Solve $\log_{1/3}(x^2-3x+3)≥0$ I want to solve $$\log_{1/3}(x^2-3x+3)≥0$$
Now I know the result is: $x ∈ <1;2>$, but i am not sure how to get it.
My thoughts: $\frac{1}{3}$ to the power of positive number $= (x^2-3x+3)$, now I would solve $x^2-3x+3$ with the help of discriminant to get the points where $x$ is zero, and say the answer is for the positive intervals. Is this correct / is there easier way to do it?
 A: $$\log_{1/3}(x^2-3x+3) \ge 0$$
This occurs when:
$$x^2-3x+3 \le 1$$
$$x^2-3x+2 \le 0$$
Factorizing, we get:
$$(x-1)(x-2) \le 0$$
Solving the inequality, we get:
$$1 \le x \le 2$$
ALTERNATE SOLUTION
Consider the following equation:
$$\log_{1/3}(x^2-3x+3) = 0$$
If we solve, we get:
$$x^2 -3x + 2 = 0$$
$$(x-1)(x-2) = 0$$
Now, if we have $x < 1$, $x^2 - 3x + 3 \ge 1$
This means that a possible value is $\log_{1/3}5$, which would a value less than $0$.
Similarly if $x > 2$, $x^2 - 3x + 3 \ge 1$.
When $1 < x < 2$, we have that $0 < x^2 - 3x + 3 < 1$, meaning only fractional answers exist.
This means that a possible value is $\log_{1/3}\frac{1}{3}$, which would a value greater than $0$.
A: Raise $\frac{1}{3}$ to the power of both sides, and flip the inequality because $\frac{1}{3}^x$ is a decreasing function:
$$
\begin{aligned}
\log_{1/3}(x^2 - 3x + 3 ) &≥ 0\\
x^2 - 3x + 3 &\le \left(\frac{1}{3}\right)^0\\
x^2 - 3x + 3 &\le 1\\
x^2 - 3x + 2 &\le 0\\
\end{aligned}
$$
Now factor to get $(x-1)(x-2) \le 0$. We need a negative and a positive sign to make a negative, hence
$$1 \le x \le 2$$
A: Hint when we solve log the inequality gets reversed thus our problem on simplifying becomes $x^2-3x+2<=0$ so $(x-1)(x-2)<=0$ thus using general wavy curve method $1<=x<=2$ its done.
A: Passing to neperian logarithms we have $$\frac{\ln(x^2-3x+3)}{\ln(\frac 13)}$$ Therefore since $\ln(\frac 13)<0$ we need to have $\ln(x^2-3x+3)\le0$. This happens when $x^2-3x+3= 1$ which give $(x-2)(x-1)=0$ $x=1$ and $x=2$ Thus
$$1\le x\le 2$$
