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Thread: 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?

    $\displaystyle x > ln(x) \forall x>0$
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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?

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

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

    QED
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