# tanh

• February 4th 2010, 01:59 AM
SubZero
tanh
Can some one please me with the following question:

Attachment 15207

• February 4th 2010, 02:30 AM
tonio
Quote:

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

Attachment 15207

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

Tonio
• February 4th 2010, 09:48 AM
SubZero
Thanks for the help.