# Prove $\log_{12}18$ is irrational

I first simplified it as $$\log_{12}18=\frac{\log_{9}18}{\log_{9}12}$$ and then proved that the numerator $\log_918$ is irrational by simplifying it to $1 + \log_9 2$ and proving $\log_9 2$ is irrational by contradiction: let $\log_9 2$ be rational so: $$\log_9 2 = \frac{m}{n}$$ $$9^m=2^n$$ which is a contradiction because it is saying that one side is odd and the other side is even. However I think this result isn't so useful because even if I do the same thing for the bottom and find out whether the denominator is rational or irrational, I still can't conclude for the original "fraction" as a whole whether it is rational or irrational. What can I do to reach a sound conclusion about the original number as a whole?

For contradiction, assume $\log_{12} 18$ is rational. Also note it's positive. Then exist positive integers $m,n$ such that $\log_{12}18=\frac{m}{n}$. By definition of $\log$, we get $12^{\frac{m}{n}}=18$. Raise both sides to the $n$'th power: $12^m=18^n$, i.e. $2^{2m}\cdot 3^m=2^n\cdot 3^{2n}$, so by Fundamental Theorem of Arithmetic, $2m=n$ and $m=2n$. But the two equations imply $4n=n$, i.e. $4=1$, contradiction.
• In this case, you can avoid unique factorization by writing it as $2^{2m-n}=3^{2n-m}$. Inverting if necessary, we get that even=odd, unless $2m-n=0$, and then $2n-m=0$. – Thomas Andrews Oct 6 '15 at 16:14