What is the inverse of y=(1-e^{-x})/(1+e^{-x)} ?

The inverse of y=(1-e^{-x})/(1+e^{-x)} is -ln((1-x)/(x+1))

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

inverse

y = \frac{1 - e ^{- x}}{1 + e ^{- x}}

- ln (\frac{1 - x}{x + 1})

Solution steps

y = \frac{1 - e ^{- x}}{1 + e ^{- x}}

Replace

x

with

y

x = \frac{1 - e ^{- y}}{1 + e ^{- y}}

Solve for

y, x = \frac{1 - e ^{- y}}{1 + e ^{- y}}

- ln (\frac{1 - x}{x + 1})

\frac{- 3 x - 9}{5 x + 15}

g (x) = \frac{x}{5}

f (x) = - \frac{4}{5} x - 12

f (x) = \frac{x ^{2} - x - 12}{2 x - 8}

f (x) = \frac{2 x + 5}{3}

What is the inverse of y=(1-e^{-x})/(1+e^{-x)} ?
The inverse of y=(1-e^{-x})/(1+e^{-x)} is -ln((1-x)/(x+1))