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To somehow prove that x>logx for all x>o

  1. Oct 1, 2011 #1
    i have been given a question ...that is to prove that x>logx for all x>0
    i know that its true.....and i have thought of a way to prove it
    i thought of drawing a graph.....but then i had a small doubt that will the graph hold true for all x>0 till infinity?
    i mean i know its true but will the graph prove that? i can only draw it for a finite no of numbers
    also i am studying differentiation and integration these days....so is thereany way of roving this using differentiation and integration? ...or any other way?
    i really appreciate anyones help....thanks
     
  2. jcsd
  3. Oct 1, 2011 #2
    In order to prove it, you can proceed by showing x-log(x)>0 for all x>0, and then showing that x-log(x) always increases (by using differentiation; it must always have a positive slope).
     
  4. Oct 1, 2011 #3
    but then again i will be doing that for a finte limit.....
    i want to do something with limits like x goes from 0 to infinity.....any other idea?
     
  5. Oct 2, 2011 #4
    Taking a limit at infinity proves nothing in this case; it will only give you information for large x behavior, and nothing about the general behavior. What I told you applies on the whole interval.
     
  6. Oct 2, 2011 #5
    You know that x>lnx for 0<x<=1. Try using derivatives to show the function x increases faster than ln(x) for all x >1.
     
  7. Oct 2, 2011 #6
    induction...?
     
  8. Oct 2, 2011 #7
    I don't think induction can be used for statements such as the OP's (the functions are continuous; they aren't just defined for natural numbers)
     
  9. Oct 2, 2011 #8
    You could also divide by x (since x > 0) and look at the behavior of log(x)/x.
     
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