# How to prove that ln &#x2061;<!-- ⁡ --> 12 </mrow>

How to prove that $\frac{\mathrm{ln}12}{\mathrm{ln}18}$ is irrational witout using the change of base rule?
I have to show that $\frac{\mathrm{ln}12}{\mathrm{ln}18}$ is irrational by using change of base rule.
At the beginning I have proved that $\mathrm{ln}r$ is irrational for any rational $r$, $r\ne 1$. Then using this we can say that $\mathrm{ln}12$ and $\mathrm{ln}18$ 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.
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diamondogsaz
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
$\frac{\mathrm{ln}12}{\mathrm{ln}18}=\frac{2\mathrm{ln}2+\mathrm{ln}3}{\mathrm{ln}2+2\mathrm{ln}3}=\frac{a}{b}$
for positive and coprime integers a and b. Then
$2b\mathrm{ln}2+b\mathrm{ln}3=a\mathrm{ln}2+2a\mathrm{ln}3$
that becomes
$\left(2b-a\right)\mathrm{ln}2=\left(2a-b\right)\mathrm{ln}3$
which tells you that $\mathrm{ln}3/\mathrm{ln}2$ is rational as well. By the change of base rule, this is the same as saying that ${\mathrm{log}}_{2}3$ is rational, so
${\mathrm{log}}_{2}3=\frac{h}{k}$
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.

Jonathan Miles
$\frac{a}{b}=\frac{\mathrm{ln}12}{\mathrm{ln}18}={\mathrm{log}}_{18}\phantom{\rule{negativethinmathspace}{0ex}}12\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\begin{array}{rl}{18}^{a/b}& =12\\ {18}^{a}& ={12}^{b}\end{array}$