Real analysis: Prove that x is in between positive and negative power, n, of 10 for natural number n

Question

Prove that if x ∈ R and x > 0, there exists an n ∈ N such that $10^{-n}$ < x < $10^n$.

Having trouble figuring out how the powers work. Do I need to use Archimedean property in any way?

Answer 1

Without appealing to the logarithm, the fact that $1/10 < 1$ implies that $(1/10)^n$ converges to $0$, so there exists an $m$ with $10^{-m} < x$ since $x >0$. Also, $10^n$ is unbounded, so there is an $N$ with $x < 10^N$. Taking $n=\max\{m, N\}$ gives the desired inequality.