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Ln(x) less than root(x)

  1. Dec 21, 2014 #1
    I was going through some important points give in my textbook and I saw this:
    ##\log_e x < \sqrt x##
    How did they get this?
    I know calculus so you can show this using differentiation, etc.
    One possible way is that they took
    ##f(x)=\sqrt x-\log_e x##
    And tried to prove it is always greater than zero.
  2. jcsd
  3. Dec 21, 2014 #2


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    You can write [itex] \sqrt x [/itex] as [itex] e^{\ln \sqrt x}=e^{\frac 1 2 \ln x} [/itex]. Now we have $$\lim_{x\to 0} \ln x=-\infty $$ and $$ \lim_{y\to -\infty} e^y=0 $$. So for the least value of [itex] \sqrt x [/itex], [itex] \ln x < \sqrt x [/itex] and because [itex] \sqrt x [/itex] is strictly increasing and [itex] \frac 1 x < \frac{1}{2\sqrt x} [/itex] for [itex] x>4 [/itex], the inequality is always satisfied.

    EDIT: Looks like rewriting [itex] \sqrt{x} [/itex] wasn't necessary, but I don't change it because it may give you a feeling for the inequality.
    Last edited: Dec 21, 2014
  4. Dec 21, 2014 #3
    For x<4 why can't there be a point of intersection? You haven't specified what happens for x<4
    I understood everything else. Thanks for the reply.
  5. Dec 21, 2014 #4


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    When [itex] x \to 0 [/itex], we have [itex] \sqrt x=0 [/itex] and [itex] \ln x \to -\infty [/itex]. In the interval [itex] (0,4] [/itex], [itex] \ln x [/itex] increases faster than [itex] \sqrt x [/itex], but that slight excess can't make up for the huge gap in such a small interval.
  6. Dec 22, 2014 #5


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    Check the derivatives - if one derivative is smaller than another and the initial values are different then you have shown that one is always smaller than the other provided this requirement is met.
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