# Thread: 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?

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

2. Originally Posted by circuscircus
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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