Is the natural logarithm actually unique as a multiplier?

Solution 1:

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 $c=\log_a(e)$.

Solution 2:

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.

Solution 3:

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 $a$.)