I've been trying to solve this many different ways but I am completely stumped. How does one go about this?
What I'm trying to figure out is something like this... (n is a natural number) How many numbers are there smaller than the n-th power of 3 ($3^n$), that are powers of 6? ($6^n$)
I've been trying to figure out an equation for this without success.
Example:
For the 2nd power of 3 ($3^2$), we have 2 occurences of a power of 6. (1 and 6). For the 4th power of 3 ($3^4$), we have 3 occurences of a power of 6. (1, 6 and 36)
Is there a formula that can be derived to establish this relationship?
The number of integers of the form $6^n$ below m is equal to $ \lfloor \log_6{m} \rfloor + 1$. So the number of powers of 6 below the nth power of 3 is equal to $\lfloor \log_6{3^n}\rfloor + 1$.
You can interpret this intuitively as being the number of times we must multiply $1$ by $3$ before we reach $m$, or just overshoot it. If we overshoot it, then the floor function ensures that we only calculate the number of powers that actually lie below $m$. We add one to each of these expressions because we want to include $1$ as a power of $6$.
This is based on your example where you include $1$ as a power, but it’s worth noting that this potentially contradicts your original description of $6^n$ ‘where $n$ is a natural number’ as the natural numbers are often, although not always, assumed to begin at $1$. Of course, you can add $1$ or not, depending on what you want.