Question

Solve the inequality: ${\log }_{1/2}(x^2-6x+12)\ge -2$

Answer 1

Given inequality

${\log }_{1/2}(x^2-6x+12)\ge -2$
$\Rightarrow x^2-6x+12\le (\frac{1}{2}{)}^{-2}=4$ ($\because$ For $0<a<1,x>0$ if ${\log }_a\left(x\right)\ge b$ then $x\le a^b)$

$\Rightarrow x^2-6x+12\le 4$
$\Rightarrow x^2-6x+8\le 0$

$\Rightarrow (x-2)(x-4)\le 0$

$\Rightarrow x\in [2,4]$
For logarithm to be defined,

$x^2-6x+12>0$

Clearly this is true for all $x\in R$
Hence, solution set is $x\in [2,4]$