I need to prove by contradiction that $\log_2(3)$ is irrational.

I'm really unfamiliar with logs to be honest, it's been awhile since I've done them and I'm unsure of how to approach this.
Any help would be appreciated.

• Suppose $\log_2 3 = p/q$ and exponentiate. Stare at both sides for a while, and then write down a proof. – user61527 Jun 27 '14 at 0:26

You want to prove that $x = \log_2(3)$ is irrational and you want to do it by contradiction. That means you assume that the statement is false and then you derive a contradiction. So say that $x$ in fact is rational.
That means $x = \frac{a}{b}$ for some integers $a$ and $b$.
Now then $$2^x = 2^{a/b}.$$ That is $$2^{\log_2(3)} = 2^{a/b} \quad \Rightarrow \\ 3 = 2^{a/b}.$$ All you have to do now is to convince yourself that there are indeed no $a$ and $b$ making this true.
• To clarify further, this would mean that $3^b = 2^a$. The LHS is odd and the RHS is even unless a = 0, but that would mean b is also 0 (for LHS = 1) and we have to take b != 0 for x to be a meaningful rational number. – Wonder Jun 27 '14 at 6:28