Question

Solve the following inequality:
${\log }_{x-3}\left(2\right(x^2-10x+24\left)\right)⩾{\log }_{x-3}(x^2-9)$

Answer 1

For $x>4$

$2(x^2-10x+24)\ge x^2-92x^2-20x+48\ge x^2-9x^2-20x+57\ge 0x\in (-\infty ,10-\sqrt{43}]\cup [10+\sqrt{43},\infty )$

Hence, $x\in (10+\sqrt{43},\infty )$

For $x\in (3,4)$

$2(x^2-10x+24)\le x^2-9x^2-20x+48\le 0x\in [10-\sqrt{43},10+\sqrt{43}]$

Hence, $x\in [10-\sqrt{43},4)$

Also,

$x^2-10x+24>0(x-6)(x-4)>0x\in (-\infty ,4)\cup (6,\infty )$ $x^2-9>0(x+3)(x-3)>0x\in (-\infty ,-3)\cup (3,\infty )$

Taking intersection, we get

$x\in (10-\sqrt{43},4)\cup (10+\sqrt{43},\infty )$