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?
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8Yes and no. There is no specific formula, but, for example $$\log(a-b) = \log(a(1-b/a)) = \log(a)+\log(1-b/a)$$ – David P Feb 08 '14 at 06:38
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5It's not possible to write $\log(a-b)=f(a)+g(b)$ because $\frac{\partial^2}{\partial a\partial b}\log(a-b)\ne0$. – anon Feb 08 '14 at 07:07
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$$\log(a-b) = \log\left(a\cdot\left(1-\frac ba\right)\right)$$
$$= \log(a) + \log\left(1-\frac ba\right)$$
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 $x=b/a$
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How come $ \log\left(1-\frac ba\right)$ is $\sum_{n=1}^{\infty} \frac{x^n}{n}$ Please explain details. thanks @anon – SAH Sep 24 '14 at 18:00
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