# Talk:List of logarithmic identities

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## Nested logarithms?

Does anyone know of any identities pertaining to nested logarithms? (e.g. ${\displaystyle \log _{a}(\log _{b}(x))}$) -- He Who Is[ Talk ] 18:27, 28 June 2006 (UTC)

I'm not aware of any identities. Nested logs quickly become undefined, when the next to last log becomes negative. Log (abs(x)) is very unstable when nested. --agr 18:52, 29 June 2006 (UTC)
Well, when the original logarithm is negative, its still defined. Its just defined as a complex number. (e.g. ${\displaystyle \log _{2}-4=2+i\pi }$) -- He Who Is[ Talk ] 22:54, 29 June 2006 (UTC)
I'm fairly certain agr meant undefined in the Reals. --ĶĩřβȳŤįɱéØ 10:13, 28 December 2006 (UTC)

## Snake rule

I don't see why it was deleted a little while ago, it's a perfectly good identity Carifio24 03:09, 26 April 2007 (UTC)

I'm not convinced it is. It appears to imply, for example, that a logarithm to a power is simply the same as the
original logarithm not raised to a power--e.g., by expanding (log2(4)) ^ 5 this "identity" gives the answer as 2, whereas
the actual value = 2 ^ 5 = 32. I think that the "special case" example appears to be a misinterpretation of the
change-of-base formula. The actual value of loga(b) is logc(b) / logc(a), not logb(c) / loga(c). Given this
and the lack of any other reference to this "identity", I'm deleting the section. —Preceding unsigned comment added by 140.180.133.97 (talk) 05:30, 26 September 2007 (UTC)

## Log (a+b+c...z) = ???

Well, just look at the subject. Any idea of a formula to solve such a monster?

Log (a+b+c...z) = ???

It seems like there should be a pattern, albeit very ugly one.

Bonkster99 03:25, 10 October 2007 (UTC)

I don't think there are any log identities for sums in arguments. For some functions, such as sinusoids, there are such properties, but not for logs. log (a+b+c+...+z) is in simplest form, I believe. —Preceding unsigned comment added by Eebster the Great (talkcontribs) 02:54, 13 December 2007 (UTC)

## Peculiar identities

Why are the identities:

${\displaystyle \log _{b}(a+c)=\log _{b}a+\log _{b}(1+b^{\log _{b}c-\log _{b}a})}$
${\displaystyle \log _{b}(a-c)=\log _{b}a+\log _{b}(1-b^{\log _{b}c-\log _{b}a})}$

... written like that? It seems rather obvious that they would be better written as:

${\displaystyle \log _{b}(a+c)=\log _{b}a+\log _{b}(1+c/a)}$
${\displaystyle \log _{b}(a-c)=\log _{b}a+\log _{b}(1-c/a)}$

respectively, since ${\displaystyle x^{\log _{x}y}=y}$ by definition, and ${\displaystyle x^{y-z}=x^{y}.x^{-z}=x^{y}/x^{z}}$, meaning that ${\displaystyle b^{\log _{b}c-\log _{b}a}=b^{log_{b}c}/b^{log_{b}a}=c/a}$ ?

Am I missing something? 130.243.139.62 17:48, 12 October 2007 (UTC)

It looks like you're right, but perhaps I'm missing it too. The only thing I can think of is that perhaps this property applies to both negative and positive arguments (real and complex logs), whereas yours would not. This leads into my question.
He is right, the identities are derived as follows,
${\displaystyle \log _{b}(a+c)=\log _{b}(a(1+c/a))=\log _{b}a+\log _{b}(1+c/a)}$
${\displaystyle \log _{b}(a-c)=\log _{b}(a(1-c/a))=\log _{b}a+\log _{b}(1-c/a)}$
The only assumption made here is that a is not zero but this is needed anyway to take the log of a so these should be re-written. —Preceding unsigned comment added by 195.112.46.6 (talk) 18:33, 1 March 2009 (UTC)
He is correct, but the way it was originally written has the only occurrences of a and c appearing in a log. This is useful in computations as computing in the log domain can help with numerical stability when the calculations are done on a computer. So his form is neater, but the original would be used often and that is likely why it was written that way. -173.33.199.131 (talk) 07:46, 31 March 2010 (UTC)

## Negative argument identities

I noticed on my calculator that some of these identities do not apply for negative arguments. For example, log5 obviously =log (-1*-5). According to your identities, that would =log(-1)+log(-5). However, my calculator gives a complex number for this. Does this mean only that that particular log identity doesn't work for complex numbers, or that the associated exponent identity does not? Are there any identities that apply to both positive and negative arguments? —Preceding unsigned comment added by Eebster the Great (talkcontribs) 02:59, 13 December 2007 (UTC)

## Citing sources

Does this article really need to have sources cited? The tag isn't needed there, or why is it there? It is a math article, all it needs is someone to add proofs. Logic should be the source from my pov. --Bobianite (talk) 05:15, 23 April 2008 (UTC)

Just to comisserate, yes there is a call for verifiability. In this case most of the assertions are just arithmetic or simple algebra. Yet you might find print materials, or a website, that has similar (probably shorter) list of identities. By getting ahold of these you can put the subject in context of growing knowledge. How long a list was there before this article ? While we can, making the link to the old print knowledge base enhances credibility of the whole WP project. Looking ahead, other means of verifying mathematical assertions will have to be developed as old knowledge fades to lower status. Just now WP has come to #7 on the Alexa web order. I'll keep my eye out too for the type of reference that will put the matter to rest.Rgdboer (talk) 21:30, 26 April 2008 (UTC)
I see now. Thanks for letting me know why. --Bobianite (talk) 22:18, 26 April 2008 (UTC)

## Poor sentence

I hate to be pedantic but I think this is a bit of a poor sentence:

"note: to say the limit of a function "equals infinity" is not strictly correct notation, as "infinity" is not a value. What is meant by the limits equations above is simply that the functions increase/decrease without bound."

For a start, there is nothing wrong with saying that a limit equals infinity because it is just notation, it has a very precice meaning in mathematics. As long as you define things well enough, there is no such thing as incorrect notation. Incidentally, the definition of tending to infinity given is also wrong, increasing without bound is not the same as tending to infinity, the sequence, 0,1,0,2,0,3,.... increases without bound, and so does xsin(x) as x tends to infinity but neither of these functions tend to infinity. —Preceding unsigned comment added by 195.112.46.6 (talk) 18:41, 1 March 2009 (UTC)

Or, if we interpret 'increases without bound' to mean monotonic increase, it's false in a different way: 1,0,10,9,100,99,1000,999,... tends to infinity but does not increase without bound. The note is entirely false: saying a limit equals infinity is strictly correct notation, infinity (in this sense) is a value (in the Extended real number line), and the given statement is not what is meant by convergence to infinity. Even if it was correct, this list is not the place for a discussion of the minutiae of limits: that's what the link to limit of a function is for. I've removed the note. Algebraist 22:26, 2 April 2009 (UTC)

## Missing log limits

${\displaystyle \lim _{x\to 0}\log {(1+x)}=x}$, but always ${\displaystyle \log {(1+x)}.

${\displaystyle \lim _{x\to \infty }\log {(1+1/x)}=x}$

I guess you mean ${\displaystyle =0}$ or the second one? That follows pretty quickly from the upper bound you gave yourself. I added some inequalities so people may deduce this themselves. Thomasda (talk) 16:19, 16 September 2015 (UTC)