Intuition behind logarithm change of base I try to understand the actual intuition behind the logarithm properties and came across a post on this site that explains the multiplication and thereby also the division properties very nicely:

Suppose you have a table of powers of 2, which looks like this: (after revision)
$$\begin{array}{rrrrrrrrrr}
0&1&2&3&4&5&6&7&8&9&10\\
1&2&4&8&16&32&64&128&256&512&1024
\end{array}$$
Each column says how many twos you have to multiply to get the number in that column.  For example, if you multiply 5 twos, you get $2\cdot2\cdot2\cdot2\cdot2=32$, which is the number in column 5.
Now suppose you want to multiply two numbers from the bottom row, say $16\cdot 64$.  Well, the $16$ is the product of 4 twos, and the $64$ is the product of 6 twos, so when you multiply them together you get a product of 10 twos, which is $1024$.

I found that very helpful to understand the actual proofs for this property.
I still struggle to get the idea behind the change of base rule. I'm familiar with the proof that goes like:
$$\log_a x = y \implies a^y = x$$
$$\log_b a^y = \log_b x$$
$$y \cdot \log_b a = \log_b x$$
$$y = \frac{\log_b x}{\log_b a}$$
But can somehow provide a explanation in the style of the quoted answer why this actually works?

Intuition is always tricky to get across, but I can try.
$\log_bx$, as you noted, tells you how many $b$s you need to multiply together to get $x$. Now if you need $\log_ba$ number of $b$s to multiply to get $a$, and you need $\log_ax$ number of $a$s to multiply to get $x$, we can "expand" each of those $a$s into a number of $b$s. There will be $\log_ba$ number of $b$s for each $a$, so the total number of $b$s will be $\log_ax \log_ba$. These $b$s multiply to $x$, so $\log_ax \log_ba = \log_bx$.
For example, take $b=2, a=8, x=64$. We start with $\log_ax = 2$, which tell us we need two $8$s to get $64$:
$$
8 \cdot 8 = 64
$$
We use $\log_ba = 3$, i.e., $2 \cdot 2 \cdot 2 = 8$, to expand each $8$:
$$
(2 \cdot 2 \cdot 2) \cdot (2 \cdot 2\cdot 2) = 64
$$
Now the total number of $2$s we are multiplying is $2 \cdot 3 = 6$, so $log_2 64 = 6$
