When solving $\log_6 \left(\frac{12x-1}{x-3}\right)<0$, I would raise both sides as exponents to a base of $6$, resulting in $\frac{12x-1}{x-3}<1$, then move the $1$ to the left side, make a common denominator resulting in $\frac{11x+2}{x-3}<0$, and use a number line graph to solve, finding the answer as $x\in (-2/11, 3)$, restricted by the domain resulting in the final answer as $x\in(-2/11, 1/12)$.
But I have seen it solved by using a property of logs and separating $\log_6 \left(\frac{12x-1}{x-3}\right)<0$ into $\log_6 (12x-1)-\log6 (x-3)<0$, then moving $\log_6 (x-3)$ to the right side of the inequality resulting in $\log_6 (12x-1)<\log_6 (x-3)$, and then dropping the logs to make $12x-1<x-3$.
When I look at this graphically I know the second way of solving is not correct, but I am trying to figure out why?
The argument of the logarithmic must be greater than zero. So, we must have
$$\frac{12x-1}{x-3}>0$$
This implies that either $x>3$ or $x<1/12$. So, with $x$ in this domain we have
$$\frac{12x-1}{x-3}<1$$
If $x>3$, then $12x-1<x-3\implies x<-2/11$, which is extraneous. But this (i.e., $x>3$), is tacitly assumed in the second methodology.
If $x<1/12$, then
$$12x-1>x-3\implies x>-2/11$$
The admissible solution is therefore $x\in(-2/11,1/12)$