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The Wikipedia page on the natural logarithm says: 'Logarithms can be defined to any positive base other than 1, not only e. However, logarithms in other bases differ only by a constant multiplier from the natural logarithm, and are usually defined in terms of the latter.' This seems to suggest that only natural logarithms have this property, namely: $$\tag{0}log_a(x)=c.log_e(x)$$ where $x$ can vary and yet $c$ remains constant. This question follows on from this one about the result: $$\tag{1}\int\frac{c}{x}=c.log_e(x) + C$$ given by WolframAlpha, where they explicitly state that it is the natural logarithm being used. However, my reading of the Wikipedia proof of the differentiation of $ln(x)$ (which is actually base agnostic) led me to believe that the above equation can be written: $$\tag{2}\int\frac{log_a(e)}{x}=log_a(x)+C$$ If $a=e$ they do of course match; non-trivially the implication, from $(0)$, is that: $$c=log_a(e)$$ $$\frac{log_a(x)}{log_a(e)}=log_e(x)$$ (See Wikipedia, Logarithm, Change of base). Algebraically, this all checks out; however we began by assuming the property that connected the two integrals, so we must ask the question - does that property really only apply to natural logarithms? And, could the Wolfram result be rewritten without specifying a base at all?

Consider the following test:

from math import log    #log is ln in python unless a 2nd argument gives a base

print()
print('log(18,1.5)', log(18,1.5))
print('log(18,3)', log(18,3))
print('log(18,1.5)/log(18,3)', log(18,1.5)/log(18,3))
print()
print('log(14,1.5)', log(14,1.5))
print('log(14,3)', log(14,3))
print('log(14,1.5)/log(14,3)', log(14,1.5)/log(14,3))
print()

which produces the results:

7.12853    6.50872
2.63093    2.40217
2.70951    2.70951

The third and sixth results agreed exactly (i.e. in every decimal place) although one had one more place than the other. What do these results show? We are trying to test the theory that: $$log_a(x)=c.log_b(x)$$ where $x$ can vary for bases (b) other than $e$. So: $$\frac{log_a(x_1)}{log_b(x_1)}=\frac{log_a(x_2)}{log_b(x_2)}$$ because they both equal $c$. This is exactly what was being tested. Rearranging: $$\frac{log_a(x_1)}{log_a(x_2)}=\frac{log_b(x_1)}{log_b(x_2)}$$ $$log_{x2}(x_1)=log_{x2}(x_1)$$ And so $1=1$! The last step invokes the aforementioned change of base rule (which can be proven from $x=b^{log_b(x)}$ and the power law of logarithms). The upshot of all this is that it is very misleading to say that $\int c/x = c.log_e(x)+C$. It could use any logarithm at all.

EDIT: to complete the differentiation of $log(x)$ take $(2)$ above (that is how it 'comes out'), convert it to: $$\int\frac{log_a(e)}{x}=log_a(e).log_e(a).log_a(x)+C$$ from David K (2), then do $log_e(a).log_a(x)=log_e(x)$ and $log_a(e)=c$ (because $a$ is no longer related to $x$) to obtain $(1)$ above. Only the natural logarithm allows us to do this - to find the integral of $c/x$ we find $c.ln(x)$ instead of having to work out the base of $log_a(e)=c$ and then the value of $log_a(x)$. Before computers this would probably have been the deciding factor in choosing the best technique (because natural logarithms were tabulated).

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  • $\begingroup$ Why do you say that $\ln$ is base agnostic? It usually denotes the natural logarithm: $\ln x = \log_e x$. WolframAlpha does not follow this convention and uses $\log x$ to indicate the natural logarithm. $\endgroup$ – rubik Apr 29 '16 at 14:55
  • $\begingroup$ I don't think I understand the question. Are you asking about the equality $\log_a(x)=c\log_e(x)$ where $c$ depends only on $a$? Certainly one could replace $e$ with some $b$ as well. You are not stuck with the natural logarithm, it is just that you can choose to reduce any other logarithm to it, and there are various reasons why that is convenient. As for integration, yes $\int c/x dx can be written as d log_a(x) + C$, but only with $a=e$ do you have $d=c$. $\endgroup$ – Ian Apr 29 '16 at 14:59
  • $\begingroup$ Because if you follow the reasoning in the differentiation of ln(x) there is nothing to say it must be base e. It just uses the laws of logarithms. $\endgroup$ – user301988 Apr 29 '16 at 14:59
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    $\begingroup$ Could you clarify what your question is? $\endgroup$ – arctic tern Apr 29 '16 at 14:59
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    $\begingroup$ @selfawareuser Um... no, I did not. $\endgroup$ – fleablood May 9 '16 at 15:09
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Repeating the quotation from Wikipedia:

However, logarithms in other bases differ only by a constant multiplier from the natural logarithm, and are usually defined in terms of the latter.

