MATH HELP

[Farmageddon]

Soriku said:

WereKitten said:

Soriku said:

WereKitten said:

Soriku said:

lol, no one helped me last time.

Almighty math wizard, what does x equal in e^x = 0. I can't figure it out :x

No solution in real numbers, but in hyperreal it would be "minus infinite"

Negative infinity sounds fine...what about positive infinity?

It's not a solution: e^(something big)=(something bigger), thus e^(positive infinity)=(positive infinity).

Are you familiar with limits? That's the only way these statements make sense in standard analysis.

Uh, I don't think we're that far yet.

Thanks anyway, I put down negative infinity.

The rigth answer is that no number exists to satisfy e^x = 0.

"Minus infinity" is just the limit to which x goes when the expression (e^x) approaches zero. But you can never really get it to equal zero.

Think of it this way, the smaller the x  you use, the smaller and closer to 0 e^x will be, but it'll never, ever, actually get there. So you can get as close to 0 as you want by using a "big" enough negative number even if you can never quite reach 0. That's why the notion of "limits" is used. 0 would be the limit of e^x as x approaches minus infinity.

But it's by no means a number that solves your equation. If you're at highschool or something like that, I think it would be more right to simply say such a number doesn't exist, or does not pertain the Real set, or that the solution set is empty, or anything like that.

Sorry about all the redundancy. Oh, and the questions are actually pretty easy if you know what you're doing  :P