# Thread: Need help with a basic equation

1. ## Need help with a basic equation

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

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

2. i think that it must be $e^x-e^{-x}=a$. Am i correct?

3. Doh! Yes, you are right. I have edited the post.

4. Originally Posted by Sabo
Doh! Yes, you are right. I have edited the post.
$e^x+e^{-x} = a \Leftrightarrow t+\frac{1}{t} = a \Leftrightarrow t^2-at+1=0$

5. Originally Posted by Sabo
Doh! Yes, you are right. I have edited the post.
Ahaa! That equation is well know. Now, i'll give you some help:

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

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

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

Multiplying by t both sides:

$t^2+1=at$

can you finish it?

6. Well, I guess it would go something along the lines of

$t^2 -at + 1 = 0$
$(t-\dfrac{a}{2})^2 - \dfrac{a^2+4}{4} = 0$
$(t-\dfrac{a}{2})^2 = \dfrac{a^2+4}{4}$
$t-\dfrac{a}{2} = +-\sqrt{\dfrac{a^2+4}{4}}$
$t = \dfrac{a +- \sqrt{a^2+4}}{2}$

For some reason I don't think that's the end of it though...

7. now remember that $e^x>0$

8. Originally Posted by Sabo
Well, I guess it would go something along the lines of

$t^2 -at + 1 = 0$
$(t-\dfrac{a}{2})^2 - \dfrac{a^2+4}{4} = 0$
$(t-\dfrac{a}{2})^2 = \dfrac{a^2+4}{4}$
$t-\dfrac{a}{2} = +-\sqrt{\dfrac{a^2+4}{4}}$
$t = \dfrac{a +- \sqrt{a^2+4}}{2}$

For some reason I don't think that's the end of it though...
Remember that you are serching for x, no t. Remember that, if you have $e^x=y \implies x=\ln(y)$

9. Hmm...

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

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