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?
Best Answer
Here is one way of looking at it. (I'll assume that the numbers $a,b,x \in \mathbb R$ satisfy $a > 1, b > 1$, and $x > 0$.)
I dislike the name "logarithm" and I think a more descriptive name for $\log_b(x)$ is "the exponent from $b$ to $x$". We could also use the notation $[b \to x]$ instead of $\log_b(x)$. The change of base rule then tells us that the exponent from $b$ to $x$ is equal to the exponent from $b$ to $a$ times the exponent from $a$ to $x$: $$ \tag{$\spadesuit$}[b \to x] = [b \to a][a \to x] $$ or equivalently $$ [a \to x] = [b \to x]/[b \to a]. $$ In standard notation, this formula states that $$ \log_a(x) = \frac{\log_b(x)}{\log_b(a)}. $$
Note that equation $(\spadesuit)$ is obvious, because \begin{align} b^{[b \to a][a \to x]} &=(b^{[b\to a]})^{[a \to x]} \\ &= a^{[a \to x]} \\ &= x. \end{align}