I have this equation: $$ \ln\frac{2-|y-1|}{1-|y|} = \ln x $$ which becomes $$ \ln(2-|y-1|)-\ln(1-|y|) = \ln x. $$
Can the first term in LHS be written as $\ln(2)-\ln(|y-1|)\implies\ln(2)-\ln(|y|)-\ln(|1|)$ and the second term as $\ln(1)-\ln(|y|)$?
How does one solve for $y$?
On
You actually do not need to bother with logarithm for this simple question. First, you observe that if $x \leq 0$ the RHS makes no sense, so there are no solutions. If instead $x > 0$, then the equation can be rewritten as $$ \frac{2-|y-1|}{1-|y|} = x, $$ since the logarithm is an injective function. Now you should be able to solve it by yourself. For instance, try to plot a graph of the LHS.
Remember that you have to discard solutions which give non-positive values for $x$.
No. In general $$\ln(a-b)=\ln(a)-\ln (b)$$ if and only if $$a-b=\frac ab,\ a-b\gt 0,\ a\gt 0,\ b\gt 0.$$
To solve $$\frac{2-|y-1|}{1-|y|}=x$$ for $y$, separate it into three cases as $y\le 0,0\lt y\le 1,1\lt y.$ But note that $$1-|y|\not=0\iff y\not=\pm1.$$
Case 1 : If $y\le 0$, then we have $$\frac{2-(-(y-1))}{1-(-y)}=x\Rightarrow x=1.$$ Note that $y\le 0$ can be any real number except $-1$.
Case 2 : If $0\lt y\lt 1$, then we have $$\frac{2-(-(y-1))}{1-y}=x\Rightarrow y+1=x(1-y)\Rightarrow y(1+x)=x-1\Rightarrow y=\frac{x-1}{1+x}$$because $x\gt 0\Rightarrow 1+x\not=0$.
Case 3 : If $y\gt 1$, then we have $$\frac{2-(y-1)}{1-y}=x\Rightarrow -y+3=x(1-y)\Rightarrow y(-1+x)=x-3\Rightarrow y=\frac{x-3}{x-1}$$ because $x\not=1$.