Property of Natural Log- Inequality equation

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Discussion Overview

The discussion revolves around the properties of the natural logarithm, specifically the inequality involving the natural logarithm and the expression [x/(1+x)]. Participants are exploring how to prove the inequality [x/(1+x)] < ln(1 + x) < x, and whether it holds true under all conditions.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant references a property of the natural logarithm from Wikipedia and seeks a proof for the inequality [x/(1+x)] < ln(1 + x).
  • Another participant asserts that the inequality is incorrect, stating it should be smaller or equal, and suggests considering the function f(x) = [x/(1+x)] - ln(1 + x) to analyze the claim.
  • A different participant provides a counterexample using x = 0, demonstrating that the inequality does not hold as stated, since 0 < 0 < 0 is contradictory.
  • One participant acknowledges the correction and suggests that plotting the function may not be necessary for proving the inequality.
  • Another participant recommends performing a standard analysis of the first and second derivatives of the function f to draw conclusions about the inequality.

Areas of Agreement / Disagreement

Participants disagree on the validity of the inequality [x/(1+x)] < ln(1 + x). Some argue it is incorrect or only holds under certain conditions, while others suggest methods for analysis without reaching a consensus on the proof.

Contextual Notes

The discussion highlights the need for careful consideration of conditions under which the inequality holds, as well as the importance of derivative analysis in proving such claims. There are unresolved aspects regarding the behavior of the function f(x) across its domain.

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

Consider the function

f:(-1,+\infty)\longrightarrow \mathbb{R}

f(x) = \frac{x}{x+1} - \ln (1+x)

Plot it.
 
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.
 
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??
 
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.
 
Thanks a lot..!
 

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