# Find an $a$ for which $e^x>x^5 \forall x > a$

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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Everywhere except between $a$ and $b$, where $a$ is a bit bigger than $1$, and $b$ is a bit bigger than $12$. For better approximations to $a$ and $b$, use a numerical method, like Newton-Raphson. –  André Nicolas Sep 18 '12 at 1:58
You have two options: 1) Use the Lambert W-Function (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. –  Amzoti Sep 18 '12 at 1:59

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 $20.51$.
sorry, I don't follow. I (think) I get $e^x \gt \frac{x^6}{6!}.$ because $\frac{X^6}{6!} > x^5$ and thus the same reasoning for $e^x\gt \frac{x^7}{7!}$, but I am not getting the part $$x^2\gt 7!$$ –  yiyi Sep 18 '12 at 4:02
@MaoYiyi: $e^x\gt \frac{x^7}{7!}$. So if $\frac{x^7}{7!}\gt x^5$, everything is fine. Rewrite this as $x^{7}\gt (7!)x^5$, then cancel $x^5$ to get $x^2\gt 7!$. –  André Nicolas Sep 18 '12 at 4:17
Nice! Also, I notice that there seems to be a relationship. $$x \gt \sqrt[n-5]{n!}$$ –  yiyi Sep 18 '12 at 4:23
@MaoYiyi: Yes, we could continue. But this is just a quick method of getting some sort of bound. For fairly tight bounds for "$a$" in the inequality $e^x \gt x^n$, I would take a different approach. –  André Nicolas Sep 18 '12 at 4:30