<s>[INST] 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.
 [/INST] $$\begin{align}\ln (a + b) &= \ln \left(\frac{b}{b - 1} + b\right)\\&=\ln\left(\frac{b + b^2 - b}{b - 1}\right)\\&=\ln\left(\frac{b^2}{b - 1}\right)\\&= \ln \left(\frac{b}{b - 1} \cdot  b\right)\\&= \ln\left(\frac{b}{b - 1}\right) + \ln b\\&= \ln a + \ln b\end{align}$$
The trick is that the $b$ terms in the numerator gets eliminated after merging both $a$ and $b$ into a single fraction.
</s>