Question

Solve the following inequality:
$\frac{1}{{\log }_3(x+1)}<\frac{1}{2{\log }_3\sqrt{x^3+6x+9}}$

Answer 1

$\frac{1}{{\log }_3(x+1)}<\frac{1}{2{\log }_3\sqrt{(x^2+6x+9)}}$

$\frac{1}{{\log }_3(x+1)}<\frac{1}{2{\log }_3(x+3)}$

$(x+1)>0$ & $(x+3)>0x\ne 0,-2$

$x>-1x>-3$

For all $x>-1,{\log }_3(x+3)>0$

For $xϵ(-1,0)$

$\frac{1}{{\log }_3(x+1)}<\frac{1}{2{\log }_3(x+3)}$

${\log }_3(x+1)<2{\log }_3(x+3)$

$(x+1)<{(x+3)}^2$

$(x+1)<x^2+6x+9$

$x^2+5x+8>0$

$xϵR$

Hence, $xϵ(-1,0)$

For $xϵ(-1,\infty )$

$\frac{1}{{\log }_3(x+1)}<\frac{1}{2{\log }_3(x+3)}$

$2{\log }_3(x+3)<{\log }_3(x+1)$

${(x+3)}^2<(x+1)$

$x^2+6x+9-x-1<0$

$x^2+5x+8<0$

$xϵ\varphi$

Hence $xϵ\varphi$

Taking intersection and union we get $xϵ(-1,0)$