Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x^{1000}$.

share|improve this question
4  
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
1  
$2^x$ 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
1  
You can see that $x=2^{16}$. Since $x>16000$, $2^x>2^{16000}=x^1000$ –  Thomas Andrews Feb 22 '12 at 4:15
1  
@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
add comment

2 Answers

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 $a\le 1$.)

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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