Can some one please me with the following question:

Attachment 15207

Thank you in advance

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- Feb 4th 2010, 01:59 AMSubZerotanh
Can some one please me with the following question:

Attachment 15207

Thank you in advance - Feb 4th 2010, 02:30 AMtonio

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

Now answer your question(Happy)

Tonio - Feb 4th 2010, 09:48 AMSubZero
Thanks for the help.