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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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    $\begingroup$ Yes and no. There is no specific formula, but, for example $$\log(a-b) = \log(a(1-b/a)) = \log(a)+\log(1-b/a)$$ $\endgroup$ – David Peterson Feb 8 '14 at 6:38
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    $\begingroup$ It's not possible to write $\log(a-b)=f(a)+g(b)$ because $\frac{\partial^2}{\partial a\partial b}\log(a-b)\ne0$. $\endgroup$ – anon Feb 8 '14 at 7:07
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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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    $\begingroup$ (Need to write $x=b/a$.) Why was this downvoted? $\endgroup$ – anon Feb 8 '14 at 7:44
  • $\begingroup$ How come $ \log\left(1-\frac ba\right)$ is $\sum_{n=1}^{\infty} \frac{x^n}{n}$ Please explain details. thanks @anon $\endgroup$ – Electricman Sep 24 '14 at 18:00
  • $\begingroup$ @Electric http://en.wikipedia.org/wiki/Mercator_series $\endgroup$ – anon Sep 24 '14 at 23:56

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