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 $x^{1000}$.

  • 4
    $\begingroup$ If we have learned anything over the past few days, it's that apparent patterns sometimes fail $\endgroup$ Feb 22, 2012 at 4:00
  • 3
    $\begingroup$ $2^x$ will certainly, eventually surpass $x^{1000}$; the initial behavior is irrelevant. $2^x$ overtakes $x^{1000}$ somewhere around $x\approx 13,750$. $\endgroup$ Feb 22, 2012 at 4:03
  • 1
    $\begingroup$ You can see that $x=2^{16}$. Since $x>16000$, $2^x>2^{16000}=x^1000$ $\endgroup$ Feb 22, 2012 at 4:15
  • 1
    $\begingroup$ @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 $x^1000$ $\endgroup$ Feb 22, 2012 at 4:29

2 Answers 2


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 $a\le 1$.)


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.