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Pre-calc question: e?

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Solve algebraically: e^x + e^-x = 9

Any help would be much appreciated. Please explain so that I can do similar problems on my own. Thanks and 10 points to the best answer.

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  1. X e^x eq becomes

    e^2x + 1 = 9e^x

    let u = e^x, eq becomes

    u^2 -9u + 1 = 0

    use quad formula to solve for u

    then find x


  2. (1) First, multiply both sides by e^x to obtain the following:

    e^(2x) + 1 = 9*e^x, a quadratic equation in variable e^x.

    (2) subtract 9*e^x from both sides, resulting in the "classic" form of a quadratic equation

    e^(2x) - 9*e^x + 1 = 0

    (3) Solve this quadratic for the two solutions of e^x:

    e^x = [9 + sqrt(81 - 4)]/2 and e^x = [9 - sqrt(81 - 4)]/2

    (4) The two results are: e^x = 8.88748 and e^x = 0.112518

    (5) Taking the natural log (ln) of these numbers yields

    x = + 2.1846

    and

    x = - 2.1846. [Note: Remember that ln(e^x) = x* ln(e) = x]

  3. this ends up being a quadratic equation...

    remember that e^-x = 1/e^x

    multiply everything by e^x ==> (e^x)^2 + 1 = 9e^x

    but (e^x)^2 = e^(2x)

    now you have e^(2x)- 9e^x + 1 = 0

    let e^x = u, giving u^2 - 9u + 1 = 0

    using the QF:

    u = [9 +/- sqrt(81 - 4)] / 2

    u = [9 +/- sqrt(77)] / 2

    since u = e^x, then x = ln u

    so x = ln{[9 +/- sqrt(77)] / 2}

    both values inside the ln are positive, so there are two solutions:

    x = ln{[9 + sqrt(77)] / 2} and x = ln{[9 - sqrt(77)] / 2}
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