How to simplify $\ln |x-1| = 2\ln |\frac {y}{x}-1| - 3\ln |\frac{y}{x} - 2| + C; C = const$ to $(y-2x)^3 = C(y-x-1)^2$? How to simplify $\ln |x-1| = 2\ln |\frac {y}{x}-1| - 3\ln |\frac{y}{x} - 2| + C; C = const$ to $(y-2x)^3 = C(y-x-1)^2$?
I'm trying to solve $(2x - 4y + 6)dx + (x+y-3)dy = 0$. The two lines intersect at $(1,2)$. After substituting to $(\tau = x-1, \mu = y-2)$ and then to $(\tau, u = \mu/\tau)$ I get $-\dfrac{d\tau}{\tau} = \dfrac{(u+1)du}{u^2 - 3u +2}$ and then I integrate. On the left I have $- \ln |x-1| + C$. On the right, as the denominator is the same as $(u-3/2)^2 - 1/4$ I substitute with $z = (u-3/2)$ and finally get $-2\ln |z + 1/2| + 3 \ln |z - 1/2| + C \equiv -2 \ln |\frac{y}{x} - 1| + 3\ln |\frac{y}{x} -2| + C $. After raising to powerof $e$ I get $|x-1| = \dfrac{(\frac{y}{x} - 1)^2}{|\frac{y}{x} - 2|^3}*e^C \equiv \dfrac {|x|(y-x)^2}{|y-2x|^3}*e^C$ but I'm not able to obtain the form in the answer key.

Don't complete the square; instead from
$$-\frac1\tau\,d\tau=\frac{u+1}{u^2-3u +2}\,du$$
decompose $\frac{u+1}{u^2-3u+2}$ through partial fractions, integrate and temporarily leave the equation in terms of $\tau,u$ to get
$$K-\ln\tau=3\ln(u-2)-2\ln(u-1)$$
Then simplify:
$$K=\ln\frac{\tau(\mu/\tau-2)(\mu/\tau-2)^2}{(\mu/\tau-1)^2}=\ln\frac{(\mu-2\tau)^3}{(\mu-\tau)^2}$$
$$C=\frac{(\mu-2\tau)^3}{(\mu-\tau)^2}=\frac{(y-2x)^3}{(y-x-1)^2}$$
$$(y-2x)^3=C(y-x-1)^2$$
