1. Help....calc Help !

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.

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

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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).

3. Originally Posted by Luckyjoshua
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.
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

4. 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.

5. 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!