- #1

mattmns

- 1,118

- 6

I am to prove that [tex]\log_{2} 7[/tex] is irrational. So I started by saying that what if [tex]\log_{2} 7[/tex] is rational. Then it must be in the form of [tex]\frac{m}{n}[/tex] where m and n are integers. So now [tex]\log_2 7 = \frac{m}{n}[/tex] So I took the 2^ up of each and now [tex]7 = 2^{\frac{m}{n}}[/tex] Then [tex]7 = \sqrt[n]{2^m}[/tex] But now I seem to be lost. Do I now try to prove that [tex]\sqrt[n]{2^m}[/tex] is irrational, or what do I need to do. Any ideas? Maybe proove that [tex]2^{\frac{m}{n}} \neq 7[/tex] by 2^{anything rational} must be something?

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