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Math Help - hyperbolic functions inverse

  1. #1
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    Exclamation 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.
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  2. #2
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    Quote 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.
    3=e^x+\frac{1}{e^x}
    Thus,
    3e^x=e^{2x}+1
    Thus,
    e^{2x}-3e^x+1=0
    Let,
    \chi=e^x
    Then,
    \chi^2-3\chi +1=0
    Now you can solve quadradic.

    ----
    This should not be called a calculus thread. Because it has nothing to do with calculus (unless you are doing hyperbolic functions).
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  3. #3
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    Quote 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...

    3 = e^x + e^{-x}

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

    3 = 2 cosh(x)

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

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

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark
    I'm not sure if this is what you are looking for...

    3 = e^x + e^{-x}

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

    3 = 2 cosh(x)

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

    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.
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  5. #5
    Forum Admin topsquark's Avatar
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    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 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!
    Last edited by topsquark; July 27th 2006 at 02:02 PM. Reason: Minus sign missing! and post script!
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