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I was working on some problems for fun and came upon this one:

The problem is part for me is isolating the $x$-term $e^x>x^5$

I get to some form of $x > 5\ln(x)$ I know that you can do some algebraic manipulation; however, none seem to help me, get $x$ by itself.

I do get $\frac{x}{\ln x}>5$ but what is next?

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    You have two options: 1) Use the Lambert W-Function (http://mathworld.wolfram.com/LambertW-Function.html) and 2) Use numerical methods to find where the inequality is true (it helps to look at$a$plot of exp(x) and x^5. Let us know if that helps.2012-09-18

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The modified post now only asks for an $a$ such that $e^x \gt x^5$ if $x \gt a$. If you know the power series for $e^x$, we can arrive quickly at some answers. We have $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{51}+\frac{x^6}{6!}+\cdots.$ In particular, if $x$ is positive, then $e^x \gt \frac{x^6}{6!}.$ It follows from the above inequality that $e^x\gt x^5$ if $x\gt 6!$.

Of course $6!$ is kind of a big number. But the same power series tells us that when $x$ is positive, then $e^x\gt \frac{x^7}{7!}.$ From this we obtain that $e^x \gt x^5$ if $x^2\gt 7!$, that is, if $x\gt \sqrt{7!}$. The number $\sqrt{7!}$ is a bit under $71$.

Similarly, we find that our desired inequality holds if $x\gt \sqrt[3]{8!}$. Similarly, we find that everything is OK for $x$ past $\sqrt[4]{9!}$, for $x$ past $\sqrt[5]{10!}$, and so on. The number $\sqrt[5]{10!}$ is about $20.51$.

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    You could always add to your answer.2012-09-18