# Prove that $\log_{36} 30$ is irrational number.

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?

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What does $NWD(p,q)$ stand for? – Andrés E. Caicedo Sep 1 '13 at 17:35
@AndresCaicedo it is Polish for $\operatorname{gcd}(p,q)$; pl.wikipedia.org/wiki/Najwi%C4%99kszy_wsp%C3%B3lny_dzielnik – oldrinb Sep 1 '13 at 17:37
@Andres Caicedo I mean GCD. But in polish it is NWD ;) – Thomas Sep 1 '13 at 17:38
Note $\log_{36}(30)=\frac12\log_6(30)=\frac12\log_6(6)+\frac12\log_6(5)=\frac12+ \frac{1}{2}\log_6(5)$. So if it streamlines the argument at all, you could just focus on whether $\log_6(5)$ is irrational. – alex.jordan Jan 26 '15 at 22:30

So, we have $(6^2)^p=(6\cdot 5)^q\implies 6^{2p-q}=5^q$

As $(6,5)=1$ we need $q=2p-q=0$ which is impossible as $q\ne0$

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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

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Nice answer, a lot better than mine. – André Nicolas Sep 1 '13 at 19:34

Hint: Note that if $q\ge 1$, then $5$ divides $30^q$.

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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.

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