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 $5a$.

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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 if we wanted!)

More generally, we can see that the essential fact that makes this proof work 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 $x^k \le y$.