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

[LearninDaMath]

Homework Statement

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

If this is the correct answer, is it the only answer?

 

[D H]

That is not the correct answer.

 

[gb7nash]

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

 

[romsofia]

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

 

[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?

 

[romsofia]

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

 

[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)?

 

[eumyang]

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.

 

[dextercioby]

[...]
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.