Prove that $\log_52$ is irrational

Question

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?

Answer 1

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.

Answer 2

The fundamental theorem of arithmetics is that every number can be uniquely written as the product of prime factors.

Now, $2^n$ and $5^m$ can be uniquely written as product of factors; hence, the representations:

$$2^n = 2 \times 2 \times \cdots \times 2$$

$$5^m = 5 \times 5 \times \cdots \times 5$$

$n$ times and $m$ times respectively, are unique.

What is the conclusion?

Answer 3

$$\log_5 (2)= m/n\implies m,n\ne 0 $$ and WLOG $m,n\gt 0$ But $5^m$ is always odd while $2^n$ is always even so they can not be equal.