Question

The inequality ${\log }_a(x^2-x-2)>{\log }_a(-x^2+2x+3)$ is known to be satisfied for $x=\frac{4}{9}$. Find all solutions of this inequality.

Answer 1

${\log }_a(x^2-x-2)>{\log }_a(-x^2+2x+3)$

${\log }_a(x^2-x-2)-{\log }_a(-x^2+2x+3)>0$

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

${\log }_a\left[\frac{(x+1)(x-2)}{-(x+1)(x-3)}\right]>0$

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

For $aϵ(0,1)$

$\frac{-(x-2)}{(x-3)}<1$

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

${(x-3)}^2+(x-2)(x-3)>0$

$(x-3)(x-3+x-2)>0$

$(x-3)(2x-5)>0$

$xϵ(-\infty ,\frac{5}{2})\bigcup (3,\infty )$

Hence, $xϵ(0,1)$

For $aϵ(1,\infty )$

$\frac{-(x-2)}{(x-3)}>1$

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

${(x-3)}^2+(x-3)(x-2)<0$

$(x-3)(x-3+x-2)<0$

$(x-3)(2x-5)<0$

$xϵ(\frac{5}{2},3)$

Hence, $xϵ(\frac{5}{2},3)$

Also $(x^2-x-2)>0-x^2+2x+3>0$

$(x-2)(x+1)>0x^2-2x-3<0$

$xϵ(-\infty ,-1)\bigcup (2,\infty )(x-3)(x+1)<0$

$xϵ(-1,3)$

$xϵ(2,3)$

Taking intersection here we get $xϵ(\frac{5}{2},3)$