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Math Help - tanh

  1. #1
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    tanh

    Can some one please me with the following question:

    tanh-untitled.jpg

    Thank you in advance
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  2. #2
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    Quote Originally Posted by SubZero View Post
    Can some one please me with the following question:

    Click image for larger version. 

Name:	untitled.JPG 
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ID:	15207

    Thank you in advance

    y=\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1}\Longrightarrow e^{2x}(y-1)=-(y+1) \Longrightarrow e^{2x}=\frac{1+y}{1-y} \Longrightarrow x=\frac{1}{2}\,\ln\left(\frac{1+y}{1-y}\right)=\ln \sqrt{\frac{1+y}{1-y}} /

    Now answer your question

    Tonio
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  3. #3
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    Thanks for the help.
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