TITLE: How does calculus without Euler's number (e) look?
QUESTION [18 upvotes]: I'm starting to study calculus and I've become very interested in Euler's number ($e$). I understand that the property of being its own derivative makes it the "natural" base to work on for studying rates of change.
However, I was wondering what would happen if we pretended not to know about the existence of $e$. Would trying to find the derivative of something like $a^x$ lead us into finding the definition of $e$ or is it possible to avoid $e$ altogether?
In this video it says that not using $e$ in calculus leads to some pretty crazy math. What does that math look like?

REPLY [4 votes]: Nothing would change really, because $e$ isn't that useful anyway.
That may seem a bit of a crazy statement, but bear with me. $x\mapsto e^x$ is of course an extremely useful function, just, it doesn't actually hinge much on Euler's number.
Starting from algebra, it doesn't make that much sense to calculate the derivative $\frac{\partial}{\partial x} b^x$ in the first place, for any basis $b\neq1$, because $b^x$ simply isn't defined for $x\in\mathbb{R}$, and you need real arguments in order for derivatives to make sense†.
You can, however, leave “iterated-multiplication exponentiation” completely aside and just define the function $\exp$ with the specific goal of fulfulling $\frac{\partial}{\partial x} (\exp x) = \exp x$. After all, that is a Lipschitz-continuous ordinary differential equation; hence if we choose $\exp 0 = 1$ as the starting condition, the Picard-Lindelöf theorem tells us that this uniquely determines the function $\exp$ for all real arguments.
You can then (using Taylor expansion or integral formulas) go on to observe that $\exp$ behaves in every regard as if it was a function of the form $x\mapsto b^x$. Namely, you have
$$
    \exp(p+q) = \exp p \cdot \exp q
$$
and can (using the starting condition) therefore, at least for rational $n$,  always write
$$
  \exp n = (\exp 1)^n
$$
(which we know as $\exp n = e^n$, but that's just a trivial shorthand definition of $e = \exp 1$.)
By introducing also logarithms as the inverse to $\exp$, you can then write any power function in terms of $\exp$ and that readily allows you to differentiate any such function. But that doesn't at all require that you've ever introduced the convention of what $e^x$ means. So, you'd just get slightly longer formulas because you'd occasionally need to write out $\exp 1$, but the maths as such wouldn't change at all.

†This isn't in fact that much of a problem because $b^x$ is readily defined on all rationals, these are dense in $\mathbb{R}$, and you can show that $b^x$ maps Cauchy sequences to Cauchy sequences, which can also be used to define arbitrary-basis realexponential functions as the unique continuous extension of the rational exponentials. Problem is, the limits of the resulting Cauchy sequence are not in $\mathbb{Q}$, and computing them isn't very practical without using $\exp$ as a proxy.