<s>[INST] Between any two powers of $5$ there are either two or three powers of $2$ 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?
 [/INST] Basically, it boils down to the fact that $5$ is between $2^2 = 4$ and $2^3 = 8$. Here's a proof, though.
Let $5^a$ and $5^{a+1}$ be the two consecutive powers of $5$. Let $2^b$ be the smallest power of $2$ that exceeds $5^a,$ and $2^c$ the largest below $5^{a+1}$.
Then we have $2^{b-1} \leq 5^a < 2^b$ and $2^c < 5^{a+1} \leq 2^{c+1}$.
From this we get
$$\frac{2^{c}}{2^{b}} < \frac{5^{a+1}}{5^a}  \leq \frac{2^{c+1}}{2^{b-1}} $$
or, equivalently,
$$2^{c-b} < 5 \leq 4 \cdot 2^{c-b}.$$
This inequality can be rewritten as $ \frac{5}{4} \leq 2^{c-b} < 5$, which proves that $c - b$ is either $1$ or $2$. Thus the powers of $2$ between $5^a$ and $5^{a+1}$ are either $2^b, 2^{b+1} = 2^c$, or $2^b, 2^{b+1}, 2^{b+2} = 2^c.$ This is what you wanted.
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