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Thread: How do i solve the following problem?

  1. #1
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    How do i solve the following problem?

    Prove the following statement:

    There is a real number x such that x>1 and $\displaystyle 2^x$ >x^(10)

    I used a graphing calculator to find an X that works.. Determined that to be 59.. because 2^59 is giving me 5.8E17 vs x^10 which is giving me 5.1E17.. However, how do i solve this problem without the use of the calculator?
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  2. #2
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    Re: How do i solve the following problem?

    $\displaystyle \begin{aligned}2^x > x^{10} \implies x \log 2 &> 10 \log x \\ {\log 2 \over 10} &> \frac1x \log x = -\frac1x \log \frac1x \\ -\tfrac1{10}\log 2 &< \frac1x \log \frac1x \\ \text{Write } y = -\log x = \log \tfrac1x \text{, so that} \qquad -\tfrac1{10}\log 2 & < y\mathrm e^y \\ W\left(-\tfrac1{10}\log 2\right) & < y = -\log x \\ x & > \mathrm e^{-W\left(-\tfrac1{10}\log 2\right)} \end{aligned}$
    Where $\displaystyle W$ is the Lambert W function.

    I'm not sure that this is correct in all particulars, but the method ought to work.
    Last edited by Archie; Nov 10th 2015 at 01:09 PM.
    Thanks from ajaved
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  3. #3
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    Re: How do i solve the following problem?

    Thank you
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