Prove that a logarithm is irrational

Question

I’m stuck with the following problem:

Prove that $\log_{2} 3 \in \mathbb{R} - \mathbb{Q} $ .

Thanks in advance!

Answer 1

Assuming that $$\log_2(3)=\frac{p}{q},$$ you get: $$ 3 = 2^{p/q}$$ or: $$ 3^q = 2^p, $$ contradiction.

Answer 2

Suppose that $\log_23$ is rational, i.e. $$\log_23=\frac{n}{m}$$ where $m,n$ are positive integers.

Then, we have $$\log_23=\log_22^{\frac{n}{m}}\Rightarrow 3=2^{\frac nm}\Rightarrow 3^m=2^n.$$ Here, the LHS is odd and the RHS is even, which is a contradiction.