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Climbing Powers



Antoine from Lycée St Genes submitted the following solution:

I checked with the calculator and saw that $9^{10}>10^9$.
Obviously we can't do that for the next questions.

I decided to deal with logarithms:

$log\, 99^{100}=100\, log\, 99=199.56$
$log\, 100^{99}=99\, log\, 100=198$
$log\, 99^{100}>log\, 100^{99}\,$

Therefore, $99^{100}>100^{99}$.

Same for $999^{1000}$ and $1000^{999}$:

$log\, 999^{1000}=1000\, log\, 999=2999.56$
$log\, 1000^{999}=999\, log\, 1000=2997$
$log\, 999^{1000}>log\, 1000^{999}$

Therefore, $999^{1000} > 1000^{999}$.

Michael sent in this more general approach: 

My approach not only works for large numbers, but can also be applied to smaller numbers too.
$$x^y>y^x \iff ln\, x^y > ln\, y^x \iff y\, ln\, x > x\, ln\, y \iff \frac{ln\, x}{x} > \frac{ln\, y}{y}$$
Let $f(n) = \frac{ln\, n}{n}$. Hence $f(x) > f(y) \iff x^y > y^x$.

Differentiating $f(n)$ with respect to $n$ using the Quotient Rule,
$$f'(n) = \frac{n \times \frac{1}{n} - ln\, n \times 1}{n^2} = \frac{1-ln\, n}{n^2}$$
Hence the function has a stationary point at $ln\, n=1$, i.e. at $n = e$, $f(n) = e^{-1}$, and no others.
More importantly, it shows that $f(n)$ is a decreasing function for $n>e$, since this is where $f'(n) = \frac{1-ln\, n}{n^2}$ is negative.
Hence $$e \leq x < y \Rightarrow f(x) > f(y)$$
And so $$e \leq x < y \Rightarrow x^y > y^x$$
So $$9^{10} > 10^9,$$
$$99^{100} > 100^{99},$$
$$999^{1000} > 1000^{999},$$
$$(a\, billion\, 9s)^{(1\, and\, a\, billion\, 0s)} > (1\, and\, a\, billion\, 0s)^{(a\, billion\, 9s)},$$
and so on.

Here is another neat way of thinking about the problem:

It can be shown that $e$ is the limit as $n \to \infty$ of the increasing sequence $(1+\frac{1}{n})^n$. So we have $$e>(1+\frac{1}{n})^n.$$
Then if $n>e$, $$n>(1+\frac{1}{n})^n$$
So $$n>\frac{(n+1)^n}{n^n}$$
and $$n^{n+1}>(n+1)^n$$
So for all natural numbers $n$ greater than $e$, we can see straight away that $n^{n+1}>(n+1)^n$.