Question

if x= e^y-e^-y/e^y+e^-y show that ,y=1/2log1+x/1-x base e

Answer 1

x=(e^y-e^-y)/(e^y+e^-y)

x (e^y+e^-y)=e^y-e^-y

x.e^y+xe^-y=e^y-e^-y

x^e^y-e^y=-e^-y-xe^-y

(1-x)e^y=e^-y (1+x)

e^y/e^-y=(1+x)/(1-x)

e^2y=(1+x)/(1-x)

take both side log (base e )

loge^2y=log (1+x)/(1-x)

2y=log (1+x)/(1-x)

y=1/2log (1+x)/(1-x)

Answer 2

Answer:

Step-by-step explanation: