The set of real values of $x$ satisfying the given equation are?

Question

$\log_{\frac 12}(x^2-6x+12) \ge -2$. 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.

Answer 1

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]

Answer 2

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 $$2\le x\le 4$$

Answer 3

Hint 1:

$$ -2 = \log_{\frac{1}{2}}4 $$

Hint 2:

$$ \log_{\frac{1}{2}}x $$ is a monotone decreasing function