Prove $\log_4 6$ is irrational

Question

Prove $\log_{4}6$ is irrational.

I've seen proofs of this which boil down to: $$4^{m} = 6^{n}$$

But how does this prove that it is irrational? For example it is possible to have $n$ be: $$x=\frac{\log 4}{\log 6}$$

Answer 1

It proves it's irrational because for it to be rational would require integers $n,m$ that make the equation you've written true. However, that equation decomposes into:

$$ 2^{2m} = 3^n 2^n $$

or if you like:

$$ 2^{2m-n} = 3^n $$

For this to be true, an integer power of $3$ would have to be equal to some other integer power of $2$. Since prime decompositions are unique, and therefore any power of $3$ (or $2$) cannot be decomposed into any other primes, this is impossible.

Answer 2

We will prove by contradiction. Assume $\log_{4}(6)$ is rational, and equal to $\frac{m}{n}$ for integers $m$ and $n$. Then, we have:

$6 = 4^{\frac{m}{n}}$

$6^{n} = 4^{m}$

$2^{n}3^{n} = (2^{m})^{2}$

$2^{n} = 3^{m}$

But this is impossible, as a number cannot simultaneuosly be a power of $2$ and a power of $3$. Thus, $\boxed{\log_{4}(6)\text{ is irrational.}}$