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    excaly's Avatar
    excaly Posts: 3, Reputation: 1
    New Member

    May 12, 2010, 06:06 PM
    Quadratic: Solve (e^x - e^-x)/2 = 2 (sinhx = 2)
    I know x = 1.443 but I need to solve it now using algebra, not my calculator.

    Pauls Online Notes : Algebra - Solving Exponential Equations
    Algebra II: Exponential and Logarithmic Functions - Math for Morons Like Us

    These are 2 of the sites I've been on, I've also gone through my textboox but I can't find any information for when you have two e's and two uses of x in the equation.

    This is as far as I got.
    (e^x - e^-x)/2 = 2
    e^x - e^-x = 4
    ln4= lne^x - lne^-x

    Thanks in advance, I'm really struggling here.
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
    Ultra Member

    May 13, 2010, 04:49 AM

    Are you familiar with Newton's Method?
    excaly's Avatar
    excaly Posts: 3, Reputation: 1
    New Member

    May 13, 2010, 06:46 AM
    No I am not
    ebaines's Avatar
    ebaines Posts: 12,130, Reputation: 1307

    May 13, 2010, 07:13 AM

    Newton's method is an approximation technique - very effective at getting an estimate of the answer to as many decimals of accuracy as you like. However, there is a way to get an exact solution here. Multiply through by e^x and rearrange, and you get:

    Now substitute :

    Solve using the quadratic equation, but remember that only the positive solution of u is valid. Then convert back to x using

    You will find that

    One other point - in your original attempt you wrote this:

    But that is incorrect. If you take the log of the left hand side, you must also take the log of the right hand side as a whole:

    But log(a-b) does NOT equal log(a) - log(b). So your last step is invalid.
    excaly's Avatar
    excaly Posts: 3, Reputation: 1
    New Member

    May 13, 2010, 07:18 AM

    Wow thank you very much ebaines and galactus.

    Very very helpful.

    Ill spend a while on here tomorrow and see if I can help out any other members :)

    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
    Ultra Member

    May 13, 2010, 11:36 AM

    Ebaine's solution I think is best, but here is the Newton's method I mentioned. It comes in handy when the equations are hard to solve algebraically. This one can be though.

    The derivative of is 2cosh(x)

    So, we have:

    Use an initial guess of say, 1.5

    Plugging that in the above we get


    Now, iterate by plugging this into the formula. Then, we get:


    There. Close enough.

    That is the solution.

    We could also have tried another guess of , say, 1. It would converge to the same solution, though not quite as fast.

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