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.