Problem:
Find all positive solutions of this equation:
x^y = y^x
An obvious solution is x=y and x=2, y=4.
Spoiler:
The Mutuabola [Rowe]
-------------
Recurrent interest has been expressed in the relationship
given in such forms as:
x^(1/x) = y^(1/y) (1)
x log y = y log x (2)
x^y = y^x (3)
which are mutual in the sense that the relationship is
unchanged by an interchange of the variables.
For one such form, substitute y=kx in (3) and solve for x,
transforming the implicit (3) to the explicit parametric
relations
x = k ^ (1/(k-1)) and y = k ^ (k/(k-1)).
For another such form, transform to the parameter u = 1/(k-1),
whence the parametric relations
x = (1 + 1/u)^u and y = (1 + 1/u)^(u+1).
Observe that
(-2)^(-4) = (-4)^(-2) = 1/16
i^(-i) = (-i)^i = exp(pi/2) = 4.81047738096535165541
This solution is the k = -1 case,
with i = exp(i*pi/2) and -i = exp(-i*pi/2).
Reference:
- R. C. Archibald;
Problem notes, No. 9,
American Mathematical Monthly 28 (1921) 141-143
- H. Martyn Cundy;
x^y = y^x : an investigation,
Mathematical Gazette 131 June 1987, Note 71.18
- Ulrich Dammer;
Untersuchung der Gleichung x^y = y^x,
MNU 54:4 (Juni 2001) 215-216
- L. E. Dickson;
History of the Theory of Numbers,
New York, Vol II,
- Geneva and Lausanne, 1748, p687
- Angela Dunn;
Mathematical Bafflers,
Dover Publ. (1980, 1964 repr.),
Problem: An Exponential Equation, p197, p213
- solutions of x^y = y^x
- Leonhard Euler;
Introductio in analysin infinitorum,
Tome II, Lausanne 1748, Cap. 21, Section 519, p294
- Martin Gardner;
Fractal Music, Hypercards and More Math. Recreations from SA Magazin
Freeman (1991) New York
Chap. 18: Pi and Poetry: Some Accidental Patterns
- what is bigger e^pi or pi^e
- solutions of x^y = y^x
- A. Hausner;
Algebraic number fields and the Diophantine equation m^n = n^m,
American Mathematical Monthly 68 (1961) 856-861
- he solves A^N(B) = B^N(A) with N(x) the norm function
- P. Hohler, P Gebauer;
Kann man ohne Rechner entscheiden, ob e^pi oder pi^e gr\"osser ist?
Elemente der Mathematik 36 (1981) 131-134
- R. A. Knoebel;
Exponentials reiterated,
American Mathematical Monthly 88 (1981) 235-252
- Dieter K\"otter;
F\"ur welche Zahlenpaare ist das Potenzieren kommutativ?
http://home.t-online.de/home/d.koetter/potenz.html
- Y. S. Kupitz, H. Martini;
On the equation x^y = y^x,
Elemente der Mathematik 55 (2000) 95-101
- I. Lehmann;
Wie symmetrisch sind die sieben Grundrechenarten,
Mathematik in der Schule 34 (1996) 344-356
- E. J. Moulton;
The real function defined by x^y = y^x,
American Mathematical Monthly 23 (1916) 233-237
- R. Robinson Rowe;
The Mutuabola,
Journal of Recreational Mathematics 3 (1970) 176-178 (Errata, 4 p69)
- Daihachiro Sato;
Algebraic solution of x^y = y^x (0 < x < y),
Proceedings of the American Mathematical Society 31:1 (Jan. 1972) 316
- Thomas W. Shilgalis;
Graphical Solution of the Equation a^b = b^a,
The Mathematics Teacher, 66 (1973) 235
- Marta Sved;
On the rational solutions of x^y = y^x,
Mathematics Magazine 63:1 (1990) 30-33
Zbl. 0713.11027
- Contour Plot of z = x^y - y^x,
http://home.netcom.com/~hjsmith/Display/xyyx.html
- alt.algebra.help FAQ
What are the solutions to x^y = y^x?
http://oakroadsystems.com/aah/yxxy.htm
- all real solutions (for all quadrants).
0 <= x = y
0 < x < y : x = (1 + 1/t)^t, y = (1 + 1/t)^(1+t), t > 0
0 > x > y : x = -(1 + 1/t)^t, y = -(1 + 1/t)^(1+t), t > 0, t odd
x < 0 < y : x = -(1/t - 1)^(1-t), y = (1/t - 1)^(-t), 0 < t < 1
- When Exponentiation Commutes
Puzzle 37 (April 1998)
http://www.mathcad.com/library/LibraryContent/puzzles/puzzle.asp?num=37
http://www.mathcad.com/library/LibraryContent/puzzles/soln37/soln37.html
- The graph of x^y = y^x consists of two curves.
1. Where do the two curves intersect?
2. The points (4, 2) and (9/4, 27/8) fall on one of the curves.
What is the length of the arc connecting these two points?
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From - Mon May 26 12:40:49 1997
From: jmccarty@sun1307.spd.dsccc.com (Mike McCarty)
Newsgroups: sci.math
Subject: Re: x^y = y^x
Date: 23 May 1997 21:19:26 GMT
Organization: DSC Communications Corporation
)Can anyone tell me if there are any other solutions to
)this equation:
)
) x^y = y^x
)
)apart from the obvious x=2, y=4 ?
(-2)^(-4) = (-4)^(-2)
sqrt(3)^sqrt(27) = sqrt(27)^sqrt(3)
Many years ago I did a complete analysis of this equation, finding all
complex solutions. It is not a simple equation to solve. However, I can
point out the way (and maybe even resurrect enough interest to redo the
analysis).
Let y = px. Then
x^(px) = (px)^x
or
px log(x) = x log(px)
Thus x = 0 is a possible solution with y = 0, if you choose to define 0^0.
o.w.
p log(x) = log(p) + log(x)
or
(p-1)log(x) = log(p)
Thus p = 1 is a possibility (from which y = x)
and
log(x) = log(p)/(p-1) { p <> 1 }
or
x = p^[1/(p-1)]
y = px = p^[1+1/(p-1)] = p^[p/(p-1)]
This gives you parametric equations for x, y > 0. There are other real
points not in the first quadrant, but you get into complex
considerations at that point.
If you are interested further, I'll be glad to discuss with you via
e-mail.
Mike