I have to calculate log2 ratio for odds and don't know how to do it.

 Group A      Group B
 0.031571     0.0170071

There are occurrences of event in GroupA and Group B - I want to calculate how much Group A has this specific event compared to Group B, hence I need log2 for odds ratio.

I was thinking that such ratio should be calculated like this:

log2(GroupA/GroupB) = log2(0.031571/0.0170071) = 0.892463

However in this stackoverflow answer they calculate it like this:

(GroupA/GroupB)/log(2) = (0.031571/0.0170071)/log(2) = 2.67814

My question is - how to calculate log2 ratio; and what is the difference between these two approaches?

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    $\begingroup$ Remember that the log of a ratio is equal to the difference of the logs: $\operatorname{log}_{k}(a/b)=\operatorname{log}_{k}(a)-\operatorname{log}_{k}(b)$, where $k$ is the base ($k=2$ in your case). Maybe that helps you figuring out if this is what you want to calculate. $\endgroup$ – COOLSerdash Jun 16 '13 at 12:38
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    $\begingroup$ Apparently, in the link I gave they tried to calculate log of a ratio in awk and awk can't calculate log2(A/B) and one has to log(A/B)/log(2). $\endgroup$ – PoGibas Jun 16 '13 at 12:54
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    $\begingroup$ I added a comment to that SO answer pointing out the mistake. I think it's just a typo, but because it resulted in hugely erroneous example output, I have also downvoted that answer pending a correction. $\endgroup$ – whuber Jun 16 '13 at 14:15

The log odds ratio is the log of the odds ratio, not the odds ratio divided by a log. I don't know what problem the link you gave was trying to solve, but it wasn't this one. You take the log of the OR because the OR is bounded by 0 and infinity and is multiplicatively symmetric around 1; while the log(OR) is unbounded and additively symmetric around 0.

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