Prove that $\log_5(2) \in \mathbb{R}\setminus \mathbb{Q}$ (irrational numbers).
I know there is a question out there already for this but my problem is that I need to prove this using the fundamental theorem of arithmetic. I'm not too good with proving via this method and so would appreciate clarification and help.
What I have so far:
Suppose by way of contradiction that $\log_5 (2)$ is rational,
then it can be written as $\log_5 (2)= m/n$ for some $m$,$n$ that are integers
Then $5^{m/n} = 2$ which is equivalent to $5^m = 2^n$. Now I'm stuck here. Where does the fundamental theorem of arithmetic come in and how can I use it to show that $5^m = 2^n$ cannot be true?
Best Answer
The fundamental theorem of arithmetic tells you that $5^m$ and $2^n$ each have a unique prime factorization, which you have displayed. If they were equal, you would have one number with two factorizations.