There is nothing here that says this property of the natural logarithm is unique. Indeed, if \begin{align} \log_a(x) &= A\log_e(x) &&\text{and}& \log_b(x) &= B\log_e(x), \end{align} then $$ \frac{\log_a(x)}{\log_b(x)} = \frac{A\log_e(x)}{B\log_e(x)} = \frac AB $$ and therefore $$ \log_a(x) = \left(\frac AB\right) \log_b(x). $$ That is, $A/B$ is the constant multiplier by which the logarithms in base $a$ and base $b$ differ, and there will always be some such constant multiplier no matter which two bases are used.

So far, this agrees with your conclusions, keeping in mind that nothing we have found contradicts the quote from Wikipedia.

The reason for choosing the natural logarithm as the one that "usually" appears in the definitions of other logarithms has to do with the fact that many mathematicians find the natural logarithm generally convenient for things that many mathematicians usually like to do. (The word "usually" in the Wikipedia entry should be a clue that the definitions do not need to be made this way. It's also not true that the natural logarithm is everyone's favorite logarithm all the time. In some applications, other logarithms are sometimes preferred.)

But the natural logarithm is unique in the way it solves the integral you were looking at, $$ \int\frac cx\,dx = c \log_e(x) + C. \tag1 $$ The natural logarithm is the only logarithm for which you can correctly write that equation with the same multiplicative constant $c$ on both sides, without introducing any other multipliers. You can of course use the "every logarithm is proportional to every other logarithm" property to write $$ \int\frac cx\,dx = (c \log_e(a)) \log_a(x) + C. \tag 2 $$ That's a simple formula, but not as simple as Equation $(1)$, because we need to insert a factor of $\log_e(a)$ that Equation $(1)$ does not require; in general, $c \log_e(a) \neq c$ except when $a = e$.

Up to this point, the constant $c$ in Equations $(1)$ or $(2)$ can be anything we want: $c=1$, $c=2$, $c=-10$, $c=e$, or $c=100017$, any of those will work. But suppose we give up all that freedom and commit to the statement that $c = \log_a(e)$. Then Equation $(2)$ becomes \begin{align} \int \frac{\log_a(e)}{x}\,dx &= (\log_a(e) \log_e(a)) \log_a(x) + C \\ &= \log_a(x) + C. \end{align}

So we've eliminated $e$ from the right-hand side, but now look, there it is on the left-hand side where there was no $e$ before. Furthermore, what happened to our so versatile formula to solve the integral $\int \frac cx\,dx$? How can we solve $\int \frac 3x\,dx$? We can't simply set $c=3$, because we've already committed to setting $c = \log_a(e)$. OK, there's a way around this: choose $a$ so that $\log_a(e) = 3$. We can do this by setting $a = \sqrt[3]e$. Then \begin{align} \int \frac3x \,dx &= \int \frac{\log_{\sqrt[3]e}(e)}{x}\,dx \\ &= \log_{\sqrt[3]e}(x) + C \end{align} and there's that pesky base-of-the-natural-logarithm $e$ again; it's back on the right-hand side now, but used in a way that's much weirder and harder to understand than before.

So happiness in integrating any multiple of $1/x$ is best achieved by remembering and using Equation $(1)$ exactly as written, without trying to get rid of the natural logarithm that "naturally" appears there.


I will grant you that some of the statements on the Wikipedia page are not so defensible. For example,

Another sense in which the base-e-logarithm is the most natural is that it can be defined quite easily in terms of a simple integral or Taylor series and this is not true of other logarithms.

This is leaning rather heavily on a preconceived notion of what is "simple". As we saw in Equation $(2)$, it just takes one more multiplicative factor in order to define any logarithm in terms of a simple integral. Granted, that one factor happens to be the natural logarithm of our other logarithm's base; but it's just one number; we do not have to develop the entire natural logarithm function to derive it. It's better (in my opinion) to say that the definition of the natural logarithm by these means is simpler than the corresponding definitions of other logarithms.

