# Proof: log(x) is irrational

#### 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?

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#### Hurkyl

Staff Emeritus
Gold Member
mattmns said:
Do I now try to prove that $$\sqrt[n]{2^m}$$ is irrational
You just need to prove it's not 7.

#### mattmns

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

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

Homework Helper
mattmns said:

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

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

Homework Helper
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.

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