# hyperbolic functions inverse

• Jul 27th 2006, 01:32 PM
Luckyjoshua
Help....calc Help !
ahh...i need help on solving this problem...must be done algabraiacally..not by a graph... :eek:

3= e^(x) + e^(-x)

Of course solving for x.
• Jul 27th 2006, 01:36 PM
ThePerfectHacker
Quote:

Originally Posted by Luckyjoshua
ahh...i need help on solving this problem...must be done algabraiacally..not by a graph... :eek:

3= e^(x) + e^(-x)

Of course solving for x.

$\displaystyle 3=e^x+\frac{1}{e^x}$
Thus,
$\displaystyle 3e^x=e^{2x}+1$
Thus,
$\displaystyle e^{2x}-3e^x+1=0$
Let,
$\displaystyle \chi=e^x$
Then,
$\displaystyle \chi^2-3\chi +1=0$

----
This should not be called a calculus thread. Because it has nothing to do with calculus (unless you are doing hyperbolic functions).
• Jul 27th 2006, 01:40 PM
topsquark
Quote:

Originally Posted by Luckyjoshua
ahh...i need help on solving this problem...must be done algabraiacally..not by a graph... :eek:

3= e^(x) + e^(-x)

Of course solving for x.

I'm not sure if this is what you are looking for...

$\displaystyle 3 = e^x + e^{-x}$

$\displaystyle 3 = 2 \cdot \frac{e^x + e^{-x}}{2}$

$\displaystyle 3 = 2 cosh(x)$

$\displaystyle cosh(x) = \frac{3}{2}$

$\displaystyle x = cosh^{-1} \left ( \frac{3}{2} \right )$

-Dan
• Jul 27th 2006, 01:48 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
I'm not sure if this is what you are looking for...

$\displaystyle 3 = e^x + e^{-x}$

$\displaystyle 3 = 2 \cdot \frac{e^x + e^{-x}}{2}$

$\displaystyle 3 = 2 cosh(x)$

$\displaystyle cosh(x) = \frac{3}{2}$

$\displaystyle x = cosh^{-1} \left ( \frac{3}{2} \right )$

-Dan

That is not the ideal method. Because the inverse hyper-cosine is a function it only returns one value. There might be more (which is the case here). Thus, you only solved have of the problem.
• Jul 27th 2006, 02:00 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
That is not the ideal method. Because the inverse hyper-cosine is a function it only returns one value. There might be more (which is the case here). Thus, you only solved have of the problem.

You have a good point. So, since cosh(x) is symmetric about the y-axis the solution transforms into $\displaystyle x = \pm cosh^{-1} \left ( \frac{3}{2} \right )$.

Do you (TPH) have another method to get to this point directly? (If so I would be interested to see it. Really!)

-Dan

Oops! Didn't see your post under mine! :) Gotcha!