TITLE: Between any two powers of $5$ there are either two or three powers of $2$
QUESTION [13 upvotes]: Is this statement true?

Between any two consecutive powers of $5$, there are either two or
  three powers of $2$.

I can see that this statement is true for cases like
$$5^1 < 2^3 < 2^4 < 5^2$$
or
$$5^3 < 2^7 < 2^8 < 2^9 < 5^4$$   
But I am having a trouble figuring out the proof through generalization.
Could somebody help me?

REPLY [2 votes]: It's easy enough to see that this is indeed true, even without using logarithms.
Let $a$ be some power of $5$, and let $b$ be the least power of $2$ not less than $a$.
Then the next power of $5$ after $a$ is obviously $5a$, while the next three powers of $2$ after $b$ are $2b$, $4b$ and $8b$.
Since, by definition, $a \le b$, it follows that $5a \le 5b < 8b$.  Thus, there can be at most three powers of $2$ ($b$, $2b$ and $4b$) between $a$ and $5a$.
Conversely, since $b$ is the least power of $2$ not less than $a$, it follows that $\frac12 b < a$.  Thus, equivalently, $2b < 4a < 5a$, so there must be at least two powers of $2$ ($b$ and $2b$) between $a$ and $5a$.

BTW, if you look closely at the proof above, you may note that it doesn't actually use the assumption that $a$ is a power of $5$ anywhere.  Thus, in fact, we've proven a more general result: for any positive number $a$, there are either two or three powers of $2$ between $a$ and $5a$.
(In fact, the proof doesn't really use the assumption that $b$ is a power of $2$, either, so we could generalize the result even further in this direction if we wanted!)
Also, you may notice that the key observation behind the result above is that the number $5$ lies strictly between $2^2 = 4$ and $2^3$ = 8.  Thus, by essentially the same logic as above, we can prove a similar result for other bases:

Between any two consecutive powers of $x$ there are at least $k$ and at most $k+1$ powers of $y$, where $x$ and $y$ are any numbers greater than $1$, and $k$ is the largest integer such that $y^k \le x$.