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Math Help - How to prove x > ln(x) For all x>0?

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    How to prove x > ln(x) For all x>0?

    How would I prove this?

    x > ln(x) \forall x>0
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by circuscircus View Post
    How would I prove this?

    x > ln(x) \forall x>0
    Consider the real-valued function f(x) = x - \ln x. Note that f(x) is continuous on (0, \infty)

    Since \ln x < 0 for x \in (0,1), we have that f(x) > 0 on (0,1). When x = 1, f(x) = 1>0 also. Now we consider x > 1. Note that f'(x) = 1 - \frac 1x. Clearly this quantity is positive for all x>1. But that means that f(x) is increasing on (1,\infty) for all x. Thus, we have that f(x) > 0 for all x>0. But that means that x - \ln x > 0 for all x > 0. Adding \ln x to both sides of this inequality, we obtain x > \ln x for all x>0, as desired.

    QED
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