For constants $a$ and $b$, I know that I can break up $\log(a/b)$ into $\log(a) - \log(b)$.
Can I conveniently break up $\log(a - b)$ somehow into several terms?
$\endgroup$ 21 Answer
$\begingroup$$$\log(a-b) = \log\left(a\cdot\left(1-\frac ba\right)\right)$$
$$= \log(a) + \log\left(1-\frac ba\right)$$
if $\displaystyle\left|\frac ba\right| \lt 1$ then it can be written as
$$= \log(a) -\sum_{n=1}^{\infty} \frac{x^n}{n}$$
Where $x=b/a$
$\endgroup$ 3
