Question

Solution set of the inequality $lo{g}_{0.8}\left(lo{g}_6\frac{x^2+x}{x+4}\right)<0$

• A. (-4,-3)
• B. $(-3,4)\cup (8,\infty )$
  
• C. $(-3,\infty )$
  
• D. $(-4,-3)\cup (8,\infty )$

Answer 1

The correct option is D

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

Since, the base of $lo{g}_{0.8}\left(lo{g}_6\frac{x^2+x}{x+4}\right)<0$
is less than 1,
$lo{g}_6\left(\frac{x^2+x}{x+4}\right)>1\Rightarrow \frac{x^2+x}{x+4}>6\Rightarrow \frac{x^2+x}{x+4}-6>0$
$\Rightarrow \frac{x^2+x-6x-24}{(x+4)}>0\Rightarrow \frac{(x-8)(x+3)}{(x+4)}>0,x\ne -4.$
$\Rightarrow -4<x<-3$ or $x>8$.
Thus, required solution set is $(-4,-3)\cup (8,\infty )$