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

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

How would I prove this?

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

2. Originally Posted by circuscircus
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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