# $$f(x) = x^x$$Given this function, defined, let's say for

1. Jun 19, 2004

### Paul

$$f(x) = x^x$$
Given this function, defined, let's say for all real numbers, is there any way to tell when x is rational versus irrational for integer values of f(x)?
e.g.
$$x^x = 4$$
x = 2
$$x^x = 27$$
x = 3
$$x^x = 3$$
x = 1.825455054...

Thanks!

Last edited: Jun 19, 2004
2. Jun 19, 2004

### Tom Mattson

Staff Emeritus
You mean other that simply taking the xth root of x?

3. Jun 19, 2004

### Paul

Do you mean the xth root of f(x)? Or am I misunderstanding? And yes, I meant other than simply looking at specific argument and function values.

4. Jun 19, 2004

### Tom Mattson

Staff Emeritus
Whoops, I sure do.

I'm not sure off the top of my head, but let me play with it.

5. Jun 19, 2004

### abertram28

3^3 doesnt equal, so x=3 is irrational. sorry to nitpick.

interesting problem though. i bet there isnt a way besides applying the interger set for x and assuming all others will be irrational. is there any value for f(x) that results in a non interger rational number? if so, that negates using the interger set to find the y.

6. Jun 19, 2004

### Paul

Thanks, I corrected the typo. And I guess that would put us on the path to a solution. Essentially, given that x is not an element of Z (the integer set), is $$x^x \in Z$$ possible?

7. Jun 19, 2004

### Nexus[Free-DC]

Okay, suppose x=a/b, where a,b are integers. Assume that the fraction is reduced, ie. gcd(a,b)=1. Then x^x= (a/b)^(a/b)=$$(\frac{a^a}{b^a})^\frac{1}{b}$$

But $$gcd(a^a,b^a)=1$$, and therefore $$(\frac{a^a}{b^a})^\frac{1}{b}$$ is irrational. It follows that if x^x is an integer, then either x is an integer or transcendental.

8. Jun 19, 2004

### abertram28

so there are also no solutions that are non intergers but rational, like 1/3 and 1/4?
i understand that x^x is an interger if x is an interger.

that tells us that the set of intergers for x gives us the solution set of y? there are no fractional xs for interger ys? sorry im dumb. i want to learn though!

9. Jun 20, 2004

### StonedPanda

I was wondering the same thing about taking the logarithm or ln of a function.

10. Jun 20, 2004

### Zurtex

I was interested in this quite some time ago. I finally found an article saying there is no known way of rearranging the equation:

$$y=x^x$$

In to some function of y in terms of x. But I did find an iterative formulae so you could approximate to as much accuracy as you wanted if you had y and wanted to know x. I'll see if I can find it again.