The following is a nice way to think about logarithms: \(\log_b a \equiv\) The number of times \(a\) is divided into \(b\) parts such that each part becomes \(1\).

For example, \(1000\) needs to be divided into 10 parts \(3\) times to get to 1, so \(\log_{10}1000 = 3\).

Reaching arbitrary numbers by repeated division

Now, we may ask—how many times should \(a\) be divided into \(b\) parts to reach an arbitrary number \(x\)? We can split \(a\) reaching \(1\) to \(a\) reaching \(x\) and then \(x\) reaching \(1\).

Let \(k \equiv\) The number of times \(a\) is divided into \(b\) parts to reach \(x\).

We have, \(\log_b a = k + \log_b x \\ \implies k = \log_b a - \log_b x = \log_b \left (\frac a x \right )\)

Intuition for the change of base rule

We will use this to build intuition for the following identity: \(\log_qa = \log_pa . \log_qp\)

Informally, the above asks, how do we relate the number of times we divide \(a\) in \(q\) parts to reach \(1\) and the number of times we divide \(a\) in \(p\) parts to reach \(1\)?

Let us look at the first step of calculating \(\log_a p\). We divide \(a\) into \(p\) parts, and each part is of size \(a/p\). Just to perform this first step, if I was only allowed to divide by \(q\), how many divisions would I need to reach \(a/p\) ?

Well, this is the question we asked in the previous section! The number of divisions is equal to \(\log_q \frac{a}{a/p} = \log_q p\). Notice this is independent of \(a\), so this holds true to reach any \(x/p\) from \(x\), in other words, it holds for every step of the calculation of \(\log_a p\).

So, we have \(\log_q p\) divisions [in the process where you only divide by \(q\) to reach \(1\)] for each of the \(\log_p a\) divisions [in the process where you only divide by \(p\) to reach \(1\)].

Which gives us: \(\log_q a = \log_p a. \log_q p\)

Yayyy!

I would love to be able to formalize the intuition I have here, and also extend it for other logarithm identities.