How to prove the following? Let $k \in \mathbb{N}$, so that $k \neq 10^n, n \in \mathbb{N}$. Prove then, that $\log(k)$ is irrational. I'm fairly new to proofs, so I've taken a look on the internet which gave me the following:
Assume that $\log(k) = \frac{m}{n}$ is rational number, with $m, n$ being positive integers. Then $$10^\frac{m}{n} = k$$ $$10^m = k^n.$$ It is obvious that equation does not hold, because $k \neq 10^n$.
Is this a right approach?
You have the right approach, but why is it "obvious" that $10^m=k^n$ is impossible? For instance, $4^3=8^2$ even though $8$ is not an integer power of $4$. To make this rigorous, you can think about the prime factorizations of $10^m$ and $k^n$.