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


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

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

    $\displaystyle \displaystyle 2x = \ln 4$

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



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


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

    Let $\displaystyle e^x = y$

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

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

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

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

    Resubstituting $\displaystyle y$ for $\displaystyle e^x$:

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

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

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

    OR

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

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

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

    This still isn't it.
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  5. #5
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    $\displaystyle e^x - e^{-x} = 4
    $

    multiply every term by $\displaystyle e^x$ ...

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

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

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

    $\displaystyle e^x = 2 + \sqrt{5}$

    $\displaystyle x = \ln(2 + \sqrt{5})$
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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 $\displaystyle e^x$ in the numerator

    Otherwise $\displaystyle e^x - e^{ - x} - 4 = e^x - \frac{1}{e^{x}}- 4$
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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 $\displaystyle e^x$ ... 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 $\displaystyle e^x$ ... and subsequently easier to solve.
    Thanks, I see it now
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