Prove that there exists one and only $n\in\mathbb{Z}$ so that $10^{n}\leq x<10^{n+1}$

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 $$x$$: $$a\leq x when $$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?

• Any positive real number $x$ has representation $\lfloor x\rfloor=\sum a_i\cdot10^i$ so the statement is trivial. – TheSimpliFire Apr 18 '20 at 15:59

$$10^n \leq x < 10^{n+1} \iff n \leq \log_{10} x < n+1$$. 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 $$(0,\infty) \to \mathbb{R}$$.