# Thread: Solving Unknowns in a Logistic Growth Model

1. ## Solving Unknowns in a Logistic Growth Model

Hello! I have the model y= A/1+Be^(xC) (where C= -Ak and e is the natural logarithm) and, first, I wanted to know if anyone is familiar with this particular model.

Second, I have been trying to solve for the unknowns A, B, and C by plugging in three points on a graph that I have, thusly:

16.4 = A/1+Be^(2C), 16.8 = A/1+Be^(10C), and 21.6 = A/1+Be^(19C)

I decided to set up a system of equations and solve for each variable one at a time, so first I isolated A in the first two equations:

A = 16.4 + 16.4Be^(2C)
A = 16.8 + 16.8Be^(10C)

and set them equal: 16.4 + 16.4Be^(2C) = 16.8 + 16.8Be^(10C)

From this point, I ran into a snag because I wasn't sure how I wanted to bring the exponents down without eliminating B entirely. I thought about simplyifying by dividing by B so that I have 16.4e^(2C) = .4/B + 16.8e^(10C), but I'm not sure that helps.

My qustion is, really, what is my next step so that I can finish solving for the system?

2. Originally Posted by bee.talbot
Hello! I have the model y= A/1+Be^(xC)
I assume you mean A/(1+ Be^(xC)). What you wrote is A+ Be^(xC).

(where C= -Ak and e is the natural logarithm) and, first, I wanted to know if anyone is familiar with this particular model.

Second, I have been trying to solve for the unknowns A, B, and C by plugging in three points on a graph that I have, thusly:

16.4 = A/1+Be^(2C), 16.8 = A/1+Be^(10C), and 21.6 = A/1+Be^(19C)

I decided to set up a system of equations and solve for each variable one at a time, so first I isolated A in the first two equations:

A = 16.4 + 16.4Be^(2C)
A = 16.8 + 16.8Be^(10C)

and set them equal: 16.4 + 16.4Be^(2C) = 16.8 + 16.8Be^(10C)
From this point, I ran into a snag because I wasn't sure how I wanted to bring the exponents down without eliminating B entirely. I thought about simplyifying by dividing by B so that I have 16.4e^(2C) = .4/B + 16.8e^(10C), but I'm not sure that helps.
You have used the first two equations to eliminate A, but still have both B and C in one equation. You need to use the third equation with either the first or second equation to eliminate A also so that you have two equations in B and C. Then eliminate either B or C to have one equation in a single variable.

My qustion is, really, what is my next step so that I can finish solving for the system?