What is the inverse of (1.05)/(1+100e^{-0.285t)} ?

The inverse of (1.05)/(1+100e^{-0.285t)} is -(ln(\frac{1.05-t)/(100t))}{0.285}

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inverse of (1.05)/(1+100e^{-0.285t)}

inverse

\frac{1.05}{1 + 100 e ^{- 0.285 t}}

- \frac{ln ( \frac{1.05 - t}{100 t} )}{0.285}

Solution steps

y = \frac{1.05}{1 + 100 e ^{- 0.285 t}}

Replace

t

with

y

t = \frac{1.05}{1 + 100 e ^{- 0.285 y}}

Solve for

y, t = \frac{1.05}{1 + 100 e ^{- 0.285 y}}

- \frac{ln ( \frac{1.05 - t}{100 t} )}{0.285}

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

f (x) = lo g_{2} (x + 1) + 3

g (x) = \frac{x + 5}{3}

f (x) = - x - 2

- \frac{6}{7} x + \frac{32}{7}

What is the inverse of (1.05)/(1+100e^{-0.285t)} ?
The inverse of (1.05)/(1+100e^{-0.285t)} is -(ln(\frac{1.05-t)/(100t))}{0.285}