Question

Solve the following inequality:
${\log }_{\frac{3x}{x^2+1}}(x^2-2.5x+1)⩾0$

Answer 1

We know that

$x^2-2.5x+1>0x^2-2x-0.5x+1>0(x-2)(x-0.5)>0x\in (-\infty ,0.5)\cup \left(2.\infty \right)$

Also $\frac{3x}{x^2+1}>0x>0x\in (0,\infty )$

Taking intersection, we get $x\in (0,0.5)\cup (2,\infty )$

$\frac{3x}{x^2+1}=1\Rightarrow 3x=x^2+1\Rightarrow x^2-3x+1=0$

$x=(\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2})$

$\frac{3x}{x^2+1}>1$

For $x\in (\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2})$

For $x\in (\frac{3-\sqrt{5}}{2},0.5)\cup \left(\frac{2,3+\sqrt{5}}{2}\right)$

$\log (x^2-2.5x+1)\ge 0x^2-2.5x+1\ge 1x(x-2.5)\ge 0x\in (-\infty ,0]\cup [2.5,\infty )$

Hence, $x\in [2.5,\frac{3+\sqrt{5}}{2})$

For $x\in (0,\frac{3-\sqrt{5}}{2})\cup (\frac{3+\sqrt{5}}{2},\infty )$

$\log (x^2-2.5x+1)\le 0x^2-2.5x+1\le 1\Rightarrow x(x-2.5)\le 0x\in [0,2.5]$

Hence, $x\in (0,\frac{3-\sqrt{5}}{2})$

Taking union, we get $x\in (0,\frac{3-\sqrt{5}}{2})\cup [2.5,\frac{3+\sqrt{5}}{2})$