Question

Find the range of $x$ satisfying ${\log }_{3x}\left(\frac{x-2}{3}\right)<1$

Answer 1

${\log }_{3x}\left(\frac{x-2}{3}\right)<1$

${\log }_{3x}\left(\frac{x-2}{3}\right)$ is defined for $x\ne 0$ and $x>1$

$\Rightarrow$Solution$1:x\in (1,\infty )-\left\{0\right\}$

${\log }_{3x}\left(\frac{x-2}{3}\right)$ is defined for $\frac{x-2}{3}>0$

$\Rightarrow x\ge 2$

or Solution$2:x\in [2,\infty )$

Again,${\log }_{3x}\left(\frac{x-2}{x}\right)<1$

$\Rightarrow \frac{x-2}{3}<3x$

$\Rightarrow x-2<9x$

$\Rightarrow -8x-2<0$

$\Rightarrow -8x<2$

$\Rightarrow x>\frac{-1}{4}$

$\Rightarrow$Solution$3:x\in (-\frac{1}{4},\infty )$

Final Solution$=x\in (1,\infty )-\left\{0\right\}\cap [2,\infty )\cap (-\frac{1}{4},\infty )$

$=x\in (-\frac{1}{4},0)\cup (0,1)\cup (1,2]\cup (2,\infty )$