What is the inverse of f(x)=(100)/(1+20e^{-0.05x)} ?

The inverse of f(x)=(100)/(1+20e^{-0.05x)} is -(ln(\frac{100-x)/(20x))}{0.05}

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inverse of f(x)=(100)/(1+20e^{-0.05x)}

inverse

f (x) = \frac{100}{1 + 20 e ^{- 0.05 x}}

- \frac{ln ( \frac{100 - x}{20 x} )}{0.05}

Solution steps

y = \frac{100}{1 + 20 e ^{- 0.05 x}}

Replace

x

with

y

x = \frac{100}{1 + 20 e ^{- 0.05 y}}

Solve for

y, x = \frac{100}{1 + 20 e ^{- 0.05 y}}

- \frac{ln ( \frac{100 - x}{20 x} )}{0.05}

- \frac{3}{4} x + 6

f (x) = \frac{- 2 x + 7}{6 x - 10}

f (x) = 10 - 6 x^{3}

f (x) = 4 - \frac{2 x}{3}

f (x) = \frac{x + 5}{x - 7}

What is the inverse of f(x)=(100)/(1+20e^{-0.05x)} ?
The inverse of f(x)=(100)/(1+20e^{-0.05x)} is -(ln(\frac{100-x)/(20x))}{0.05}