TITLE: Why the $\log$ is so special?
QUESTION [5 upvotes]: When I first learn about the logarithm function $\log$ or $\ln$. My professor said that $\log x$ is a function that when we derive we get the inverse function $1/x$. This $\log$ becomes very popular and "changes the world". I would ask why the $\log$ function is so special? What is the history behind such function?
I mean there are many other functions that we do not know their antiderivatives, e.g., $e^{x^2}$. Why we do not say that for example a function called '$\textrm{something}$' and when we derive it we get $e^{x^2}$. May be I give the bad example here because $e^{x^2}$ has something to do with Gauss but I mean why exactly $1/x$, $\log x$, and $\exp x$?

REPLY [3 votes]: Logarithms convert multiplication (difficult) into addition (easy). What follows is not the historical origin of $\log$, just one way of constructing $\log$ and $\exp$ using only tools of elementary calculus.
If we define
$$
\log x = \int_{1}^{x} \frac{dt}{t},\quad x > 0,
$$
then for all $a$, $b > 0$,
\begin{align*}
\log(ab) = \int_{1}^{ab} \frac{dt}{t}
  &= \int_{1}^{a} \frac{dt}{t} + \int_{a}^{ab} \frac{dt}{t} \\
  &= \int_{1}^{a} \frac{dt}{t} + \int_{1}^{b} \frac{a\, dt}{at} \\
  &= \int_{1}^{a} \frac{dt}{t} + \int_{1}^{b} \frac{dt}{t} \\
  &= \log a + \log b.
\end{align*}
The key fact is that the differential $dt/t$ is invariant under scaling $t \mapsto at$. (This calculation is not the historical origin of $\log$, but it's an elementary yet important application of change of variables.)
Here are two senses in which addition is easier than multiplication. First, if adding or multiplying two digits is regarded as an "operation", then adding two $n$-digit numbers entails $O(n)$ operations, while multiplying two $n$-digit numbers requires $O(n^{2})$ operations. Second, addition can be performed with an analog device by concatenating lengths of intervals; a slide rule performs multiplication in exactly this way.
Further, logarithms convert exponentiation into multiplication. Inductive arguments using
$$
\log(ab) = \log a + \log b
$$
show $\log(a^{n}) = n \log a$ for all integers $n$. Since $(a^{n/m})^{m} = a^{n}$, it follows that
$$
m \log(a^{n/m}) = \log (a^{n}) = n \log a,\quad\text{i.e.}\quad
\log(a^{n/m}) = (n/m) \log a.
$$
One might be led this way to define real exponentiation so that $\log(a^{x}) = x \log a$ for all real $x$. (In more detail, $\log$ is differentiable on $(0, \infty)$ and has positive derivative, so $\log$ is strictly increasing, and therefore has a differentiable inverse function $\exp$. (IIRC, the chain rule applied to the identity $\log\bigl(\exp(x)\bigr) = x$ shows $\exp' = \exp$.) Particularly, the function
$$
a^{x} = \exp(x \log a)
$$
is differentiable and extends the algebraic definition "$a^{n/m}$ is the $m$th root of $a^{n}$" to arbitrary real exponents.)