Solve the equation $\displaystyle e^x - e^{-x} = a$ where a is an arbitrary real number.

I have got a hint, set $\displaystyle e^x = t$, but unfortunately that does not help me much.

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- Feb 10th 2010, 06:10 AMSaboNeed help with a basic equation
Solve the equation $\displaystyle e^x - e^{-x} = a$ where a is an arbitrary real number.

I have got a hint, set $\displaystyle e^x = t$, but unfortunately that does not help me much. - Feb 10th 2010, 06:18 AMfelper
i think that it must be $\displaystyle e^x-e^{-x}=a$. Am i correct?

- Feb 10th 2010, 06:26 AMSabo
Doh! Yes, you are right. I have edited the post.

- Feb 10th 2010, 06:28 AMdanielomalmsteen
- Feb 10th 2010, 06:28 AMfelper
Ahaa! That equation is well know. Now, i'll give you some help:

$\displaystyle e^x+e^{-x}=a$

Let be $\displaystyle e^x=t \implies e^{-x}= \dfrac{1}{t}$ then the equation transform to

$\displaystyle t+\dfrac{1}{t}=a$

Multiplying by t both sides:

$\displaystyle t^2+1=at$

can you finish it? - Feb 10th 2010, 06:48 AMSabo
Well, I guess it would go something along the lines of

$\displaystyle t^2 -at + 1 = 0$

$\displaystyle (t-\dfrac{a}{2})^2 - \dfrac{a^2+4}{4} = 0$

$\displaystyle (t-\dfrac{a}{2})^2 = \dfrac{a^2+4}{4}$

$\displaystyle t-\dfrac{a}{2} = +-\sqrt{\dfrac{a^2+4}{4}}$

$\displaystyle t = \dfrac{a +- \sqrt{a^2+4}}{2}$

For some reason I don't think that's the end of it though... - Feb 10th 2010, 06:52 AMdanielomalmsteen
now remember that $\displaystyle e^x>0$

- Feb 10th 2010, 06:53 AMfelper
- Feb 10th 2010, 07:21 AMSabo
Hmm...

$\displaystyle t = e^x = \dfrac{a+-\sqrt{a^2+4}}{2} \implies x = \ln(\dfrac{a+-\sqrt{a^2+4}}{2})$

I guess that can be considered a solution? Or? If it is, can we simplify it further?