Minecraft Village

Given that

$\begin{aligned} \frac {dx}{dt} & = k x & \small \blue{\text{where }k \text{ is a constant.}} \\ \frac {dx}x & = k \ dt & \small \blue{\text{integrate both sides}} \\ \int \frac 1x\ dx & = \int k\ dt \\ \ln x & = kt + \blue C & \small \blue{\text{where }C \text{ is the constant of integration.}} \\ x(t) & = \blue A e^{kt} & \small \blue{\text{where }A = e^C} \\ x(t) & = \blue{x(0)}e^{kt} & \small \blue{\text{Putting }t=0 \text{, we have }A=x(0)} \end{aligned}$

$x(0)$ is the current population we need to find. We know $x(6) = x(0) e^{6t} = 2x(0) \implies e^{6t} = 2$ and

$\begin{aligned} x(0)\blue{e^{9k}} & = 10000 \\ \blue{2\sqrt 2}x(0) & = 10000 \\ \implies x(0) & = \frac {10000}{2\sqrt 2} \approx \boxed{3536} \end{aligned}$

The population is given by the equation $P(t) = P_0e^{\alpha t}$ for some constant $\alpha,$ where $P_0$ is the original population. We are given that $P(6) = 2P_0$ and $P(9) = 10000.$ So $e^{6\alpha} = 2,$ so $e^{9\alpha} = 2^{3/2},$ so $10000 = P_0 \cdot 2^{3/2},$ so $P_0 = 2500\sqrt{2} \approx \fbox{3536}.$

Let the proportionality constant be $k$ . Then $\dfrac{dx}{dt}=kx\implies \dfrac{dx}{x}=kdt\implies x=x_0e^{kt}$ , where $x_0$ is the initial population. So $2x_0=x_0e^{6k}\implies k=\dfrac{\ln 2}{6}\implies x=x_0(e^{\ln 2})^\dfrac{t}{6}=x_0\times 2^{\dfrac{t}{6}}$ . Hence $10000=x_0\times 2^{\dfrac{9}{6}}=2\sqrt 2x_0\implies x_0=2500\sqrt 2\approx \boxed {3536}$ .