How do you isolate X in the following:

[CrazCo]

e^x+e^-x=3

 

i tried xlne+-xlne=ln3

x(1)+-x(1) = ln3

x-x = ln3

 

 

but that doesn't work

 

btw the answer is +- 0.96

[Bignose]

have you learned about the hyperbolic trig functions yet?

[CrazCo]

no :S

[Cap'n Refsmmat]

If you have this:

 

[math]e^x + e^{-x} = 3[/math]

 

taking the natural log of both sides gives you this:

 

[math]\ln (e^x + e^{-x}) = \ln 3[/math]

 

not what you wrote above. See what you can do using the properties of logarithms to tease x out of there.

[CrazCo]

tbh i have never seen anything like that.

[D H]

[math]\ln (e^x + e^{-x}) = \ln 3[/math]

 

See what you can do using the properties of logarithms to tease x out of there.

There is no teasing that x out of there. Bignose suggestion to look into the hyperbolic functions is exactly right.

[Cap'n Refsmmat]

Indeed. I just realized the method I had done quickly before breaks a few rules as well... but it does give me an answer of 0.97, which is confusingly close.

[ed84c]

My suggestion would be first to multiply through by e^x and move everything to one side (ie. you get ...... = 0).

 

Then take a leap of faith as this this is a very similar function to solve to some you have been solving for years.

 

(Hint: The answer to these is often = +/- .... likeyou have there)

[D H]

My suggestion would be first to multiply through by e^x and move everything to one side (ie. you get ...... = 0).

 

Then take a leap of faith as this this is a very similar function to solve to some you have been solving for years.

 

(Hint: The answer to these is often = +/- .... likeyou have there)

Yep. That will do, also, and that is probably what is wanted for this (what appears to be homework) problem.

 

Of course saying acosh(1.5) is much easier.

[ed84c]

Yup true.

 

And I guess even if homework Qs arelooking for use of hyperbolics, it could be good practice to do it the long way round as you end up effectively deriving the equation for acosh,

[D H]

The hint ed84c gave is a bit cryptic (and appropriately so). Several hours have passed.

 

A bit less cryptically, the idea is the solve for [math]e^x[/math]. Set [math]u=e^x[/math]. The equation in the original post becomes [math]u+1/u=3[/math].

 

Now multiply both sides by u. What kind of equation results from this?

[CrazCo]

LIKE THIS?

 

e^x + 1/e^x = 3

 

e^2x -3(e^x)+1=0

 

Assume A = e^x

 

A^2 - 3A + 1 = 0

 

Quadratic Formula

 

3+- sqrt 9-4/2 = A

 

3+- sqrt5/2 = A

 

Replace

 

3+sqrt5/2 = e^x

 

x = ln(3+sqrt5/2) = .96

X = ln(3-sqrt5/2) = -.96

 

 

im proud i got it!!!!

[D H]

There you go!