Ask Uncle Colin: Comparing powers

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, ${(1+\frac{1}{1000})}^{1000}$ 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 ${\left(\frac{1000}{1001}\right)}^{1000}\times 1000$.

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

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 $f^{\prime }(x)=\frac{\ln (x)-1}{(\ln (x){)}^2}$, 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.)

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I hope one of those methods makes sense to you!

- Uncle Colin