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Thread: Hyperbolic Function (tanh & sech)

  1. #1
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    Hyperbolic Function (tanh & sech)

    Q: Solve the equation $\displaystyle 4 \mathrm{tanh}x - \mathrm{sech} x = 1$ giving your solution in logarithmic form.
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    My Method:

    $\displaystyle 4 \mathrm{tanh}x - \mathrm{sech} x = 1$

    $\displaystyle \therefore4 \left( \frac{e^x - e^{-x}}{e^x + e^{-x}} \right) - \left( \frac{2}{e^x + e^{-x}} \right) = 1$

    $\displaystyle \therefore \frac{4 (e^x - e^{-x}) -2}{e^x + e^{-x}} = 1$

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

    $\displaystyle \therefore 3e^x - 5e^{-x} - 2 = 0$

    $\displaystyle \therefore 3e^{2x} - 2e^x - 5 = 0$

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    However, this does not factorise and give an nice $\displaystyle \ln$ form answer so where have I gone wrong on my method? The answer is supposed to be $\displaystyle \ln \left( \frac 53 \right)$.
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    Quote Originally Posted by Air View Post
    $\displaystyle \therefore 3e^{2x} - 2e^x - 5 = 0$

    __________________
    However, this does not factorise and give an nice $\displaystyle \ln$ form answer so where have I gone wrong on my method? The answer is supposed to be $\displaystyle \ln \left( \frac 53 \right)$.
    Hello,

    This factorises

    $\displaystyle y=e^x$

    $\displaystyle 3e^{2x} - 2e^x - 5 = 0 \Longleftrightarrow 3y^2-2y-5=0$

    $\displaystyle \Delta=4+4\cdot5\cdot3=8^2$

    -> the solutions are...

    (note that y>0 since y=e^x)
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    Oh yes, It factorised to $\displaystyle (5e^x - 3)(e^x +1)$ and $\displaystyle e^x$ cannot be negative hence $\displaystyle (e^x +1)$ is ignored. Thanks.
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