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Solutions to X^Y = Y^X

Date: 12/21/2000 at 02:29:38
From: Barry Pelz
Subject: x^y = y^x and the Lambert W Relation

Hello Dr. Math,

My question is about the equation x^y = y^x. Obviously x = y works, as 
well as (2,4) and (4,2). I have solved this parametrically and have 
found the solutions (9/4,27/8) and (sqrt(3),sqrt(27)). However, there 
are several solutions that are not found using this parametric 

     y = a^(a/(a-1))
     x = (y/a)

I have been told about something called the Lambert W Relation and its 
'relation' (no pun intended) to the above problem. I am in a high 
school AP Calculus course, but I would still prefer that your 
explanation not contain unnecessarily esoteric terms or formulae.

Thanks in advance.

Date: 12/21/2000 at 06:58:07
From: Doctor Mitteldorf
Subject: Re: x^y = y^x and the Lambert W Relation

Dear Barry,

I first became aware of this equation through a paradox that was 
presented to me as a high school student. I still think exploring it 
is one of the most fertile learning experiences available:

Consider the sequence:


Each term is x power raised to the result of the previous term.

Take this function x^(x^(x^(x^x))...) and extend it out to infinity. 
Set the result equal to 2. Solve for x.

How do you solve for x? By noting that the quantity in the exponent is 
the same as the quantity as a whole. So it must be true that:

     x^2 = 2

So x is just the square root of 2. If you feel cheated, just try it.

Take x = sqrt(2) and calculate x^x = x1, then x^x1 = x2, etc. and see 
for yourself that it converges to 2 as the answer. Pretty neat. Let's 
try it with 4:

Set x^(x^(x^(x^x))...) = 4. Substitute as before, and find x^4 = 4. So 
the solution is x = 4^(1/4), the 4th root of 4.

Check again if you want, by seeing if the powers converge toward 4. 
But in the process you realize: the 4th root of 4 is the same as the 
square root of 2. Surely the process can't converge to both numbers. 
In fact, you've just checked it and it converges to 2, not to 4.

Why does it work for 2 but not for 4?

- Doctor Mitteldorf, The Math Forum   

Date: 12/21/2000 at 13:02:28
From: Doctor Rob
Subject: Re: x^y = y^x and the Lambert W Relation

This equation can be manipulated into the form:

     ln(x)/x = ln(y)/y.

This means that x and y are two values that give the same value of the 
function f(z) = ln(z)/z. If you plot w = f(z) versus z in the 
zw-plane, you will find a horizontal asymptote is w = 0, and a 
vertical asymptote is z = 0. A horizontal line w = c intersects this 
curve in exactly two points if 0 < c < 1/e, one point (e,1/e) if c = 
1/e or c <= 0, and no points if c > 1/e. The z-coordinates of the two 
points give you the solutions (x,y) where x and y are not equal. If x 
and y are not equal, then y/x = a will be different from 1. That means 
that x = y/a, and then:

       ln(y/a)/(y/a) = ln(y)/y
     a*[ln(y)-ln(a)] = ln(y)
         (a-1)*ln(y) = a*ln(a)
               ln(y) = [a/(a-1)]*ln(a)

                   y = a^[a/(a-1)]
                   x = a^[1/(a-1)]

Every solution with x and y unequal does correspond to one of these 
values of a, namely a = y/x.

See also this page:

   Solutions to x^y = y^x^y-x^y   

If you think you have other solutions not parametrized by the above 
equations, please think again, or send them to us.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Basic Algebra
High School Sequences, Series

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