TITLE: What's so "natural" about the base of natural logarithms?
QUESTION [82 upvotes]: There are so many available bases. Why is the strange number $e$ preferred over all else?
Of course one could integrate $\frac{1}x$ and see this. But is there more to the story?

REPLY [10 votes]: I'm surprised I never answered this; maybe I was deterred by the fact that several other answers are here.
One short answer is this: An exponential function $y=a^x$ grows at a rate proportional to its present size, but only when the base is $e$ does it grow at a rate equal to its present size.  In other words
$$
\frac{d}{dx} a^x = \left(\text{constant}\cdot a^x \right),
$$
but only when $a=e$ is the "constant" equal to $1$.
The number $a=2$ is too small for this to happen.  To see that consider
$$
\frac{d}{dx} 2^x = \lim_{h\to0} \frac{2^{x+h}-2^x}{h} = \lim_{h\to0}2^x\frac{2^h-1}{h}
$$
This last limit is equal to $\displaystyle 2^x \lim_{h\to0}\frac{2^h-1}{h}$.  That step can be done because $2^x$ is a "constant", but in this instance, "constant" means "not depending on $h$".  Then observe that $\displaystyle\lim_{h\to0}\frac{2^h-1}{h}$ is a "constant", where "constant" now means "not depending on $x$".
So
$$
\frac{d}{dx} 2^x = \left(\text{constant}\cdot 2^x\right).
$$
But what number is this "constant"?  Notice that as $x$ increases from $0$ to $1$, $2^x$ increases from $1$ to $2$, so the average slope on that interval is $\dfrac{2-1}{1-0}=1$.  Since the curve gets steeper as $x$ increases, it's not yet that steep at $x=0$.  Its slope at $x=0$ is $\left.\dfrac{d}{dx}2^x\right|_{x=0}=\left(\text{constant}\cdot2^0\right)$, so that "constant" must be less than $1$.
A similar argument shows that if $4$ is used as the base, the "constant" is more than $1$.  This is done by using the interval from $-1/2$ to $0$ instead of the interval from $0$ to $1$.
So $2$ is too small, and $4$ is too big, to be the natural base. $e$ must be somewhere between $2$ and $4$.  In a similar way one can show that $3$ is to big, but that's where the previously simple arithmetic gets messy.  Use the interval from $-1/6$ to $0$ for that.
Similarly with logarithms:
\begin{align}
\frac d {dx} \log_6 x & = \frac{\text{constant}} x, \\[12pt]
\text{ and } \quad \frac d {dx} \log_e x & = \frac{\text{constant}} x, \text{but this time, the constant is 1.}
\end{align}
What is natural about $e$ is the same thing that is natural about radians:
$$
\frac d {dx} (\text{sine of $x$ degrees}) = \Big((\text{cosine of $x$ degrees}) \times \text{constant} \Big)
$$
but only when radians are used is the "constant" equal to $1$. $($With degrees the constant is $\pi/180.)$