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Help with Inequality with Exponents question

  1. Mar 22, 2010 #1
    How do you prove the following?

    [tex]7^{\sqrt{5}} > 5^{\sqrt{7}}[/tex]

    Without calculators or working out [tex]{\sqrt{5}}[/tex], [tex]{\sqrt{7}}[/tex], of course.
  2. jcsd
  3. Mar 22, 2010 #2
    Well, consider both numbers and raise them both to the power of square root of 5.

    So we have
    7^5 \ and \ 5^{\sqrt{35}}

    Now, consider the following:

    6 = \sqrt{36} > \sqrt{35}

    So we can just compare 7^5 and 5^6. I'm sure you can calculate those by hand.
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