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Use the unique factorization for integers theorem and the definition of logarithms to prove that $\log_3 (7)$ is irrational.

I am taking a beginners fundamental mathematics module, no advanced stuff please. Thanks!

My attempt.

Suppose for a contradiction that it is rational, that is $\log_3(7)=\frac{a}{b}$ for some $a,b\in R$ where $b\neq0$. Therefore, by the definition of logarithms, $7=3^{\frac{a}{b}}$. By the theorem of unique factorization, $7=1*7$ is unique.

Ok I'm plain stuck! Any help please?

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If $7=3^{a/b}$ we have $7^b=3^a$, that is a contradiction by the UFT, since $3$ and $7$ and different primes. –  Jack D'Aurizio Oct 30 '12 at 15:35
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@JackD'Aurizio Make this an answer. –  Graphth Oct 30 '12 at 15:39
    
Awesome thanks yeah that. –  Yellow Skies Oct 30 '12 at 15:45
    
Here is a more general result –  The Chaz 2.0 Oct 30 '12 at 15:51
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In your beginning of solution, it should be "for $a,b\in\mathbb{Z}$. We get $7^b=3^a$. Without loss of generality $b\gt 0$. Then $a\gt 0$, else $3^a$ is too small. Finally, can use Unique Factorization, or the fact that the prime $7$ divides $7^b$, so $7$ divides $3^a$. But if a prime divides a product it divides one of the terms. –  André Nicolas Oct 30 '12 at 16:08
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1 Answer

up vote 1 down vote accepted

Use that $$\log_3(7)=\frac{a}{b} \iff b\log(7)=a\log(3) \iff \log(7^b)=\log(3^a) \iff 7^b=3^a ,$$ contradiction.

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This is precisely the answer Jack provided in the comments. You should give Jack a few hours to post his comments as an answer. –  JavaMan Oct 30 '12 at 18:27
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