Question

If ${\log }_{x-3}(x^2-10x+24)\ge {\log }_{x-3}(x^2-6),$ then

• A. $x\in (\sqrt{6},6)$
• B. $x\in (\sqrt{6},4)$
• C. $x\in (\sqrt{6},3)$
• D. $x\in (3,4)$

Answer 1

The correct option is D $x\in (3,4)$

${\log }_{x-3}(x^2-10x+24)\ge {\log }_{x-3}(x^2-6)$
$\log$ function is defined when
$\left(i\right)x^2-10x+24>0$
$\left(ii\right)x^2-6>0$
$\left(iii\right)x-3>0\text{and}x-3\ne 1$

$\left(i\right)\Rightarrow (x-4)(x-6)>0$
$\Rightarrow x\in (-\infty ,4)\cup (6,\infty )\cdots \left(1\right)$

$\left(ii\right)x^2-6>0$
$\Rightarrow x\in (-\infty ,-\sqrt{6})\cup (\sqrt{6},\infty )\cdots \left(2\right)$

$\left(iii\right)\Rightarrow x\in (3,\infty )-\left\{4\right\}\cdots \left(3\right)$

Now, $\left(1\right)\cap \left(2\right)\cap \left(3\right)$ will give
$x\in \cup (3,4)\cup (6,\infty )\cdots \left(4\right)$

$\text{Case}1:x-3>1\Rightarrow x>4$
${\log }_{x-3}(x^2-10x+24)\ge {\log }_{x-3}(x^2-6)$
$\Rightarrow (x^2-10x+24)\ge (x^2-6)$
$\Rightarrow x\le 3$
which is not possible as $x>4$

$\text{Case}2:0<x-3<1\Rightarrow x\in (3,4)\cdots \left(5\right)$
${\log }_{x-3}(x^2-10x+24)\ge {\log }_{x-3}(x^2-6)$
$\Rightarrow (x^2-10x+24)\le (x^2-6)$
$\Rightarrow x\ge 3\cdots \left(6\right)$
From $\left(5\right)$ and $\left(6\right)$
$x\in (3,4)\cdots \left(7\right)$

From $\left(4\right)$ and $\left(7\right)$
$x\in (3,4)$