# Math Help - tanh

1. ## tanh

Can some one please me with the following question:

Thank you in advance

2. Originally Posted by SubZero
Can some one please me with the following question:

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

3. Thanks for the help.