# Domain of f(x,y)=x^y

1. Nov 10, 2012

### JuanSolo

1. The problem statement, all variables and given/known data
What is the domain of f(x,y)=x^y

3. The attempt at a solution
I thought it would be all real pairs except (0,0)
but it is x>0 and y all real numbers?

2. Nov 10, 2012

### Vorde

If $z=x^y$, then $ln(z)=yln(x)$

Do you see why x cannot be less than zero?

3. Nov 10, 2012

### Dick

How do you define (-1)^(1/2) or even worse (-1)^sqrt(2)?

4. Nov 10, 2012

### Zondrina

Look at this. Now where is ln(x) defined? Is it discontinuous anywhere? Are there any points where the function isn't defined...

5. Nov 10, 2012

### JuanSolo

Sorry, I don't think I fully understand
Would it not be defined when y is something like 2 or whatever, even if x is negative?

6. Nov 11, 2012

### SammyS

Staff Emeritus
How about when x < 0, if y = 1/2 or y = √(2). What then?

7. Nov 11, 2012

### loy

see this If z=x^y, then ln(z)=yln(x) , x cannot be less than zero since x is not defined if it is less than 0 . if you dont believe, you can press your calculator to see whether it has a result when you put in x a value which is less than 0.

8. Nov 11, 2012

### loy

but as long as x>0 , y can be defined in any value in R.
given x>0,
IF y>0 , f(x,y)=x^y .
IF y<0 , f(x,y)=x^(-y) = 1/(x^y)
IF y=0 , f(x,y)=1.
BUT when x<0 , the function is not defined , meaning , you can't get any answer, the function is discontinuous in x<0.
the domain of x is (0,∞) ,while y is (-∞,∞)

9. Nov 11, 2012

### vela

Staff Emeritus
It depends on your definition of exponentiation. If you were to restrict y to the integers, then yes, you could consistently define xy for negative values of x. On the other hand, for real numbers x and y, xy is often defined by xy=ey log x, which isn't defined if x<0.

10. Nov 11, 2012

### JuanSolo

I see
This is the first time I'm hearing of such a definition, I'll have to learn more about it
But thank you, I think I understand now

11. Nov 11, 2012

### SammyS

Staff Emeritus
$\displaystyle \text{If }\ y<0\,,\ \text{ then }f(x,y)=x^{-|y|} = \frac{1}{x^{|y|}}\ .$​