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$\log(a - b) - \log(a - c)$

Does this have a simpler form? Perhaps one where the $a$s have cancelled out? I know it can also be expressed as a log of the fraction: $\log\frac{(a-b)}{(a-c)}$, but the same question applies.

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You cannot cancel the $a$s in any reasonably general way for precisely the same reason that you cannot in general find any nicer expresssion for $\frac{a-b}{a-c}$; the $a's$ just don't cancel there. As Ross writes, you can trade the two $a$s for two $b$s instead, or you could trade them for two $c$s (by taking $1+\frac{c-b}{a-c}$ instead), but you'll have two of one of the variables any which way. –  Arturo Magidin Feb 17 '11 at 4:55

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It depends what you think is simpler. $\log(a - b) - \log(a - c)=\log \frac{a-b}{a-c}=\log \left(1-\frac{b-c}{a-c}\right)$

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Like Ross said, it depends what you think is simpler. You can try out something like,

$$ \log(a-b) - \log(a-c) = \log(a) + \log(1-\frac{b}{a}) - \log(a) - \log(1 - \frac{c}{a}) = \log(1-\frac{b}{a}) - \log(1 - \frac{c}{a})$$ where $\log(a)$ is cancelled out. Then expand it to get the results that you want. But since your question is not clear as to what results you are seeking, nor does it say anything about a,b and c, I am not sure if this might be of help.

Anyways, this expression wont have a simpler form, like you said, where the 'a's are cancelled out. It needs some extra condition to actually cancel out 'a' from the expression.

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