Dear Uncle Colin,

Which is larger, $x=1000^{1001}$ or $y=1001^{1000}$?

- Calculating Oversize Massive Powers And Realising… Eurgh

Hi, COMPARE, and thanks for your message!

These feel like they ought to be a similar size! In fact, the first one, $x$ is the larger. Let’s show that in several ways.

Numbers nearby

Let’s think about the ‘nearby’ number $A = 1000^{1000}$. Clearly, both of our numbers are bigger than this.

Now, $x = 1000A$, and $y = 1.001^{1000}A$. A sensible question is, is $1.001^{1000}$ bigger than 1000? No - in fact, $\br{1 + \frac{1}{1000}}^{1000}$ is very close to $e$.

So $x$ is the larger, by a factor of about $1000/e$.

What’s $\frac{x}{y}$?

If we divide the pair, we get $\frac{1000^{1001}}{1001^{1000}}$, which is $\br{\frac{1000}{1001}}^{1000}\times 1000$.

Take logs: this is $1000 \ln\br{1 - \frac{1}{1000}} + \ln(1000)$.

Since $\ln(1-t) \approx -t$, this is roughly $-1 + \ln(1000)$, and $\ln(1000) > 1$.

(This is morally the same as the other way.)

Multivariable calculus

I’m not recommending this way, but it popped into my head.

Consider the function $f(X,Y) = X^Y = e^{Y\ln(X)}$

Differentiating partially:

  • $\frac{\delta}{\delta X}f = \frac{Y}{X}f(X,Y)$
  • $\frac{\delta}{\delta Y}f = \ln(X)f(X,Y)$

When $X=Y=1000$, these work out to be $1000^{1000}$ and $1000^{1000}\ln(1000)$.

The derivative with respect to $Y$ is much larger, so increasing the power makes the value of the function increase far more quickly than increasing the base – again, $x$ is the larger.

One more way

(This one is @realityminus3’s idea, adapted slightly.)

Consider $f(x) = \frac{x}{\ln(x)}$. Its derivative is $f’(x) = \frac{\ln(x) - 1}{(\ln(x))^2}$, which is positive for all $x > e$.

That implies $f(1001) > f(1000)$, so $\frac{1001}{\ln(1001)} > \frac{1000}{\ln(1000)}$.

In turn, $1001 \ln(1000) > 1000 \ln(1001)$, and as a result $1000^{1001} > 1001^{1000}$.

(Trust RM3 to point me towards the neatest way. Good rat.)

I hope one of those methods makes sense to you!

- Uncle Colin