Can the logarithm of a prime number be rational?

[mattmns]

I am to prove that $$\log_{2} 7$$ is irrational. So I started by saying that what if $$\log_{2} 7$$ is rational. Then it must be in the form of $$\frac{m}{n}$$ where m and n are integers. So now $$\log_2 7 = \frac{m}{n}$$ So I took the 2^ up of each and now $$7 = 2^{\frac{m}{n}}$$ Then $$7 = \sqrt[n]{2^m}$$ But now I seem to be lost. Do I now try to prove that $$\sqrt[n]{2^m}$$ is irrational, or what do I need to do. Any ideas? Maybe proove that $$2^{\frac{m}{n}} \neq 7$$ by 2^{anything rational} must be something?

 

[Hurkyl]

Do I now try to prove that $$\sqrt[n]{2^m}$$ is irrational

You just need to prove it's not 7.

 

[mattmns]

How about:

$$7 = 2^{\frac{m}{n}}$$

So, $$7^n = 2^m$$

which is a contradiction because, $$7^n$$ is always odd while $$2^m$$ is always even, for n and m as integers and $$n \neq 0$$

 

[Curious3141]

How about:

$$7 = 2^{\frac{m}{n}}$$

So, $$7^n = 2^m$$

which is a contradiction because, $$7^n$$ is always odd while $$2^m$$ is always even, for n and m as integers and $$n \neq 0$$

In this case, that approach is fine. But it wouldn't work, for example if the question asks you about base 3 logs. In that instance, use the uniqueness of prime factorisation.

 

[Tide]

More generally, if you wanted to prove it for a different base (e.g. 3) you could simply invoke the prime factorization theorem at that last step.