Intuition behind logarithm change of base

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}

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 $log_2 64 = 6$

It might help to crank though some easy examples

$4096 = 2^{12} = (2^2)^6 = (2^3)^{4}$

$\log_2 4096 = 12\\ \log_4 4096 = \frac {12}{2} = \frac {\log_2 4096}{\log_2 4}\\ \log_8 4096 = \frac {12}{3} = \frac {\log_2 4096}{\log_2 8}\\ \log_{16} 4096 = \frac {\log_2 4096}{\log_2 16} = 3 \implies (2^4)^3 = 2^{12}$

Motivated by @dxiv's comment, we can adapt your table to understand the change of base rule, although I'm just going to illustrate changing base from $2$ to $4$ and prove the formula in the (equivalent) form $$ \log_2 x = log_2 4 \times \log_4 x. $$ Let's add a row to your table:

$$\begin{array}{rrrrrrrrrr} \text{exponent}& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ \text{power of 2}&1& 2& 4& 8& 16& 32& 64& 128& 256& 512& 1024\\ \text{power of 4}&1& 4& 16& 64& 256& 1024& 4096& 16384& 65536& 262144& 1048576 \end{array}$$

The top row, labeled exponents, is where you read off the logarithms, and they can be to the base $2$ or the base $4$ depending on which lower row you are looking at.

Imagine that there are counters in the bottom two rows, and they both start in the first column, on their $1$. Start moving the bottom counter forward, one column at a time. Now suppose the 'powers of 2' counter starts moving forward and it tries to stay on the same number as the 'powers of 4' counter is on. So when 'powers of 4' jumps to '4', 'powers of 2' has to move two columns to get to its 4. When 'powers of 4' then moves to 16, 'powers of 2' again has to move two columns to get to its 16. I.e. the powers of 2 counter has to move twice as fast as the powers of 4 counter. That means when you read off the logarithms from the top row $$\log_2 x = 2 \times \log_4x.$$

If you added a 'powers of 8' row you could easily see that $\log_2 = 3 \times \log_8x$, and then generalize it to compare $\log_a$ and $\log_{a^n}$.