[\log \left(\frac{x+3}{x-3}\right)\right]}^2-{\left(\log 2\right)}^2]>0$

$\log \left(\frac{x+3}{x-3}\right)[\log \left(\frac{x+3}{x-3}\right)+\log 2][\log \left(\frac{x+3}{x-3}\right)-\log 2]>0$

$\log \left(\frac{x+3}{x-3}\right)ϵ(-\log 2,0)\bigcup (\log 2,\infty )$

$\frac{1}{2}<\frac{x+3}{x-3}<12<\frac{x+3}{x-3}<\infty$

$\frac{x+3}{x-3}>\frac{1}{2}\frac{x+3}{x-3}<1$

$2(x+3)(x-3)>{(x-3)}^2(x+3)(x-3)<{(x-3)}^2$

$2x^2-18>x^2+9-6x{x}^2-9<x^2+9-6x$

$(x^2+6x-27)>06x<18$

$(x+9)(x-3)>0x<3$

$xϵ(-\infty ,-9)\bigcup (3,\infty )xϵ(-\infty ,3)$

$\frac{x+3}{(x-3)}>2\frac{x+3}{(x-3)}<\infty$$(x+3)(x-3)>2{(x-3)}^2(x+3)(x-3)<\infty$

$x^2-9>2x^2+18-12xx+3$

$x^2+27-12x<0xϵR-\left\{3\right\}$

$(x-9)(x-3)<0$

$xϵ(3,9)$

Taking intersection we get

$xϵ(-\infty ,-9)\bigcup (3,9)$

Also, $\frac{x+3}{(x-3)}>0(x+3)(x-3)>0$

$xϵ(-\infty ,-3)\bigcup (3,\infty )$

and

$\left(\right)>0\sqrt{}\ge 0$

$x>3x+3\ge 0$

$xϵ(3,\infty )x\ge -3$

$xϵ(-3,\infty )$

Taking intersection we get,

$xϵ(3,9)$