# How can we determine the value of a, b and c in logistic function: f(x) = A / (1 + Be –Cx)? Consider we have the value of f(x) and x.

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### 1 Answer

A is the horizontal asymptote as x->oo.

If we know the maximum value, of the function, we can find A.

If we have 2 other values we can find B and C.

Lets say we have (x_1, y_1) and (x_2, y_2)

`A = y_1(1+Be^(-Cx_1))= y_1 + y_1Be^(-Cx_1)`

`A = y_2(1+Be^(-Cx_2)) = y_2 + y_2Be^(-Cx_2)`

`Ay_2e^(-Cx_2) = y_1y_2e^(-Cx_2) + y_1y_2Be^(-Cx_1)e^(-Cx_2)`

`Ay_1e^(-Cx_1) = y_1y_2e^(-Cx_1) + y_1y_2Be^(-Cx_2)e^(-Cx_1)`

Subtract

`A(y_2e^(-Cx_2)-y_1e^(-Cx_1)) = y1_y_2(e^(-Cx_2)-e^(-Cx_1))`

`Ay_2e^(-Cx_2)-y_1y_2e^(-Cx_2) = Ay_1e^(-Cx_1)-y_1y_2e^(-Cx_1)`

`(Ay_2-y_1y2)e^(-Cx_2) = (Ay_1 - y_1y_2)e^(-Cx_1)`

`e^(-C(x_2-x_1)) = (Ay_1 - y_1y_2)/(Ay_2-y_1y_2)`

`-C(x_2 - x_1) = ln(Ay_1-y_1y_2) - ln(Ay_2-y_1y_2)`

`C = (ln(Ay_2 - y_1y_2) - ln(Ay_1-y_1y_2))/(x_2-x_1)`

We can find B from one of the original equations `y_1=A/(1+Be^(-Cx_1))` .

`B = (A - y_1)/(y_1e^(-Cx_1))`