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Property of Natural Log- Inequality equation

  1. Nov 26, 2011 #1
    Hi.

    I just saw on wikipedia that natural logarithm has such a property:
    [x/(1+x)] < ln (1 + x) < x

    (http://en.wikipedia.org/wiki/Natural_logarithm)

    Can anyone pls tell me how to prove this?

    Proving [x/(1+x)] and ln (1 + x) less than 'x' is easy.. But how abt [x/(1+x)] < ln (1 + x) ??

    Thanks in advance..
     
  2. jcsd
  3. Nov 26, 2011 #2

    dextercioby

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    It's wrong. It's smaller or equal.

    Consider the function

    [tex] f:(-1,+\infty)\longrightarrow \mathbb{R} [/tex]

    [tex] f(x) = \frac{x}{x+1} - \ln (1+x) [/tex]

    Plot it.
     
  4. Nov 26, 2011 #3
    Following what dextercioby said, the following inequality does not always hold true.

    [x/(1+x)] < ln (1 + x) < x

    For example, let x = 0. We the have:

    0/(1+0) < ln(1+0) < 0
    0 < 0 < 0

    This makes no sense.
     
  5. Nov 26, 2011 #4
    Thanks for response.

    Yes its smaller or equal (Sorry couldn't rite it..)..

    Is plotting the function f(x) = [x/(1+x)] - ln (1 + x) best way 2 prove it??
     
  6. Nov 26, 2011 #5

    dextercioby

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    You needn't plot it, just make the standard analysis of 1st and 2nd derivatives of f to be able to draw the desired conclusion.
     
  7. Nov 26, 2011 #6
    Thanks a lot..!!
     
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