# E^(x^2) = 0

1. Nov 18, 2011

### LearninDaMath

1. The problem statement, all variables and given/known data

e^(x^2) = 0

2. Relevant equations

How do you solve this for x? Never had to solve a power to a power before.

3. The attempt at a solution

e^x^2 = 0

lne^x^2 = ln0

log(base e)of e^(x^2) = 1

e^(1) = e^(x^2)

1 = x^(2)

x = √1

2. Nov 18, 2011

### D H

Staff Emeritus
That is not the correct answer.

3. Nov 18, 2011

### gb7nash

Think about it. Can a positive constant raised to any number equal to 0?

4. Nov 18, 2011

### romsofia

Think about what gb7 said, also just so you know ln(0) is undefined, it's not equal to 0!

5. Nov 18, 2011

### LearninDaMath

Ohh!! so there must be no solution then, right?

And double Ohh!!, ln0 is undefined...its ln1 that equals zero!!

Is my assumption of no solution correct?

6. Nov 18, 2011

### romsofia

That is correct :D

7. Nov 18, 2011

### LearninDaMath

Awsome!! thanks.

So if f(x) = (2x^(2) + 1)(e^(x^2))

and f(x) = 0

so that (2x^(2) + 1)(e^(x^2)) = 0 and I solve for x,

e^(x^2) = 0 is no solution

and (2x^(2) + 1) = 0 is:

x^2 = -1/2 so

x = √-1/2

And since √-1/2 has a negative on the inside, it's a complex number and has no solution also, right?

Therefore, are there no real roots for (2x^(2) + 1)(e^(x^2)) = 0 ?

And since this happens to be a first derivative function, does that then mean there are no critical points on the original function xe^(x^2)?

8. Nov 18, 2011

### eumyang

That's right.

9. Nov 19, 2011

### dextercioby

Yes, the function's increasing as x takes all real values from -infinity to +infinity.