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## #1 2005-08-26 09:27:56

mikau
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### logarthimic operations : order of precedance?

The rules for logarithms are:

log m + log n = log mn

log m - log n = log n/n

x log m = log of m to the x'th power.

But which comes first? Sometimes you get long logarithmic equations and it doesn't come out because I didn't use these simplifications in the correcto order.

Which comes first?

A logarithm is just a misspelled algorithm.

## #2 2005-08-26 09:39:43

ajp3
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### Re: logarthimic operations : order of precedance?

if you use them correctly, there is no preferred order; you should be able to work any equation using any approach (i.e. order) you wish. just make sure that you always follow the order of operation rules for arithmetic!

if you would post a specific question where you ran into trouble, I would be more than happy to explain what I'm talking about using your situation as an example...

oh and btw the second one should be: (log m) - (log n) = log (m/n)

## #3 2005-08-26 09:45:26

mikau
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### Re: logarthimic operations : order of precedance?

ok, I guess I messed up when that happend. So it doesn't matter which somes first? Not even the exponant rule?

A logarithm is just a misspelled algorithm.

## #4 2005-08-26 09:52:40

MathsIsFun
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### Re: logarthimic operations : order of precedance?

Parentheses, Exponents, Multiply and Divide, Add and Subtract

And I would say, for example, that "log x²" would really mean "log (x²)", not "(log x)²"

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #5 2005-08-26 11:20:16

ajp3
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### Re: logarthimic operations : order of precedance?

right, I missed that ambiguity!

correct, mikau: it doesn't matter which comes first, not even the exponent rule... the only ordering you have to obey is PEMDAS, as MathsIsFun said.

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