# Solving e^(x^2) = 0: Is There Only One Answer?

• LearninDaMath

e^(x^2) = 0

## Homework Equations

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

## 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

That is not the correct answer.

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

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

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?

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?

That is correct :D

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)?

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 ?
That's right.

[...]
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)?

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