Suppose y=tanh inverse x , show y=1/2 ln(1+x)/(1-x)" ?

Question

Suppose y=tanh inverse x , show y=1/2 ln(1+x)/(1-x)" ?

Answer 1

y=tanh inverse x => tanh(y) = x

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

Or {(e^y + e^-y) + (e^y -- e^-y)} / {(e^y + e^-y) -- (e^y -- e^-y)} = (1+x)/(1--x)

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

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

Or 2y = ln{(1+x) / (1--x)}

Or y = (1/2)ln{(1+x) / (1--x)}

Answer 2

y = arctanh x

tanh y = x

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

Multiply top and bottom of fraction by e^y

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

e^2y - 1 = x*e^2y + x

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

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

2y = ln[(1 + x)/(1 - x)]

y = (1/2)ln[(1 + x)/(1 - x)]