$\log_{\frac 12}(x^2-6x+12) \ge -2$. I am unable to understand the last two steps of the given solution.
We observe that $0$ does not satisfy the inequality. If the given solution is incorrect, how do I arrive at the correct answer? Please post the correct method for solving such types of questions.
The inequality is equivalent to $$\frac{\ln(x^2-6x+12)}{\ln(\frac{1}{2})}\geq -2$$ or $$\ln(x^2-6x+12)\le-2\ln(\frac{1}{2})$$ so $$\ln(x^2-6x+12)\le\ln((\frac{1}{2})^{-2})$$ this gives $$2\le x\le 4$$
Hint 1:
$$ -2 = \log_{\frac{1}{2}}4 $$
Hint 2:
$$ \log_{\frac{1}{2}}x $$ is a monotone decreasing function
Let's raise the equation by using base 1/2. X^2-6x+12=<4 (because the right hand side changes sign, we flip the inequality)
X^2-6x+8=<0. (x-4)(x-2)=<0
For x between 2 and 4, the left hand side is negative, so these do satisfy the inequality. For all other x, they don't satisfy the inequality.
So our solution set is [2,4]
