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Thread: [SOLVED] Prove log x < e^x

  1. #1
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    [SOLVED] Prove log x < e^x

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

    $\displaystyle \log x < e^x$

    ?

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

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

    $\displaystyle \log x < e^x$

    ?

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

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