Question

Solve the following inequality:
${\log }_{\frac{1}{3}}(x+1)>{\log }_3(2-x)$

Answer 1

${log}_{\left(1/3\right)}(x+1)>{log}_3(2-x)$

$-{log}_3(x+1)>{log}_3(2-x)$ Also,

${(x+1)}^{-1}>(2-x)$ (x+1) > 0 & (2-x) > 0

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

$(x+1)>2+3\times -x^3$ $xϵ(-1,\infty )\&xϵ(-\infty ,2)$

$x^3-2x-1>0$

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

$x^2(x+1)-{(x+1)}^2>0$

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

$xϵ(-1,\frac{1-\sqrt{5}}{2})\cup (\frac{1+\sqrt{5}}{2},\infty )$

Taking intersection we get $xϵ(-1,\frac{1-\sqrt{5}}{2})\cup (\frac{1+\sqrt{5}}{2},2)$