Everything Is Logarithms

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Everything Is Logarithms

May 25, 2026

Some connections between things, which I have not seen elsewhere. Maybe they mean something?

1. The Baseless Logarithm

Normally one writes a logarithm with a base, \(\log_b (x)\), to mean

\[y = \log_b (x) \Lra b^y = x\]

And then you can change the base of the logarithm with

\[\log_b (x) = \frac{\log_a (x)}{\log_a(b)}\]

Which follows from rearranging \(\log_a (x) = \log_a (b^{\log_b x}) = \log_b (x) \times \log_a (b)\).

One way of thinking about what this formula does is that it is a change of units. Similar to writing \(2 \text{ km} = 2000 \text{ m} / \frac{1000 \text{ m}}{1 \text{ km}}\) or \(5 \text{ bytes} = 40 \text{ bits}/\frac{8 \text{ bits}}{1\text{ byte}}\). It says: how many copies of \(b\) are in \(x\)? It’s the number of copies of \(a\) in \(x\), divided by the number of copies of \(a\) that are in \(b\).

This is perfectly simple, but for some reason it’s hard to think about logarithms that way. The notation kind of… obfuscates things? Specifically it is hard to read \(\log_b x\) as “how many copies of \(b\) are in \(x\)”, because that English expression should correspond to the notation \(x/b\), not \(\log_b x\).

I found a way of thinking about logarithms which I think makes this clearer, but you have to allow a sort of odd object that I am call the baseless logarithm. It is simply a logarithm without a base:

\[\log N\]

which we regard as an abstract object, not a number. Then we write our normal “based” logarithm as a ratio of two of these baseless logarithms:

\[\log_2 N = \frac{\log N}{\log 2}\]

Note, this is already sort of a thing people colloquially do, e.g. leaving out the base of logarithms in asymptotic formulas. But I do not mean it as a shorthand. It is useful to regard it as an actual algebraic object.

We interpret \(\log 2\) as being the unit “bits”. To write \(\log N\) in bits is to factor it as a multiple of \(\log 2\):

\[\log N = \frac{\log N}{\log 2} \log 2 = \log_2 (N) \log 2 = \log_2 (N) \text{ bits}\]

Then the change-of-base for logarithms follows from just writing the same geometric quantity in different units. For example \(\log e\) as a unit is sometimes called “nats”:

\[\begin{aligned}<br>\log N = \frac{\log N}{\log 2} \log 2 = \log_2 (N) \text{ bits} = \frac{\log N}{\log e} \log e = \ln (N) \text{ nats}<br>\end{aligned}\]

The baseless \(\log N\) is sort of the multiplicative version of an object that might be familiar from discussions of vectors. It is common with vectors to distinguish between points and displacements: a displacement vector \(\b{v}\) is given by the difference of two points \(\v = (b) - (a)\). When we write think of points as having coordinates, this involves an explicit choice of origin \(\O\), such that \(\b{a} \equiv (a) - \O\) and \(\b{b} \equiv (b) - \O\). Then a displacement vector is constructed by subtracting off the factors of \(\O\), \(\b{v} = \b{b} - \b{a} = ((b) - \O) - ((a) - \O) = (b) - (a)\). The baseless logarithm implemens the same thing but with multiplication: the value \(\log N\) may be thought of as \(\log N / \log \O\) for an unspecified choice of origin; turning it into an actual numeric value involves dividing two such logarithms to cancel out the origin, \(\log_M N = \log N / \log M = (\log N / \log \O) / (\log M / \log O)\). I think of \(\log N\) as the point corresponding to \(N\) and \(\log N / \log \O\) as its corresponding displacement vector once you pick a coordinate system. I prefer to think of the point as more fundamental.

You might ask: if we have a baseless logarithm \(\log N\), do we also have a “baseless exponential”? Normally \(b^{\log_b N}\) can be written as something like \(b^{\log_b N} = b^{\ln N / \ln b} = e^{\ln N} = N\); is there any way to do this without actually choosing a base? I think the answer has to be “no”. All we can say is that we have split the one object, a logarithm \(\log_b N\) which is the solution of \(b^y = N\), into two objects, \(\log N\) and \(\log b\), each of which on their own are without “units” and so have no numerical meaning. It is just like points in space: a point on its own has no operation of addition and does not have a length. We can subtract points to produce vectors (relative to a symmetry group) but not add them, and the usual operations in coordinates all require a choice of origin.

In fact there are many surprising similarities between logarithms and vectors.

2. Logarithms are Vectors

When doing vector algebra and differential geometry in a properly covariant way, we distinguish between abstract vectors and vectors in a particular coordinate system. My personal convention for this is to refer to the abstract vectors as “geometric” vectors and always write them...