Why can $\ln|y-1| = \ln|x+3| + C$ be replaced by $\ln(1-y) = \ln(x+3) + C$?

Question

In this case $C = -\ln(2)$. I just can't quite follow the logic on why the replacement works.

Answer 1

Recall the definition of absolute value, v.gr. for $\ln|y-1|$ :

$\ln|y-1|=\left\{\begin{matrix} y-1& \text{if }y-1>0\Leftrightarrow y>1& \\ -(y-1)=1-y&\text{if }y-1<0\Leftrightarrow y<1 & \end{matrix}\right.$

as indicated by the text $x = -1$ and $y=0$ then the case in which $y <1$ in the definition previously exposed.

Answer 2

Because $y-1$ in this case is negative, so $|y-1|=(1-y)>0$.