TITLE: Prove that $\log_{36} 30 $ is irrational number.
QUESTION [6 upvotes]: Prove that $\log_{36} 30 $ is irrational number.
We can suppose that $\log_{36} 30 $ is rational number. So we have that $\log_{36} 30 = \frac{p}{q}$ where $\gcd(p,q) = 1$. By definition of logarithm we have  $36^{\frac{p}{q}} = 30$ thus $36^p=30^q$. And now I have to prove that these numbers $p,q$ doesn't exists. How can I do it?

REPLY [8 votes]: In the equality 
$$36^p = 30^q $$ 
as 
1) $36$ as it ends in $6$ , $36$ to the power anything always ends in $6$ , 
2)$30$ as it ends in $0$ , $30$ to the power anything always ends in $0$ , 
so differing units digit itself proves that for no integer this equality is possible except for p and q as $0$ , but $q\neq0$ as per definition of rationality , hence proved that p,q doesn't exist for this equality

REPLY [2 votes]: One way is to rely on the fact that prime factorizations of integers are unique.  The number $30^q$ includes $5$ among its prime factors, but $36^p$ does not.