Question

Solve the following inequality:
${\left(\frac{1}{10}\right)}^{{\log }_{x-3}(x^2-4x+3)}⩾1.$

Answer 1

${\left(10\right)}^{-{\log }_{(x-3)}(x^2-4x+3)}\ge 1-{\log }_{(x-3)}(x^2-4x+3)\ge 0\frac{\log (x^2-4x+3)}{\log (x-3)}\le 0For,x>4\log (x^2-4x+3)\le 0x^2-4x+3\le 1x^2-4x+2\le 0x\in [2-\sqrt{2},2+\sqrt{2}]Hence,x\in \varphi For,x\in (3,4)\log (x^2-4x+3)\ge 0x^2-4x+3\ge 1x^2-4x+2\ge 0x\in (-\infty ,2-\sqrt{2}]\cup [2+\sqrt{2},\infty )Hence,x\in [2+\sqrt{2},4)Also$

$x^2-4x+3>0(x-1)(X-3)>0x\in (-\infty ,1)\cup (3,\infty )$ and $x-3>0x>3x\in (3,\infty )$

Taking intersection, we get $x\in [2+\sqrt{2},4)$