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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?

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    $\begingroup$ Hint: use logs. $\endgroup$
    – lulu
    Apr 19, 2019 at 20:29

1 Answer 1

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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.

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  • $\begingroup$ It's that simple? I feel like a dumbass. Thanks. $\endgroup$ Apr 19, 2019 at 20:34
  • $\begingroup$ I hope so! I may be wrong... Someone please correct me if so. $\endgroup$ Apr 19, 2019 at 20:34
  • $\begingroup$ Actually, we need to account for the zeroth power of 6 (being 1), so we need to add one to both of these expressions, which I have now done. $\endgroup$ Apr 19, 2019 at 20:36
  • $\begingroup$ It is pretty common to consider $0$ to be a natural number. Unfortunately it is also pretty common to consider $0$ not to be a natural number. In general just the appearance of the term "natural number" is not enough to tell you which meaning the writer had in mind. $\endgroup$ Apr 19, 2019 at 21:08
  • $\begingroup$ I suppose that’s true. I’m not trying to be biased one way or the other actually, it’s just that the comparatively lower level mathematics of which I am most familiar so far always begins the naturals at one. $\endgroup$ Apr 19, 2019 at 21:10

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