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Math Help - Natural log problem

  1. #1
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    Natural log problem

    Show that x2 - 12x + 11 + 18ln(x) > 0, for x > 1.


    My attempt:

    I start by evaluating x2 - 12x + 11 = p.
    The expression equals zero at x = 1, x = 11. It is positive for all x > 11 and negative on 11 > x > 1.

    Since ln(x) > 0 for every x > 1, and the expression p is positive when x > 11; the inequality holds for x > 11.

    I'm guessing the next step is to show that (x-11)(x-1) > -18ln(x) for 11 > x > 1.

    But I don't know how to proceed from here
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  2. #2
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    Re: Natural log problem

    Quote Originally Posted by Cinnaman View Post
    Show that x2 - 12x + 11 + 18ln(x) > 0, for x > 1.
    I suggest that you consider f(x)=x^2-12x+11+18\ln(x).

    Note that f(1)=0 and if x>1 then f'(x)>0.
    Thanks from Cinnaman
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