Question

Solve the following inequality:
${\log }_{0.5}\left({\log }_6\frac{x^2+x}{x+4}\right)<0$

Answer 1

${\log }_2\left({\log }_6\left(\frac{x^2+x}{x+4}\right)\right)>0$

$0<{\log }_6\left(\frac{x^2+x}{x+4}\right)and{\log }_6\left(\frac{x^2+x}{x+4}\right)>1$

$\frac{x^2+x}{x+4}>6$

$\frac{x^2+x-6x-24}{x+4}>0$

$\frac{x^2-5x-24}{x+4}>0$

$\frac{(x-8)(x+3)}{x+4}>0$

$Therefore,x\epsilon (-4,-3)\cup (8,\infty )$

$and\frac{x^2+x}{x+4}>0$

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

$Therefore,x\epsilon (-4,-1)\cup (0,\infty )$

There fore we can say that,

$x\epsilon (-4,-3)\cup (8,\infty )$