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Could someone provide, please, a proof of the theorem below?

"Being $x$ and $b$ integers greater than $1$, which can not be represented as powers of the same basis (positive integer) and integer exponent, then the logarithm of $x$, in base $b$, is an irrational number."

Well... Assuming that the logarithm is a rational number $p/q$ ($p$ and $q$ are relatively prime integers), I know I can write $x^q = b^p$, but I can not conclude from this fact that $b$ and $x$ are powers of a same integer (and with integer exponent).

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It looks like homework, if so you should use the homework tag. –  AD. Nov 6 '10 at 13:03
    
I re-edit the issue. The common base should be a positive integer –  Paulo Argolo Nov 6 '10 at 15:22
    
@J.M.:Could you complete the resolution? I'm having trouble finishing it. Thanks! –  Paulo Argolo Nov 7 '10 at 0:19
1  
What sort of trouble, exactly? Edit your question to show what you're trying and where you are unable to finish. –  J. M. Nov 7 '10 at 0:22
    
@J.M.:I just edit the question. Help me more! –  Paulo Argolo Nov 7 '10 at 0:50

1 Answer 1

up vote 7 down vote accepted

Hint:

If $\log_b\;x$ were a rational number $\frac{p}{q}$, then

$$x^q=b^p$$

Now, go back to your initial assumption...

Edited to add: This might be of interest.

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Some hint :D (I suggest you remove the identity) –  AD. Nov 6 '10 at 13:03
    
@AD: Well if it isn't, to use Arturo's words, screaming at him yet, then I don't know what else to do. ;) –  J. M. Nov 6 '10 at 14:17

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