# Logistic Equation through three points

I am trying to solve a logistic equation $$\frac{dP}{dt}=rP\left ( 1-\frac{P}{K} \right )$$ for the values of $$r$$ and $$K$$ through the points $$(0,100),(10,130),(20,160)$$. I have derived a general solution to be $$P(t)=\frac{P_0K}{P_0+(K-P_0)e^{-rt}}$$ so solving with first initial conditions gives $$P(t)=\frac{100K}{100+(K-100)e^{-rt}}$$. From here, how would I go about solving for the other variables $$r$$ and $$K$$ with the other initial conditions?

WolframAlpha calculated $$K=260$$ and $$r=\frac{\ln \left ( \frac{5}{8} \right )}{10}$$, but how would I solve it analytically?

I think there is an error in your posted data and you meant to write: $$(0,100),(10,130),(\color{red}{20},160)$$

We use the two additional values to set up two equations in two unknowns as

$$\dfrac{100 K}{(K-100) e^{-10 r}+100}=130 \\ \dfrac{100 K}{(K-100) e^{-20 r}+100}=160$$

Solving for $$K$$ and $$r$$ yields

$$K=260, ~~ r = \dfrac{\ln \left ( \dfrac{\color{red}{8}}{\color{red}{5}} \right )}{10}$$

Note: We can solve the system many ways, for example, using the first equation to solve for $$r$$

$$r = \dfrac{1}{10} \ln \left(\dfrac{13 (K-100)}{10 (K-130)}\right)$$

We can now substitute that in the second equation and then solve for $$K$$

$$\dfrac{169 (K-100)}{K-91}=160 \implies K = 260$$