ahh...i need help on solving this problem...must be done algabraiacally..not by a graph...
3= e^(x) + e^(-x)
Of course solving for x.
$\displaystyle 3=e^x+\frac{1}{e^x}$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).
I'm not sure if this is what you are looking for...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
You have a good point. So, since cosh(x) is symmetric about the y-axis the solution transforms into $\displaystyle x = \pm cosh^{-1} \left ( \frac{3}{2} \right )$.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!