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Will $2^x$ take over $x^{1000}$ ?

I thought that exponential functions had the fastest growth rate, however, graphing it on wolfram alpha made it seem as if the initial behaviors of the two functions implied $2^x$ never overtook $x^{1000}$.

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If we have learned anything over the past few days, it's that apparent patterns sometimes fail – The Chaz 2.0 Feb 22 '12 at 4:00
$2^x$ will certainly, eventually surpass $x^{1000}$; the initial behavior is irrelevant. $2^x$ overtakes $x^{1000}$ somewhere around $x\approx 13,750$. – Arturo Magidin Feb 22 '12 at 4:03
You can see that $x=2^{16}$. Since $x>16000$, $2^x>2^{16000}=x^1000$ – Thomas Andrews Feb 22 '12 at 4:15
@ThomasAndrews: $x$ cannot have a specific value. One can choose any large enough $x$, and yours is sufficient. If you want multicharacter exponents, put them in brackets: x^{1000} renders as $x^{1000}$ instead of x^1000 which shows as $x^1000$ – Ross Millikan Feb 22 '12 at 4:29

Yes. Expontentials will always squash powers. Do solve 2^x=x^1000 on WA and it will return the approximate form $x\approx 13746.8$. Now input plot 2^x / x^1000 from x=13745 to x=13748:

$\hskip 2in$ picture

You can see clearly that it goes from $\le1$ to $\ge1$ in this interval, so this is where $2^x$ overtakes $x^{1000}$.

One way to see that exponentials always overtake powers is through l'Hospital's rule applied to the fraction $x^n/a^x$. Differentiate numerator and denominator $n$ times to obtain

$$\frac{n!}{(\log a)^n a^x} $$

which clearly goes to $0$ as $x\to\infty$ when $a>1$. (Naturally this isn't the same story if $a\le 1$.)

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The answer is yes. If you are familiar with calculus, then you can see this be taking derivatives: the $1001$-st derivative of $2^x$ is $(\ln 2)^{1001}2^x$, while the $1001$-st derivative of $x^{1000}$ is $0$. Thus the $1000$-th derivative of $2^x$ will eventually overtake the $1000$-th derivative of $x^{1000}$, the $999$-th derivative the $999$-th, etc. until we eventually get $2^x>x^{1000}$. If you want an explicit $x$ such that $2^x>x^{1000}$, consider $x=1,000,000$. Then using the fact that $2^{1000}>1,000,000$ (which should not be hard to see) we have $$2^x=2^{1000\cdot 1000}=(2^{1000})^{1000}>(1,000,000)^{1,000}=x^{1000}$$ and furthermore, for $x>1,000,000$ we have $2^x>x^{1000}$ as can be seen by taking derivatives.

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