# Logistic function?

Solve for the logistic function with initial condition = 14, limit to growth = 42 and passing through (1, 28).

### 2 Answers

- Ian HLv 71 month ago
With finite resources an initial population Po grows nearly exponentially at first, but the population P eventually approaches a sort of saturation level, known as the carrying capacity K. This is the population level at which the birth and death rates of a species precisely match, resulting in a stable population over time.

Such a logistics growth model is derived by assuming dP/dt= rP(1 - P/K)

Details of the derivation can be found at

http://www.math.northwestern.edu/~mlerma/courses/m...

The solution takes the form

P = K/[1 + Ae^-rt] where A = (K - Po )/Po, ....(1)

In this example K = 42, Po = 14, so, A = 28/14 = 2 and

P = 42/[1 + 2e^-rt] and we can find r using P = 28 when t = 1

28 = 42/[1 + 2e^-r] yielding r = ln 4 and -r = ln(1/4)

e^-rt = (1/4)^t

P(t) = 42 /[1 + 2*(1/4)^t] or if you prefer

P(t) = 42 /[1 + 2*2^(-2t)]

https://www.wolframalpha.com/input/?i=P%28t%29+%3D...

Note: We could have started with the form given by Slowfinger

P(t) = K/[1 + Ab^t) which uses b = e^-r = 1/4

- SlowfingerLv 61 month ago
General form

f(x) = c / (1 + ab^x)

limit to growth is 42 => c=42

initial condition is 14 meaning f(0)=14

in above equation, substitute c=42, f(x)=14 and x=0

14 = f(0) = 42 / (1 + ab^0) = 42 / (1+a)

divide by 14

1 = 3 / (1+a)

solve for a

1+a = 3/1 = 3

a = 3 - 1 = 2

b is still unknown. Find it from condition f(1)=28

28 = 42 / (1 + 2*b^1)

28 = 42 / (1 + 2b)

solve for b

1+2b = 42/28

1+2b = 3/2

2b = 3/2 - 1 = 1/2

b = 1/4

Finally, we have a logistic equation

f(x) = 42 / (1 + 2(1/4)^x)