Solving $\ln|1+y|=\ln|x|-\ln|x+1|+c$ for $y$

Question

How do I solve the following equation for $y$? $$\ln|1+y|=\ln|x|-\ln|x+1|+c$$

I want to get rid of the absolute values in a mathematically correct way.

Sorry if the format of the equation is incorrect, I do not know how to fix it.

Answer 1

Rearrange to get $$ |1+y|=\frac{|x|}{|x+1|}e^c. $$ then if $y>-1$ you get $$ y=-1+\frac{|x|}{|x+1|}e^c $$ and if $y<-1$ you get $$ y=-1-\frac{|x|}{|x+1|}e^c $$ in which for both cases $x\neq 0,1$.

To get rid of $|x|$ you can use $|x|=\sqrt{x^2}$ or then you split the intervals of $x$ by $$ \frac{|x|}{|x+1|}=\frac{x}{x+1} $$ when $x>$ or $x<-1$ and $$ \frac{|x|}{|x+1|}=-\frac{x}{x+1} $$ when $-1<x<0$.

Answer 2

Exponentiate both sides to get

$$|1+y| = \frac{|x|}{|x+1|}e^c.$$

Since $c$ is an arbitrary constant, so is $e^c$, so rename it $A$, but note that $A$ is positive because it's equal to $e^c$. So we have

$$|1+y| = A\frac{|x|}{|x+1|}.$$

If you remove the absolute values, you get a $\pm:$

$$1+y = \pm A\frac{x}{x+1}.$$

One $\pm$ does the job for both sides. Now here the magic: But simply declaring that $A$ is no longer just positive, but can take negative values, you don't need the $\pm$ anymore. We'd say that $A$ absorbs the $\pm.$

$$1+y =A\frac{x}{x+1}.$$

Subtract $1$ from both sides and you're done.