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:
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 $\log_2(x)$.
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.
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?
This quotient counts the number of times you must multiply by $b$ to reach $x$. That, by definition, is $\log_b(x)$.