I never liked logarithms. Reps if you find x.

[Arcadio]

I don't remember the steps to solve equations like this :

e^x - e^-x = 0

It's probably something extremely simple but I just don't remember.

Anyone?

 

[busydude]

Anyone?

Base? looking at the equation it must be a natural log.

 

[jman19]

How are you going to give reps?

ln is your friend.

 

[Arcadio]

Oh, please don't tell me the answer is 0. I would feel dumb if it was 0.

 

[Brigandier]

x=0

 

[Praxis1452]

dude just look up the properties of logarithms.

Alternatively, recognize e^x=e^-x. From that equation. x=-x.

 

[rcpratt]

ln(e^(x)) = x

x - (-x) = 0

2x = 0

x = 0

 

[OOBradm]

 

[busydude]

Hehe.. I thought you were trying to find the log(e^x-e^-x).. you don't need log to solve for x.

 

[darkewaffle]

Nevermind. I suck at this stuff.

 

[yh125d]

self-ownage of the year?

 

[Matthiasa]

Yup it's 0.

 

[MikeMike]

uh...

 

[Kadarin]

Derp.

 

[jman19]

How did you manage to screw that up? Fail.

 

[Soundmanred]

I'd just write a portable app to solve it.

 

[blinblue]

There should be a rule that you must first put your math problem through Wolfram Alpha before asking for just an answer

The Answer

Besides x=0, there are unreal answers too. x=i*pi*n, where n is any integer, and i=sqrt(-1)

 

[Shadow Conception]

^how'd you find those? Is that the DeMoivre's Theorem stuff?

 

[Lifted]

LOL. Post of the day. :thumbsup:

 

[DrPizza]

It's kind of funny that you don't like logs, because
1. Logarithms are your friend in higher mathematics
2. You don't need logarithms to solve the problem. You simply need to understand exponents.

e^-x means 1/e^x.

Get a common denominator
e^2x/e^x - 1/e^x = 0
(e^2x-1)/e^x = 0

Obviously, the denominator can't be zero, so the numerator must be zero for that fraction to equal zero.
e^2x - 1 = 0
So, e^2x = 1

Hmmmmm... What's the value of the exponent so that any non-zero real number raised to that exponent is equal to 1? 0??!
So, 2x=0
x = 0

Logs? I didn't need no stinkin logs.

 

[jman19]

It's kind of funny that you don't like logs, because
1. Logarithms are your friend in higher mathematics
2. You don't need logarithms to solve the problem. You simply need to understand exponents.

e^-x means 1/e^x.

Get a common denominator
e^2x/e^x - 1/e^x = 0
(e^2x-1)/e^x = 0

Obviously, the denominator can't be zero, so the numerator must be zero for that fraction to equal zero.
e^2x - 1 = 0
So, e^2x = 1

Hmmmmm... What's the value of the exponent so that any non-zero real number raised to that exponent is equal to 1? 0??!
So, 2x=0
x = 0

Logs? I didn't need no stinkin logs.

:thumbsup:

 

[Born2bwire]

^how'd you find those? Is that the DeMoivre's Theorem stuff?

It's trivial. Use Euler's Formula and realize that it reduces to 0.5*sinh(x). So any integer multiple of i\pi will work.

Honestly OP, for a guy that claims Colombia's schools is the shiznit this is pretty simple.

 

[amdhunter]

Base? looking at the equation it must be a natural log.

I have a natural log, wanna look at it? :awe:
It has a nice base too.

 

[zCypher]

I have a natural log, wanna look at it? :awe:
It has a nice base too.

pics?

 

[TecHNooB]

It's trivial. Use Euler's Formula and realize that it reduces to 0.5*sinh(x). So any integer multiple of j\pi will work.

Honestly OP, for a guy that claims Colombia's schools is the shiznit this is pretty simple.

fixed