Question

Solve the following inequality:
${\log }_{x^2+2x-3}\frac{|x+4|-|x|}{x-1}>0$

Answer 1

${\log }_{(x^2+2x-3)}\left\{\frac{(x+4)-\left(x\right)}{(x-1)}\right\}>0$

For $\log (x^2+2x-3)$

$x^2+2x-3>0$

$(x+2)(x-1)>0$

$xϵ(-\infty ,-3)\bigcup (1,\infty )$

For $\log \left\{\frac{(x+4)-\left(x\right)}{(x-1)}\right\}$

$\frac{(x+4)-\left(x\right)}{(x-1)}>0$

$(x-1)(|x+4|-\left|x\right|)>0$

$xϵ(1,\infty )(x-1)(x+4-x)>0$

$(x-1)\left(4\right)>0x>1xϵ(1,\infty )$

For $xϵ[-4,-3)(x-1)(x+4+x)$

$2(x-1)(x+2)>0$

$xϵ(-\infty ,-2)\bigcup (1,\infty )$

Hence $xϵ[-4,-3)$

For $xϵ(-\infty ,-4)$

$(x-1)(-x-4+x)>0$

$(x-1)<0x<1$

Hence, $xϵ(-\infty ,-4)$

Taking intersection we get $xϵ(-\infty ,-3)\bigcup (1,\infty )$

$0<x^2+2x-3<1,\{\log (x^2+2x-3)<0\}$

That is, $xϵ(-\sqrt{5}-1,-3)\bigcup (1,\sqrt{5}-1)$

For $xϵ(-\sqrt{5}-1,-3)$

${\log }_{(x^2+2x-3)}\left\{\frac{(x+4)-\left(x\right)}{(x-1)}\right\}>0$

$\frac{(x+4+x)}{(x-1)}>1$

$2(x-1)(x+2)>{(x-1)}^2$

$x^2+4x-5<0$

$(x+5)(x-1)<0$

$xϵ(-5,1)$

Hence, $xϵ(-\sqrt{5}-1,-3)$

For $xϵ(1,\sqrt{5}-1)$

${\log }_{(x^2+2x-3)}\left\{\frac{(x+4)-\left(x\right)}{(x-1)}\right\}>0$

$\frac{(x+4)-x}{(x-1)}<1$

$(x-1)4<{(x-1)}^2$

$(x-1)(x-5)>0$

$xϵ(-\infty ,1)\bigcup (5,\infty )$

Hence $xϵ\varphi$

Taking intersection we get $xϵ(-\sqrt{5}-1,-3)$

For $xϵ(-\infty ,-\sqrt{5}-1)\bigcup (\sqrt{5}-1,\infty )$

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

For $xϵ(-4,-\sqrt{5}-1)$

$\frac{x+4+x}{(x-1)}>1$

$2(x-1)(x+2)>{(x-1)}^2$

$(x^2+4x-5)>0$

$xϵ(-\infty ,-5)\bigcup (1,\infty )$

Hence, $xϵ\varphi$

For $xϵ(-\infty ,-4)$

$-\frac{(x+4)+x}{(x-1)}>1$

$-4(x-1)>{(x-1)}^2$

$(x-1)(x+3)<0$

$xϵ(-3,1)$

Hence, $xϵ\varphi$

For $xϵ(\sqrt{5}-1,\infty )$

$\frac{x+4-x}{(x-1)}>1$

$4(x-1)>{(x-1)}^2$

$(x-1)(x-5)<0$

$xϵ(1,5)$

Hence $xϵ(\sqrt{5}-1,3)$

Taking intersection we get $xϵ(\sqrt{5}-1,5)$

Taking union we get $xϵ(-\sqrt{5}-1,-3)\bigcup (\sqrt{5}-1,5)$