# Thread: [SOLVED] Prove log x &lt; e^x

1. ## [SOLVED] Prove log x &lt; e^x

Hey guys. I seem to have forgotten how to work with log functions. Can someone please remind me how to show that

$\log x < e^x$

?

Either a generalized solution for all $x\in\mathbb{R}$ or else a more specific solution for $x>k$ for some constant $k$ will work fine for my purposes.

Thanks!

2. Originally Posted by hatsoff
Hey guys. I need seem to have forgotten how to work with log functions. Can someone please remind me how to show that

$\log x < e^x$

?

Thanks!
The statement $\log x < e^x$ is satisfied if and only if $e^{\log x} < e^{e^x}$, so $x < e^{e^x}$.

This is true because $y so $x < e^x$ and $e^x < e^{e^x}$ by setting $y=e^x$. But then it means $x.