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Let $x\in\mathbb{R},x>0$. Prove that there exists one and only $n\in\mathbb{Z}$ so that $10^{n}\leq x<10^{n+1}$.

I thought about using the theorem that for every $x$: $a\leq x <a+1$ when $a\in \mathbb Z$ and then by using logarithm rules get to the inequality in the question, but couldn't get to it. Also how would I prove that there is only one $n$ for this?

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  • $\begingroup$ Any positive real number $x$ has representation $\lfloor x\rfloor=\sum a_i\cdot10^i$ so the statement is trivial. $\endgroup$ – TheSimpliFire Apr 18 '20 at 15:59
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$10^n \leq x < 10^{n+1} \iff n \leq \log_{10} x < n+1$. As you note there is a unique such $n \in \mathbb{Z}$. The $\iff$ follows from the fact that $\log_{10}$ is a bijection of $(0,\infty) \to \mathbb{R}$.

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