# Saturday Puzzle 1

May 20, 2006 pm31 5:40 pm

Today is a three-puzzle Saturday. Answers will be up, posted by readers or by me, by next Friday. Here's #1, the easiest:

5 to the 3rd power is 125 and 3 to the fifth power is 243. (5^{3}=125, 3^{5}=243)

10 to the second power is 100 and 2 to the 10th power is 1024. (10^{2}=100, 2^{10}=1024)

Let's limit discussion to natural numbers {1,2,3,…} if x > y, is x to the y always less than y to the x? Are they ever equal? x > y –> x^{y } < y^{x} ??

4 Comments
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No, x^y is not always less than y^x for x>y. Consider all x>1 when y=1. (x^1) = x > (1^x) = 1. Additionally, let x=3 and y=2. (3^2) = 9 > (2^3) = 8.

Also, for x=4 and y=2, x^y = y^x = 16.

Try graphing f(x,y) = x^y – y^x. You noted that f(5,3) = 0. Additionally, you only want to look on one side of the x=y plane, the side where x>y.

I do not know of more cases than the two you propose: y = 1 , or x=3, y=2. In addition, your 2^4 = 4^2 is unique.

I meant to say f(5,3) 0.

Is there any way to prove these results in general, or investigate this function aside from graphing it?

f(x,y) = x^y – y^x.

I don’t know. There are two boundaries running through the first quadrant: x = y (x,x,0) and a second that includes (2,4,0) and (4,2,0). The second extends to some vertical asymptote (near 1???) and a symmetric horizontal asymptote.

I’m going to play and see if I can describe that 2nd boundary. I think I am going to play with derivatives (which I haven’t done except for kid level stuff in a long, long time). Let me know if you find anything before I do.