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Prove that the logarithm of 3 base 10 is irrational

The Fundamental Theorem of Arithmetic is that every integer is a product of primes.

So far I have,

Suppose $\log_{10}(5)$ is rational. Then suppose $\log_{10}(5) = \frac {p}{q}$ for some positive integers $p$ and $q$ with $\frac {p}{q}$ in lowest terms and $p< q$. Exponentiating both sides using 10 as the base we get, $5=10^{p/q}$. Take both sides to the qth power. We get $5^q=10^p=2^p*5^p$. Then we get $5^{q-p}=2^p$.

But I'm not sure if this has anything to do with the Fundamental Theorem of Arithmetic.

If you have another way of doing this that would be great too.

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    You have brought the argument to the point where it is time to use the Fundamental Theorem. You have a number which is a power of $5$, but it's also a power of $2$. The Fundamental Theorem has something to say about that.2012-12-09
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    You omitted the nontrivial part of FTA, that prime factorizations are *unique*. Apply that to $5^{p-q} = 2^p.\:$ But applying FTA is overkill since parity suffices.2012-12-09
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    Is it because since each integer has a unique prime factorization, but the above equality shows otherwise (its a number which is a power of 5 and 2). So there are no integers satisfying this. Hence, contradicting log10(5) is rational2012-12-09

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