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  • $\begingroup$ 'The natural logarithm is the only logarithm for which you can correctly write that equation with the same multiplicative constant $c$ on both sides, without introducing any other multipliers.' It's worth making the reason for that explicit: $c.log_e(x)=log_a(e).log_e(x)=log_a(x)$. The last step depends on the change of base rule, which doesn't work if $e$ is replaced with $b$. $\endgroup$ – user301988 May 3 '16 at 6:29
  • $\begingroup$ Also, for completeness: $log_e(x)=log_e(a).log_a(x)$. $\endgroup$ – user301988 May 5 '16 at 0:10
  • $\begingroup$ 'Up to this point, the constant $c$ in Equations (1) or (2) can be anything we want... any of those will work. But suppose we give up all that freedom and commit to the statement that $c=log_a(e)$.' Not too sure about that because it has to cancel with $log_e(a)$ and the value of $a$ affects the $log_a(x)$ term. $\endgroup$ – user301988 May 5 '16 at 0:22
  • $\begingroup$ In other words: $a$ and $c$ are related, and the natural logarithm is used because only then do they become truly independent i.e. $a$ is eliminated. $\endgroup$ – user301988 May 5 '16 at 0:29
  • $\begingroup$ @David K You can also Wiki edit , right? $\endgroup$ – Narasimham May 6 '16 at 3:03
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There is a mistake in your algebra. From your third equation $$\int\frac{\log_a(e)}{x}\,dx=\log_a(x)+C\tag1$$ you can deduce that for any $c$ $$\int\frac cx\,dx={c\over\log_a(e)}\log_a(x)+C.\tag2$$ (This is equivalent to the Wolfram result). Comparing (2) to your second equation, it doesn't follow that $c=\log_a(e)$.

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  • $\begingroup$ Have gone back and forth between the two integrals, still can't see it... $\endgroup$ – user301988 Apr 29 '16 at 15:19
  • $\begingroup$ Note the RHS of my last equation has $\log_a(x)$ instead of $\log_e(x)$. Concluding that $c=\log_a(e)$ is equivalent to asserting $a=e$. $\endgroup$ – grand_chat Apr 29 '16 at 15:24
  • $\begingroup$ Are you unclear how (2) follows from (1)? Or unclear why it's incorrect to conclude $c=\log_a(e)$? $\endgroup$ – grand_chat Apr 29 '16 at 15:28
  • $\begingroup$ This post follows on via the linked one to this one on $ln'(x)$. The last line in the proof (taken from Wikipedia) is $ln'(x)=\frac{ln(1+u)^{1/u}}{x}$ but as I said the proof is base independent so $c$ has to be $log_a(e)$, $u \rightarrow 0$. $\endgroup$ – user301988 May 2 '16 at 7:41
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You're just misunderstanding what the article (in Wikipedia, here) says. The quote that explains why the natural log is uniquely called "natural" is this one:

The natural logarithm can be defined for any positive real number $a$ as the area under the curve y = $1/x$ from $1$ to $a$ (the area being taken as negative when $a<1$). The simplicity of this definition, which is matched in many other formulas involving the natural logarithm, leads to the term "natural".

This is clear enough, as long as you agree that "integral of $1/x$" is simpler in some sense than "integral of $a/x$" for any other $a$. The quote you're confused about, then, is this one:

Logarithms can be defined to any positive base other than $1$, not only $e$. However, logarithms in other bases differ only by a constant multiplier from the natural logarithm, and are usually defined in terms of the latter.

Here the article is saying that, although logarithms in other bases do exist, when viewed as functions they differ in a fairly trivial way from the natural logarithm (i.e., by a multiplicative constant). It's not saying this fact about the natural logarithm makes it special (as you say, it doesn't).... the "natural" part was explained previously. It's just saying that anything fundamental there is to learn about logarithms in general can be understood by studying a particular logarithm, since they're all very closely related -- and if you're going to study just one logarithm, it may as well be the natural one. (Similarly, we could study some multiples of the trigonometric functions, but $\sin x$ and $\cos x$ are "special" in that $\sin^2 x + \cos^2 x=1$, which is simpler than $(a\sin x)^2+(a\cos x)^2=a^2$ for any other positive $a$.)

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Only homogeneous equations with logarithms are invariant to change of base.

Examples:

  • $4 \log 2 \log(xy) = 2 \log 4 \log(x) + \log 16 \log (y)$

is base independent but

  • $\log 1000000 = 6$

is true only for $\log_{10}$.

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Say you are given $$\frac{dy} {dx }=\frac{2 y} { x }$$

The integration steps are:

$$\frac{dy} {y }=\frac{2 \, dx } { x }$$

$$ \int \frac{dy} {y }=\int \frac {2 \, dx } { x }$$

$$ \log y = 2 \log x + \log C $$

Note that instead of the usual $C$ we choose to put in $ {\log C}$ for manipulative convenience because after all it is an arbitrary constant and any function of an arbitrary constant is also arbitrary.Exponentiate on both sides,

$$e^{ \log y } = e^ {2 \log x } \cdot e^{ \log C } $$ that is $$ y= C \cdot x^2 $$ where the constant of integration along with log has now become a multiplier.

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  • $\begingroup$ If you could fill in some of the intermediate steps that would be good. $\endgroup$ – user301988 May 6 '16 at 4:58
  • $\begingroup$ Steps added, better now? $\endgroup$ – Narasimham May 6 '16 at 5:54
  • $\begingroup$ Thanks, will check the logic... $\endgroup$ – user301988 May 7 '16 at 23:54

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