Defining the logarithm as an area

The logarithm can be defined without being introduced as the inverse of an exponential¹. We look at the way of defining it as the area under $y = \dfrac{1}{x}$.

Defining it by an area

Write $L(x)$ for the area from $1$ to $x$.

By the fundamental theorem of calculus $L'(x) = \dfrac{1}{x} > 0$, so $L$ increases monotonically.

A product turns into a sum

The key is how well the shape $\dfrac{1}{t}$ stands up to rescaling. Substituting $t = as$ turns $\dfrac{dt}{t}$ into $\dfrac{ds}{s}$.

Stretching the interval by a factor of $a$ leaves the area unchanged. From that, $L(ab) = L(a) + L(b)$ follows: a product turning into a sum, the defining property of a logarithm.

This is where $e$ comes from

The number $x$ with $L(x) = 1$ is what is called $e$. The position reached when the area is exactly $1$ is $e \approx 2.71828$, and $\int_1^e \dfrac{dt}{t} = 1$.

The inverse turns out to be the exponential

Writing $E$ for the inverse of $L$, the rule for differentiating an inverse turns $L'(x) = \dfrac{1}{x}$ into $E'(y) = E(y)$: a function unchanged by differentiation. Whichever of the exponential and the logarithm is put first, the same pair comes out.

The advantage of this route

It removes the need to work out in advance what $a^x$ means for an irrational exponent. Starting from the plain quantity of an integral of $\dfrac{1}{x}$, one builds the logarithm, then $e$, then the exponential, in that order.

The hyperbola on the graph is $y = \dfrac{1}{x}$, the gently rising curve representing the area is $y = \log x$, and the large dots are $(1, 0)$, where the area is $0$, and $(e, 1)$, where it reaches $1$.

1. Natural logarithm, Wikipedia