Question

Solve the following inequality:
${\log }_{0.5}\left({\log }_8\frac{x^2-2x}{x-3}\right)<0$

Answer 1

${\log }_2\left({\log }_8\left(\frac{x^2-2x}{x-3}\right)\right)>0$

${\log }_8\left(\frac{x^2-2x}{x-3}\right)>1$

$\frac{x^2-2x}{x-3}>8$

$\frac{x^2-2x-8x+24}{x-3}>0$

$\frac{x^2-10x+24}{x-3}>0$

$\frac{(x-4)(x-6)}{x-3}>0$

$x\epsilon (3,4)\cup (6,\infty )$