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Math Help - [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

    \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!
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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

    \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<e^y so x < e^x and e^x < e^{e^x} by setting y=e^x. But then it means x<e^{e^x}.
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