# Thread: How do i solve the following problem?

1. ## 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?

2. ## 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.

Thank you