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Math Help - e powers (cancel down)

  1. #1
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    e powers (cancel down)

    Hello,

    How do you cancel this equation down further?

    Thanks

    ((exp(x)exp(-x))/4) + ((exp(x)exp(-x))/4) + ((exp(-2x))/4) + ((exp(2x))/4)
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  2. #2
    MHF Contributor Unknown008's Avatar
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    \dfrac{e^x e^{-x}}{4} + \dfrac{e^x e^{-x}}{4} + \dfrac{e^{-2x}}{4} + \dfrac{e^{2x}}{4}

    Is this what you mean?

    Remember that:

    a^b a^c = a^{b+c}
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  3. #3
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    Hello, MattWT!

    Are you working with hyperbolic functions?


    How do you cancel this equation down further?

    . . \dfrac{e^x\cdot e^{-x}}{4} + \dfrac{e^x\cdot e^{-x}}{4} + \dfrac{e^{-2x}}{4} + \dfrac{e^{2x}}{4}

    Since . e^x\cdot e^{-x} \:=\:e^0 \:=\:1

    . . we have: . \displaystyle \frac{1}{4} + \frac{1}{4} + \frac{e^{-2x}}{4} + \frac{e^{2x}}{4} \;=\;\frac{e^{2x}}{4} + \frac{1}{2} + \frac{e^{-2x}}{4}

    . . . . . . . . . . \displaystyle =\;\frac{e^{2x} + 2 + e^{-2x}}{4} \;=\;\frac{(e^x + e^{-x})^2}{4}

    . . . . . . . . . . =\;\left(\dfrac{e^x + e^{-x}}{2}\right)^2 \;=\;\cosh^2\!x
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  4. #4
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    I must have messed my algebra up at the beginning then, as the top line should be equal to 1:

    The question was to dy/dx tanh(x) by writing the hyperbolic functions first.
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