How to prove that $\frac{\ln 12}{\ln 18}$ is irrational witout using the change of base rule?

Question

I have to show that $\frac{\ln 12}{\ln 18}$ is irrational by using change of base rule.

At the beginning I have proved that $\ln r$ is irrational for any rational $r$, $r\ne 1$. Then using this we can say that $\ln 12$ and $\ln 18$ are irrational.

But from here it is difficult for me to show that the fraction is irrational knowing that both the numerator and the denominator are irrational.

How to prove that $\frac{\ln 12}{\ln 18}$ is irrational without using change of base rule?

Answer 1

You can't prove it using that the quotient of irrationals is irrational for the simple reason that the statement is false.

You may instead use a different strategy: suppose $$ \frac{\ln 12}{\ln 18}=\frac{2\ln2+\ln3}{\ln2+2\ln3}=\frac{a}{b} $$ for positive and coprime integers $a$ and $b$. Then $$ 2b\ln 2+b\ln3=a\ln2+2a\ln3 $$ that becomes $$ (2b-a)\ln2=(2a-b)\ln3 $$ which tells you that $\ln3/\ln2$ is rational as well. By the change of base rule, this is the same as saying that $\log_23$ is rational, so $$ \log_23=\frac{h}{k} $$ for positive integers $h$ and $k$. Therefore $$ 3=2^{h/k} $$ or $$ 3^k=2^h $$ that's impossible because of unique factorization of integers.

Answer 2

This is in general not true. For example $a = 2\sqrt{2}, \: b= \sqrt{2}$.

then $\frac{a}{b}= 2$.

Answer 3

Let x be an irrational number.

Obviously $\frac{x}{x} = 1$.

And also, suppose any nonzero integer multiple of $x$, let $b=mx$.

$\frac{b}{x}=\frac{mx}{x}=m \in \mathbb{Z}$ is also irrational.

So it is not always true.

Answer 4

$\dfrac{a}b = \dfrac{\ln{12}}{\ln{18}} = \log_{18}{\!12}\iff\!\! \begin{align}18^{a/b} &= 12\\ 18^a&=12^b\end{align}$ $\iff \begin{align}2^a3^{2a}=2^{2b}3^b\\ 2^{a-2b}=3^{b-2a}\end{align}$ $\iff\begin{align} a-2b &= b-2a\\ 3a&=3b\,\Rightarrow\!\Leftarrow \end{align}$