Question

Find the solution set of $lo{g}_{1/4}\left(lo{g}_6\frac{x^2+x}{x+4}\right)>0$

• A. $(-3,\infty )$
• B. $(-4,3)\bigcup (8,\infty )$
• C. $(8,17)\bigcup (24,\infty )$
• D. $(-\infty ,-1)\bigcup (3,\infty )$

Answer 1

The correct option is B $(-4,3)\bigcup (8,\infty )$

$lo{g}_{1/4}\left(lo{g}_6\right(\frac{x^2+x}{x+4}\left)\right)>0$
$lo{g}_{1/4}\left(lo{g}_6\right(\frac{x^2+x}{x+4}\left)\right)>lo{g}_{1/4}\left(1\right)$
$lo{g}_6\left(\frac{x^2+x}{x+4}\right)>1$
$lo{g}_6\left(\frac{x^2+x}{x+4}\right)>lo{g}_6\left(6\right)$
$\frac{x^2+x}{x+4}>6$
$\frac{x^2+x}{x+4}-6>0$
$\frac{x^2+x-6x-24}{x+4}>0$
$\frac{x^2-5x-24}{x+4}>0$
$\frac{(x-8)(x+3)}{x+4}>0$
Therefore
$(x-8)(x+3)>0$
$x>8$ and $x>-3$ therefore $xϵ(8,\infty )$ ..(i)
$x+4\ne 0$ therefore $x\ne -4$ ...(ii)
And $(8-x)(-x-3)>0$
$8>x$ and$x+3<0$
$x<-3$ ...(iii)
Therefore from i ii and ii
$xϵ(-4,-3)\cup (8,\infty )$