What is the inverse of y= 1/(1-e^{\frac{-x){2}}} ?

The inverse of y= 1/(1-e^{\frac{-x){2}}} is ln((x^2)/(x^2-2x+1))

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inverse of y= 1/(1-e^{\frac{-x){2}}}

inverse

y = \frac{1}{1 - e ^{\frac{- x}{2}}}

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

Solution steps

y = \frac{1}{1 - e ^{\frac{- x}{2}}}

Replace

x

with

y

x = \frac{1}{1 - e ^{\frac{- y}{2}}}

Solve for

y, x = \frac{1}{1 - e ^{\frac{- y}{2}}}

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

ln (1 + x^{2})

f (x) = \frac{15 - 4 x}{4 - x}

f (x) = - (x + 5)^{2} - 4

3 tan (2 x - \frac{π}{3})

f (x) = \frac{6 x + 6}{5 x - 6}

What is the inverse of y= 1/(1-e^{\frac{-x){2}}} ?
The inverse of y= 1/(1-e^{\frac{-x){2}}} is ln((x^2)/(x^2-2x+1))