How to prove $\log_{18}12$ is irrational?

Question

I was trying to prove this through contradiction where I supposed that it was rational but it didnt seem to work out. Any help would be appreciated.

Answer 1

If it were rational, say $m/n$ then

$$18^{m/n}=12, \text{ or } 18^m = 12^n$$

Equivalently, find the respective prime factorizations

$$2^m3^{2m} = 2^{2n}3^n \iff 2^{m-2n} = 3^{n-2m}$$

This gives only the solution $n=m=0$, so the "rational" exponent is undefined.

Answer 2

@DavidPeterson beat me to answer this, but I'll answer anyway. $\log_{18}12=x$, where $18^x=12$. Assume $x$ is rational. Then $x=\frac{p}{q},p,q\in\mathbb{Z}, q\neq 0$. We thus have $18^{\frac{p}{q}} = 12\Rightarrow 18^p = 12^q\Rightarrow 2^p 3^{2p} =2^{2q}3^q\Rightarrow 2^{2q-p}=3^{2p-q}$. This gives $p=q=0$ because the bases are different, and so they can only be equal when the exponent is $0,$ which implies $2q-p=2p-q=0\Rightarrow 3q=3p=0\Rightarrow q=p=0$.

Answer 3

More generally, suppose $\log_m n $ is rational.

I will show that $m$ and $n$ are powers of the same integer.

We have $\dfrac{\log n}{\log m} =\dfrac{a}{b} $ where $a$ and $b$ are relatively prime positive integers.

Then $n^b = m^a $.

Let $n =\prod p_i^{n_i} $ and $m =\prod p_i^{m_i} $.

Then $\prod p_i^{bn_i} =\prod p_i^{am_i} $ so $bn_i = am_i $ for all $i$.

Since $(a, b) = 1$, $b | m_i$ and $a | n_i$ so $m_i = bu_i$ and $n_i = a v_i $ where $u_i$ and $v_i$ are positive integers.

Then $bn_i =abv_i =am_i =abu_i$ so $u_i = v_i$.

Therefore $n =\prod p_i^{av_i} =(\prod p_i^{v_i})^a $ and $m =\prod p_i^{bu_i} =(\prod p_i^{u_i})^b $.

Therefore both $n$ and $m$ must be powers of the same integer (since $\prod p_i^{v_i} =\prod p_i^{u_i} $).