Solve the equation $\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2$ Solve the equation $$\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2$$
We have $$D_x:\begin{cases}1-2x>0\\6x^2-5x+1>0\\1-3x>0\\1-3x\ne1\\4x^2-4x+1>0\iff(2x-1)^2>0\iff x\ne\dfrac12\end{cases}\iff x\in(-\infty;0)\cup(0;\dfrac{1}{3})$$
Also the quadratic $6x^2-5x+1$ factors as $(2x-1)(3x-1)$. The equation then becomes $$\log_{1-2x}(2x-1)(3x-1)-\log_{1-3x}(2x-1)^2=2\\\log_{1-2x}(2x-1)(3x-1)-2\log_{1-3x}(1-2x)=2,$$ as $\log_{1-3x}(2x-1)^2=2\log_{1-3x}|2x-1|,$ but we know from $D_x$ that $2x-1<0$,
$$\log_{1-2x}(2x-1)+\log_{1-2x}(3x-1)-\dfrac{2}{\log_{1-2x}(1-3x)}=2$$ I don't know what to do next.

We have that for $1-2x>0$, $1-3x>0$, $1-2x\neq 1$, $1-3x\neq 1$
$$\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2 $$
$$\iff \frac{\log((1-2x)(1-3x))}{\log (1-2x)}-2\frac{\log (1-2x)}{\log (1-3x)}=2$$
$$\iff \frac{\log (1-3x)}{\log (1-2x)}-2\frac{\log (1-2x)}{\log (1-3x)}=1$$
then by $u= \frac{\log (1-3x)}{\log (1-2x)}$ we obtain
$$u-\frac 2 u =1 \implies u = \frac{\log (1-3x)}{\log (1-2x)}=2 \implies x=\frac14$$
