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Math Help - how can i solve this equation for x

  1. #1
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    how can i solve this equation for x

    this is part of a larger question for hyperbolic functions, but i feel if i get this the rest will come into place...what is x of tanh x= 3/4, or you can solve the below equation

    3/4=(e^x - e^-x)/(e^x + e^-x)

    any help is very appreciated
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  2. #2
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    Quote Originally Posted by twostep08 View Post
    this is part of a larger question for hyperbolic functions, but i feel if i get this the rest will come into place...what is x of tanh x= 3/4, or you can solve the below equation

    3/4=(e^x - e^-x)/(e^x + e^-x)

    any help is very appreciated
    You can just run arctanh \left(\frac{3}{4}\right)

    my calculator says it is equal to 0.973 (3sf)
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  3. #3
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    im not really allowed to use my calculator...there must be some nice answer using logs?
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  4. #4
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    Quote Originally Posted by twostep08 View Post
    this is part of a larger question for hyperbolic functions, but i feel if i get this the rest will come into place...what is x of tanh x= 3/4, or you can solve the below equation

    3/4=(e^x - e^-x)/(e^x + e^-x)

    any help is very appreciated
    Yeah there is. You can use standard algebraic techniques and logs. Ignore the second solution if you only want a real answer

    3e^x+3e^{-x} = 4e^x-4e^{-x}

    e^x-7e^{-x} = 0<br />

    e^x - \frac{7}{e^x} = 0

    e^{2x} - 7 = 0

    (e^x-\sqrt{7})(e^x+\sqrt{7}) = 0

    x = \frac{1}{2}ln(7) or x = \frac{1}{2}ln(-7) = \frac{1}{2}ln(7)+ i \left(\frac{\pi}{2}\right)
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  5. #5
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    ok im following you all the way to the last step when you say e^x - (sq.rt.)7 = 1/2ln(7)

    what makes that so?
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    Quote Originally Posted by twostep08 View Post
    ok im following you all the way to the last step when you say e^x - (sq.rt.)7 = 1/2ln(7)

    what makes that so?
    It copmes from the difference of two squares when factorising the quadratic and the term is equal to 0

    e^x-\sqrt7 = 0

    From here add \sqrt7 to both sides and take the natural log

    I used the log power law to bring the 1/2 to the front
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  7. #7
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    e^x-\sqrt{7} = 0 \Rightarrow log(e^x) = x = log(\sqrt{7}) = \frac{1}{2}log7
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  8. #8
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    Quote Originally Posted by Defunkt View Post
    e^x-\sqrt{7} = 0 \Rightarrow log(e^x) = x = log(\sqrt{7}) = \frac{1}{2}log7
    Just to point out that Defunkt is also referring to the natural log ie base e

    log_e(e^x) = ln(e^x) = x
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