Show that $\ln(a+b) =\ ln(a) + \ln(b)$ when $a = \frac{b}{b-1}$ Show that $\ln(a+b) = \ln(a) + \ln(b)$ when $a = \frac{b}{b-1}$
My attempt at a solution was
$$
\ln(a+b) = \ln \left(a(1+\frac{b}{a})\right)
$$
and by setting $a = \frac{b}{b-1}$ we get
$$
\ln \left(\frac{b}{b-1}(1+(b-1))\right) = \ln \left(\frac{b}{b-1}+b \right)
$$
But how can I continue from here? 
The original question in my textbook asked if the equality $\ln(a+b) = \ln(a) + \ln(b)$ was true. 
My initial answer was that it was not true, but I missed this special case, which I think was very hard to "see". Any help would be much appreciated.
 A: $$\begin{align}\ln (a + b) &= \ln \left(\frac{b}{b - 1} + b\right)\\&=\ln\left(\frac{b + b^2 - b}{b - 1}\right)\\&=\ln\left(\frac{b^2}{b - 1}\right)\\&= \ln \left(\frac{b}{b - 1} \cdot  b\right)\\&= \ln\left(\frac{b}{b - 1}\right) + \ln b\\&= \ln a + \ln b\end{align}$$
The trick is that the $b$ terms in the numerator gets eliminated after merging both $a$ and $b$ into a single fraction.
A: $$\ln(a+b)=\ln\Bigl(\frac b{b-1}+b\Bigr)=\ln\frac{b^2}{b-1},$$
$$\ln a+\ln b=\ln(ab)=\ln\frac{b^2}{b-1}.$$
A: Seems some other people have shown it forwards. I will show how to go backwards, then.
$\ln(a) + \ln(b) = \ln(ab)$ is a logarithm law,
Now for $ab = a+b$:
$ab-a = b \Leftrightarrow a(b-1) = b \Leftrightarrow /\text{Assuming } b-1\neq 0/ \Leftrightarrow a = \frac{b}{b-1}$
A: So in general the statement 
$$
\ln(a+b)=\ln(a)+\ln(b)
$$
is not true. You're right! In the case of $a=\frac{b}{b-1}$ and everything is still well defined we have the following relation
$$
a+b=\frac{b}{b-1}+b=\frac{b+b^2-b}{b-1}=\frac{b^2}{b-1}=b\frac{b}{b-1}
$$
well and this means, since we know that $\ln(x\cdot y)=\ln(x)+\ln(y)$ holds, that we can rewrite (just in this special case)
$$
\ln(a+b)=\ln(b\cdot\frac{b}{b-1})=\ln(b)\ln(\frac{b}{b-1})=\ln(b)\ln(a)
$$
since we said that $a=\frac{b}{b-1}$ was true.
A: Since $ln(a) + ln(b) = ln(ab)$, we have
$\quad ln(a+b) = ln(a) + ln(b)$ iff 
$\quad ln(a+b) = ln(ab)$ iff 
$\quad a+b=ab$ iff 
$\quad b=a(b-1)$ iff 
$\quad a = \dfrac{b}{b-1}$
because $b=1$ can never work in the original equation.
