Question

Solve the following inequality:
${\log }_{2x}(x^2-5x+6)<1$

Answer 1

$\frac{\log (x^2-5x+6)}{\log 2x}<1$

For $x>\frac{1}{2}$

$\log (x^2-5x+6)<\log 2x{x}^2-5x+6<2x{x}^2-7x+6<0(x-1)(x-6)<0x\in (1,6)Hence,x\in (1,6)$

For $x\in (0,\frac{1}{2})$

$\log (x^2-5x+6)>\log 2x{x}^2-5x+6>2x{x}^2-7x+6>0(x-1)(x-6)>0x\in (-\infty ,1)\cup (6,\infty )Hence,x\in (0,\frac{1}{2})$

Also, $x^2-5x+6>0(x-2)(x-3)>0x\in (-\infty ,2)\cup (3,\infty )$

Taking intersection we get

$x\in (0,\frac{1}{2})\cup (1,2)\cup (3,6)$