Question

Solve ${\log }_x(x^2-1)\le 0$ for domain of $x$ which has $(m,n]$, then find $m+n$

Answer 1

${\log }_x(x^2-1)\le 0$
From the above inequality, it follows that
$x>0,x\ne 1,x^2-1>0$
$x>0,x\ne 1,(x+1)(x-1)>0$
$\Rightarrow x>0,x\ne 1,x\in (-\infty ,-1)\cup (1,\infty )$
$\Rightarrow x\in (1,\infty )$
Now, we will solve the inequality for $x>1$
$\Rightarrow x^2-1\le 1$ ($\because {\log }_{a}f\left(x\right)\le 0$ is equivalent to $f\left(x\right)\le 1$ when $a>1$)
$\Rightarrow x^2-2\le 0$
$\Rightarrow (x-\sqrt{2})(x+\sqrt{2})\le 0$
$\Rightarrow x\in [-\sqrt{2},\sqrt{2}]$
But since, $x>1$
So, the solution set of the given inequality is $(1,\sqrt{2}]$
But, given domain is $(m,n]$

$\Rightarrow m=1,n=\sqrt{2}$

$\therefore m+n=1+\sqrt{2}$