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How can prove logx< x^1/2

  1. Feb 9, 2010 #1
    How can prove logx< x^1/2 for x>1
  2. jcsd
  3. Feb 9, 2010 #2
    What have you tried so far? Perhaps induction will work
  4. Feb 9, 2010 #3
    Here's one approach:

    Define [tex]f(x) = \sqrt{x} - \ln(x)[/tex].

    See if you can show that [tex]f(1) > 0[/tex] and [tex]f'(x) \geq 0[/tex] for [tex]x \geq 1[/tex].

    That would do it. Do you see why?

    [Edit] Oops, that won't work, because it isn't true that [tex]f'(x) \geq 0[/tex] for [tex]x \geq 1[/tex]!

    So try this instead-- find the minimum value of f by solving [tex]f'(x) = 0[/tex]. If the minimum is positive (and it is), then you are done.
    Last edited: Feb 9, 2010
  5. Feb 10, 2010 #4
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