Why the $\log$ is so special?

Question

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 $\exp x$?

Answer 1

Not long ago I saw a graph showing the price of bitcoin over the past four years or so (currently in the $\$400\text{--}\$500$ range; it was between $\$2$ and $\$3$ in the fall of 2011). It looked like this: \begin{array}{r} 5000 \\[20pt] 500 \\[20pt] 50 \\[20pt] 5 \\[20pt] 0.50 \\[20pt] 0.05 \\[20pt] & & 2009 & \qquad\qquad & 2010 & \qquad\qquad &2011 & \qquad\qquad & 2012 & \qquad\qquad & 2013 \end{array} . . . with a curve, which you don't see here. Every time the price gets multiplied by $10$, a certain amound gets added to the height of the curve above the horizontal axis. In otherwords, the hieght on the vertical axis is a logarithm of the price.

It makes sense to do things this way, since if you buy $\$1$ worth of this commodity at any time, then its dollar-value at a later time depends only on how many times it got multiplied by $10$. Now suppose it went from $\$0.05$ to $\$0.50$ over some early six-month period, and in some much later six-month period of the same number of weeks it went from $\$50$ to $\$500$. In each case the logarithm increased by $1$ in the same time. How fast did the logarithm change? The rate was $1$ unit per six months. In both cases. How fast did the price change? The rate was $\$0.45$ cents per six months in the first case and $\$450$ per six months in the second case. Multiplying each by the reciprocal of the size of the price increase, we get $\$0.45\times\dfrac{1}{\$0.45}$ in the first case, i.e. $1$ unit per six months, and in the second case we get $\$450\times\dfrac{1}{\$450}$, again $1$ unit per six months. I.e. the rate at which the logarithm of the price changes is the rate of change of price times the reciprocal of the price. That's where $(d/dx)\log x = 1/x$ shows up.

A full answer to the question would be at least 100 times this long; this is just one example.

Answer 2

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.)