How to simplify logartihms with a integration constant? How can I simplify:
$$\ln \left| 1+\frac{y}{x}\right|=\ln|x|+c, $$
where $c$ is an integration constant?
I thought it would just be:
$$\left|1+\frac{y}{x}\right|=|x|+e^c , $$
where $e^c$ becomes $c$.
I know it actually simplifies to:
$$ 1+\frac{y}{x}=cx , $$
and then from here to $y=cx^2-x$.
I don't understand why it's not $+e^c$ or $+c$, and I don't understand how the absolute values got removed.

The constant $c$ can also be written as $\ln|c|$.
Note that: $\ln a + \ln b =\ln(ab)$
$\ln \lvert 1 +\frac{y}{x} \rvert= \ln|x|+\ln|c|$
$\ln \lvert 1 +\frac{y}{x} \rvert= \ln|cx|$
$1 +\frac{y}{x} = cx$
$x+y=cx^2$
$y=cx^2-x$
