The population of a mining city grows at a rate proportional to that population, in two years the population has doubled and a year later there were 10,000 inhabitants.What was the initial population?

Answer:The initial population was approximatedly 3535 inhabitants.Step-by-step explanation:The population of the city can be given by the following differential equation.[tex]\frac{dP}{dt} = Pr[/tex],In which r is the rate of growth of the population.We can solve this diffential equation by the variable separation method.[tex]\frac{dP}{dt} = Pr[/tex][tex]\frac{dP}{P} = r dt[/tex]Integrating both sides:[tex]ln P = rt + c[/tex]Since ln and the exponential are inverse operations, to write P in function of t, we apply ln to both sides.[tex]e^{ln P} = e^{rt + C}[/tex][tex]P(t) = Ce^{rt}[/tex]C is the initial population, so:[tex]P(t) = P(0)e^{rt}[/tex]Now, we apply the problem's statements to first find the growth rate and then the initial population.The problem states that:In two years the population has doubled:[tex]P(2) = 2P(0)[/tex][tex]P(t) = P(0)e^{rt}[/tex][tex]2P(0) = P(0)e^{2r}[/tex][tex]2 = e^{2r}[/tex]To isolate r, we apply ln both sides[tex]e^{2r} = 2[/tex][tex]ln e^{2r} = ln 2[/tex][tex]2r = 0.69[/tex][tex]r = \frac{0.69}{2}[/tex][tex]r = 0.3466[/tex]So[tex]P(t) = P(0)e^{0.3466t}[/tex]In two years the population has doubled and a year later there were 10,000 inhabitants.[tex]P(3) = 10,000[/tex][tex]P(t) = P(0)e^{0.3466t}[/tex][tex]10,000= P(0)e^{0.3466*3}[/tex][tex]P(0) = \frac{10,000}{e^{1.04}}[/tex][tex]P(0) = 3534.55[/tex]The initial population was approximatedly 3535 inhabitants.