GoKawiilSome 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, akin 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 \(\log_b x\). “How many factors of \(b\) are in \(x\)” is a bit better, but it still feels off.
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:
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