# How often does a power of 3 occur compared to power of 6

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?

• Hint: use logs.
– lulu
Apr 19, 2019 at 20:29

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.
• 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. Apr 19, 2019 at 21:08