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Math Help - Hyperbolic Function (tanh & sech)

  1. #1
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    Hyperbolic Function (tanh & sech)

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

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

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

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

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

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

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

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

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

    This factorises

    y=e^x

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

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

    -> the solutions are...

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