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Math Help - Solving for x in equations involving logs

  1. #1
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    Solving for x in equations involving logs

    Could somebody point me in the right direction?

    e^x + e^-^x = 2
    e^x (e^x + e^-^x) = 2*e^x
    e^2x + e = 2e^x
    e^2x - 2e^x + e = 0

    Something tells me I should be able to solve this as a quadratic, but e is a constant, and I can't figure out how to do this with a constant as the base.
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  2. #2
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    e^x+e^{-x}=2

    Times both sides by e^x

    e^{2x}+1=2 e^x

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

    make e^x = a giving

    a^{2}-2 a+1=0

    Solve from here?
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  3. #3
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    Ok, but why is e^x * e^-^x =1? Why isn't it e?
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  4. #4
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    e^x\times e^{-x}= e^{x-x}= e^0 = 1
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