Question

If ${\log }_{x-2}6+{\log }_{x+2}6>{\log }_{x-2}6\cdot {\log }_{x+2}6$, then $x\in$

• A. $(-\infty ,3)$
• B. $(-1,3)$
• C. $(\sqrt{10},\infty )$
• D. $(2,3)$

Answer 1

Answer: (C), (D)

$(2,3)$

For the inequality to be defined,
$x-2>0,x+2>0,x-2\ne 1,x+2\ne 1$
$\Rightarrow x>2,x>-2,x\ne 3,x\ne -1$
$\Rightarrow x\in (2,\infty )-\left\{3\right\}\cdots \left(1\right)$

Now, ${\log }_{x-2}6+{\log }_{x+2}6>{\log }_{x-2}6\cdot {\log }_{x+2}6$
$\Rightarrow \frac{1}{{\log }_6(x-2)}+\frac{1}{{\log }_6(x+2)}>\frac{1}{{\log }_6(x-2){\log }_6(x+2)}$

$\Rightarrow \frac{{\log }_6(x+2)(x-2)}{{\log }_6(x-2){\log }_6(x+2)}-\frac{1}{{\log }_6(x-2){\log }_6(x+2)}>0$

$\Rightarrow \frac{{\log }_6(x^2-4)-1}{{\log }_6(x-2){\log }_6(x+2)}>0$

Consider ${\log }_6(x^2-4)-1=0$
$\Rightarrow x^2-4=6$
$\Rightarrow x=\sqrt{10}\left[\text{From}\right(1\left)\right]$

${\log }_6(x-2)=0$
$\Rightarrow x-2=1\Rightarrow x=3$

${\log }_6(x+2)=0$
$\Rightarrow x+2=1\Rightarrow x=-1\text{Not possible}$


$\therefore x\in (-\infty ,3)\cup (\sqrt{10},\infty )\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
$x\in (2,3)\cup (\sqrt{10},\infty )$