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+sqrt(x^2-6x+1))/2),ln(-(-1+x-sqrt(x^2-6x+1))/2)

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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 ^{2} - 6 x + 1}{2}), ln (- \frac{- 1 + x - x ^{2} - 6 x + 1}{2})

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 ^{2} - 6 x + 1}{2}), ln (- \frac{- 1 + x - x ^{2} - 6 x + 1}{2})

f (x) = x - 1 + 1

(3.1) (- 5.3)

2 x^{3} - 3 x^{2} - 12

f (x) = lo g_{3} (x)

\frac{18 x + 80}{x ( x + 10 )}

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+sqrt(x^2-6x+1))/2),ln(-(-1+x-sqrt(x^2-6x+1))/2)