a to the b power

Since I don't know and am too lazy to find out how to make this thing write small superscript numerals, I'm using "^" here for exponentiation, OK? Well, even if it's not OK, that's what I'll be using, OK? You wanna step outside?

2^4 = 16. 4^2 = 16 also! Not counting the trivial a = b, are there any other pairs of positive integers such that a^b = b^a? How many? Not counting the trivial a = b, how many pairs of positive real numbers are there with a^b = b^a (we've already found one)?

Here's part of the answer: for every pair of positive real numbers {p, 1/p}, p <> 1, there is a pair of positive real numbers {a, b} such that a^b = b^a, and the correspondence is biunique. If p yields the pair (a,b) , then 1/p yields (b,a). So the number of pairs {a, b} with this magic property is uncountably infinite, just like the positive reals that they match.

So what are they?


  1. Not sure if you know the answer or want it. Consider the growth of the function (log x)/x

  2. Let b=ca, c>1, then a=c^(1/(c-1)) and b=c^(1+(1/(c-1))). These range from (e,e) to (2,4) for c in the range 1..2.

  3. Not sure, John, what you have in mind. Lord, you have it, with your "c" is my "p"--but p is unlimited (except p=1), not sure where (e,e) and (2,4) come in.

  4. Those are (a,b) for c, which approach (1,c) from above as c diverges. a is in the range 1..e and b in the range e..Inf, with (e,e) occurring at c 1. Another memorable point (for low number of digits) is for c 1.5, (2.25,3.375).


Post a Comment

Popular posts from this blog

Central Planning Works!

Fair's fair!

More college diversity and tolerance