The change of base formula states:

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

There exists an algebraic proof. This explanation is meant to give you more intuition for why the formula is what it is.

A multiplicative scale

Consider a number line where we reach a number \(x\) by repeatedly multiplying by some fixed number \(a > 1\). Take \(a = 2\). Starting from 1:

1 2 ×2 4 ×2 8 ×2 16 ×2 32 ×2

The number 8 is three steps from 1. The number 32 is five. The position of any number \(x\) on this scale — the number of steps from 1 — is what we call \(\log_2(x)\). For an arbitrary \(x\), the length from 1 to \(x\) is \(\log_2(x)\).

1 x log2(x)

Placing b on the scale

Now introduce another number \(b > 1\). It too has a place on our scale, at position \(\log_a(b)\). Take \(b = 8\). On the base-2 scale, 8 sits at step 3.

1 2 4 8 b x log2(b) = 3

The segment from 1 to \(b\) has a definite length on the scale: \(\log_a(b)\) steps. But \(b\) is also a perfectly good multiplier in its own right. Multiplying by \(b\) once gets you from 1 to \(b\). Multiplying again gets you from \(b\) to \(b^2\). You can traverse the same line using \(b\) as the step — in larger jumps.

The ratio

The segment from 1 to \(x\) has length \(\log_a(x)\). The segment from 1 to \(b\) has length \(\log_a(b)\). Each jump of \(\times b\) covers exactly \(\log_a(b)\) units on the scale. So we ask: how many copies of the smaller segment fit into the larger one?

1 b b² b³ x ×b ×b ×b loga(x) loga(b)
\[\frac{\log_a(x)}{\log_a(b)}\]

This quotient counts the number of times you must multiply by \(b\) to reach \(x\). That, by definition, is \(\log_b(x)\).