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Math Help - Show that x =

  1. #1
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    Show that x =

    Hi,

    My uncertainty lies in the approach to the question. Anyway, I made an attempt.

    Given \displaystyle e^x - e^{-x} = 4, show that \displaystyle x = \ln(2+\sqrt{5})


    \displaystyle \ln e^x - \ln e^{-x} = \ln 4

    \displaystyle x - (-x) = \ln 4

    \displaystyle 2x = \ln 4

    \displaystyle x = \ln 2
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  2. #2
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    e^x  - e^{ - x}  = 4\; \equiv \,e^{2x}  - 4e^x  - 1=0.
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  3. #3
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    \displaystyle\ln(e^x/e^{-x})=\ln e^x - \ln e^{-x}, but \displaystyle\ln(e^x-e^{-x})\ne\ln e^x - \ln e^{-x}.
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  4. #4
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    How did you get this part: e^{2x} - 4e^x - 1=0?
    Why did you set it as equal to zero?
    Is there any other way of doing this?



    ================================================== ===


    \displaystyle e^{2x} - 4e^x - 1=0

    Let e^x = y

    \displaystyle= y^2 - y - 1 = 0

    \displaystlye= (y - \frac{1}{2})^2 - \frac{5}{4} = 0

    y = \frac{1}{2}(1 - \sqrt{5})

    y = \frac{1}{2}(1 + \sqrt{5})

    Resubstituting y for e^x:

    e^x = \frac{1}{2}(1 + \sqrt{5})

    \ln e^x = \ln \frac{1}{2}(1 + \sqrt{5})

    x = \ln \frac{1}{2}(1 + \sqrt{5})

    OR

    e^x = \ln \frac{1}{2}(1 - \sqrt{5})

    \ln e^x = \ln \frac{1}{2}(1 - \sqrt{5})

    x = \ln \frac{1}{2}(1 - \sqrt{5})

    This still isn't it.
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  5. #5
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    e^x - e^{-x} = 4<br />

    multiply every term by e^x ...

    e^{2x} - 1 = 4e^x

    e^{2x} - 4e^x - 1 = 0

    quadratic formula (remember e^x > 0 for all x)

    e^x = 2 + \sqrt{5}

    x = \ln(2 + \sqrt{5})
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  6. #6
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    Thanks.

    What's the reason for multiplying throughout by e^x?
    No other way of doing this?
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  7. #7
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    Quote Originally Posted by Hellbent View Post
    What's the reason for multiplying throughout by e^x?
    No other way of doing this?
    It is done to make the e^x in the numerator

    Otherwise e^x - e^{ - x} - 4 = e^x - \frac{1}{e^{x}}- 4
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  8. #8
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    Quote Originally Posted by Hellbent View Post
    Thanks.

    What's the reason for multiplying throughout by e^x?
    No other way of doing this?
    to make the equation quadratic in e^x ... and subsequently easier to solve.
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  9. #9
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    Quote Originally Posted by skeeter View Post
    to make the equation quadratic in e^x ... and subsequently easier to solve.
    Thanks, I see it now
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