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.

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- Jul 27th 2006, 01:32 PMLuckyjoshuaHelp....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 PMThePerfectHackerQuote:

Originally Posted by**Luckyjoshua**

Thus,

$\displaystyle 3e^x=e^{2x}+1$

Thus,

$\displaystyle e^{2x}-3e^x+1=0$

Let,

$\displaystyle \chi=e^x$

Then,

$\displaystyle \chi^2-3\chi +1=0$

Now you can solve quadradic.

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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 PMtopsquarkQuote:

Originally Posted by**Luckyjoshua**

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

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

$\displaystyle 3 = 2 cosh(x)$

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

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

-Dan - Jul 27th 2006, 01:48 PMThePerfectHackerQuote:

Originally Posted by**topsquark**

- Jul 27th 2006, 02:00 PMtopsquarkQuote:

Originally Posted by**ThePerfectHacker**

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!