My Math Forum Help: Proving 2^x = 7 is irrational

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 July 17th, 2014, 07:33 PM #1 Newbie   Joined: Jul 2014 From: San Diego Posts: 2 Thanks: 0 Help: Proving 2^x = 7 is irrational I'm finding it difficult to start the proof. I think I'm supposed to use the limit theorem but I'm not sure. The x root is throwing me off.
 July 17th, 2014, 11:51 PM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,312 Thanks: 115 Assume x = m/n is rational and see what happens. Thanks from Deveno and topsquark
July 18th, 2014, 08:40 AM   #3
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Quote:
 Originally Posted by Pero Assume x = m/n is rational and see what happens.
but it's not x^2, we can't just apply it because then the limit may be infinity or 0, or that's what I thought.

July 18th, 2014, 01:38 PM   #4
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Quote:
 Originally Posted by summer2014 but it's not x^2, we can't just apply it because then the limit may be infinity or 0, or that's what I thought.
I don't know what limit you are thinking of. In any case 2 < x < 3.

If x = m/n (m and n relatively prime), then you are getting an nth root of a power of 2, which can't be 7. It will be irrational.

 July 18th, 2014, 01:40 PM #5 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Suppose $x = \dfrac{m}{n}$, for integers $m,n$ with $n > 0$. Then $2^x = 2^{\frac{m}{n}} = \sqrt[n]{2^m} = 7$. Take both sides to the $n$-th power, what happens? (You might want to assume $m > 0$. Why don't we need to worry about what happens when $m \leq 0$?). Thanks from topsquark

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