metalskaashw

## Answered question

2022-03-27

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

### Answer & Explanation

Esteban Sloan

Beginner2022-03-28Added 21 answers

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

Ishaan Stout

Beginner2022-03-29Added 14 answers

You can try out something like,
$\mathrm{log}\left(a-b\right)-\mathrm{log}\left(a-c\right)=\mathrm{log}\left(a\right)+\mathrm{log}\left(1-\frac{b}{a}\right)-\mathrm{log}\left(a\right)-\mathrm{log}\left(1-\frac{c}{a}\right)=\mathrm{log}\left(1-\frac{b}{a}\right)-\mathrm{log}\left(1-\frac{c}{a}\right)$
where $\mathrm{log}\left(a\right)$ 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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