# $p^q > q^p$ For what values of p and q does this hold true?

I saw a question on the internet the other day that asked which of two statements was larger. The numbers were 2^999 and 999^2. With the first being the greater number.

One way to test which was greater was to put them in to my calculator (casio fx-991ES) and see which was bigger.

Talking to my friend about it we wondered whether there was an algebraic expression to be able to deduce which is greater without the use of a calculator.

I set the expression up as p^q > q^p. and tired to manipulate it. Each time I manipulated it, I would end up with a binomial expanision to the the power of p, which I could not solve.

I was wondering if anyone could help, or if it is a problem alreay solved, or that it is already on the website, and I have not seen it.

For positive values of $q,p$, we have $$p^q>q^p$$ if and only if $$\ln p^q>\ln q^p$$ if and only if $$q \ln p > p \ln q$$ if and only if $$\frac{\ln p}{p}>\frac{\ln q}{q}$$ Hence we have a function we can look at to answer such questions, namely $f(x)=\frac{\ln x}{x}$. For particular cases of positive $p,q$, we simply evaluate $f(p)$ and $f(q)$. If $f(p)$ is larger than $f(q)$, then $p^q>q^p$; otherwise, $p^q<q^p$.
In the example in OP, $f(999)\approx 0.00591$, while $f(2)\approx 0.346$. Hence $999^2<2^{999}$.
The function $f(x)$ increases from $0$ to $e\approx 2.71828$, and then decreases after that. Hence, there are two nice rules we can use: If $p>q>e$, then $f(p)<f(q)$. If $e>p>q>0$, then $f(p)>f(q)$. However, if one of $p,q$ is less than $e$ and the other is greater, then all bets are off.
• And this works because $\ln$ is strictly increasing. – GFauxPas Jun 3 '16 at 13:46