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?

It might help to crank though some easy examples
$4096 = 2^{12} =  (2^2)^6 = (2^3)^{4}$
$\log_2 4096 = 12\\
\log_4 4096 = \frac {12}{2} = \frac {\log_2 4096}{\log_2 4}\\
\log_8 4096 = \frac {12}{3} = \frac {\log_2 4096}{\log_2 8}\\
\log_{16} 4096 = \frac {\log_2 4096}{\log_2 16} = 3 \implies (2^4)^3 = 2^{12}$
