How to solve for $3$ constants with $3$ equations

Question

I am working to solve a formula for population, where I have the following model: $P(t)= M/(1+Ae^{-Mkt})$

I have the first data points for the model: $(0,2700000), (1,3050000), (2,3450000)$

This gives me $3$ equations:

$2700000=M/(1+A),$

$3050000=M/(1+Ae^{-Mk})$

$3450000=M/(1+Ae^{-2Mk})$

However, I am not sure how to solve for each of the $3$ variables, $M, k$ and $A$? Any help would be much appreciated.

Answer 1

There are many ways to solve the problem. Here is another one : first let $e^{-Mk}=x$ and consider $$2700000=\frac M{1+A} \tag 1$$ $$3050000=\frac M{1+A x} \tag 2$$ $$3450000=\frac M{1+A x^2} \tag 3$$ $$x=e^{-Mk}\tag 4$$ $$\frac{(1)}{(2)} \implies \frac{54}{61}=\frac{A x+1}{A+1}\implies A=\frac{7}{54-61 x} \tag 5$$ $$\frac{(1)}{(3)} \implies \frac{18}{23}=\frac{A x^2+1}{A+1}\implies x=\frac{144}{161}\implies A=-\frac{1127}{90}\tag 6$$ Now, from $(1)$ $M=-31110000$.

Now, using $(4)$ $$\frac{144}{161}=\exp({31110000 k})\implies k=-\frac{\log \left(\frac{161}{144}\right)}{31110000}$$

Answer 2

Label your equations: \begin{align*} 1:& & 2\,700\,000 &= M/(1+A) \text{,} \\ 2:& & 3\,050\,000 &= M/(1+Ae^{-Mk}) \text{, and} \\ 3:& & 3\,450\,000 &= M/(1+Ae^{-2Mk}) \text{.} \end{align*}

Eliminate $k$ between 2 and 3, yielding \begin{align*} 4:& & 641\,872\,500\,000\,000 - 420\,900\,000 M + 69 M^2 \\ & &= A (-641\,872\,500\,000\,000 + 186\,050\,000 M) \end{align*}

Eliminate $M$ between 1 and 4 to obtain \begin{align*} 5:& & A &= -1127/90 \text{.} \end{align*}

Substitute 5 into 1 to find \begin{align*} 6:& & M &= -31\,110\,000 \text{.} \end{align*}

Then substitute 5 and 6 into 2 to find \begin{align*} k &= \frac{\ln \left(\frac{161}{144} \right)}{-31\,110\,000} \text{.} \end{align*}

Answer 3

Let $z=10^6.$ Using the first equation, one can solve for $A$ in terms of $M$ to get $A=\dfrac{M-2.7z}{2.7z}\;(1).$

Plugging this into the second equation gives $3.05z=\dfrac{M}{1+\frac{M-2.7z}{2.7z}e^{-Mk}}.$ Solving for $k$ in terms of $M$ yields $k=-\dfrac{1}{M}\ln\left(\dfrac{2.7z\cdot(M-3.05z)}{3.05z(M-2.7z)}\right)\; (2).$

Plug the values of $A$ and $k$ in terms of $M$ into the last equation and you will get the equation $0.0125z^2M^2+0.388875z^3M=0\Rightarrow M(0.0125z^2M+0.388875z^3)=0\Rightarrow M=-31.11z=\boxed{-31\space110\space 000}.$ Plugging this value into equations $(1)$ and $(2)$ to get that $k=\boxed{\dfrac{1}{31\space 110\space 000}\ln\left(\dfrac{144}{161}\right)}$ and $A=\boxed{-\dfrac{1127}{90}}.$