Logarithmic inequality: $\log_{1/3}^2(x^2-3x+2) - \log_{1/3}(x-1)>\log_{1/3}(x-2) +6$

Question

I need help solving this:

$$\log_{1/3}^2(x^2-3x+2) - \log_{1/3}(x-1)>\log_{1/3}(x-2) +6$$

So far I could not make sense of this, because I don't understand how to handle $\log^2$ or the $+6$ at the end... can someone explain it to me?

Answer 1

$$\log_{1/3}^2(x^2-3x+2) - \log_{1/3}(x-1)>\log_{1/3}(x-2) +6$$ $$\log_{1/3}^2(x^2-3x+2) - \log_{1/3}(x-1)-\log_{1/3}(x-2) -6 > 0$$ $$\log_{1/3}^2(x^2-3x+2) - \log_{1/3}(x-1)(x-2) -6 > 0$$ $$\log_{1/3}^2(x^2-3x+2) - \log_{1/3}(x^2-3x+2) -6 > 0$$ Now substitute $\log_{1/3}(x^2-3x+2) = t$

Answer 2

Inequallity is equivalent to \begin{align*} \log_{1/3}^2(x^2-3x+2)-\left[\log_{1/3}(x-1)+\log_{1/3}(x-2)\right]-6&>0\\ \log_{1/3}^2(x^2-3x+2)-\log_{1/3}(x^2-3x+2)-6&>0\\ \left[\log_{1/3}(x^2-3x+2)-3\right]\left[\log_{1/3}(x^2-3x+2)+2\right]&>0 \end{align*} This inequallity holds $\iff$ $\log_{1/3}(x^2-3x+2)\in(-\infty,-2)\cup(3,\infty)$ \begin{align*} \iff x^2-3x+2&\in\left(0,\frac{1}{27}\right)\cup\left(9,\infty\right) \end{align*} Since $x^2-3x+2=(x-\frac{3}{2})^2-\frac{1}{4}$: $0< x^2-3x+2<\frac{1}{27}\iff \frac{1}{4}< \left(x-\frac{3}{2}\right)^2<\frac{31}{108}\iff x\in\left(\frac{27-\sqrt{93}}{18},1\right)\cup\left(2,\frac{27+\sqrt{93}}{18} \right)$

$x^2-3x+2>9\iff \left(x-\frac{3}{2}\right)^2>\frac{37}{4}\iff x\in\left(-\infty,\frac{3-\sqrt{37}}{2}\right)\cup\left(\frac{3+\sqrt{37}}{2},\infty \right)$

Since $\log_{1/3}(x-2)$ is real defined if $x>2$ we have as Solution set: $$\left(2,\frac{27+\sqrt{93}}{18} \right)\cup\left(\frac{3+\sqrt{37}}{2},\infty \right).$$