Question

A model for the population in a small community after t years is given by P(t)=P0e^kt.

a)If the initial population doubled in 5 years, how long will it take to have tripled its initial population?
b)In addition, if the population of the community is 10,000 in 3 years, what was its initial population?

Answer 1

`\bf \textit{Amount of Population Growth}\\\\ A=Ie^(rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ I=\textit{initial amount}\\ r=rate\to r\%\to (r)/(100)\\ t=\textit{elapsed time}\\ \end{cases}`

a)

so, if the population doubled in 5 years, that means t = 5. So say, if we use an amount for "i" or P in your case, to be 1, then after 5 years it'd be 2, and thus i = 1 and A = 2, let's find "r" or "k" in your equation.


`\bf \textit{Amount of Population Growth}\\\\ A=Ie^(rt)\qquad \begin{cases} A=\textit{accumulated amount}\to &2\\ I=\textit{initial amount}\to &1\\ r=rate\\ t=\textit{elapsed time}\to &5\\ \end{cases} \\\\\\ 2=1\cdot e^(5r)\implies 2=e^(5r)\implies ln(2)=ln(e^(5r))\implies ln(2)=5r \\\\\\ \boxed{\cfrac{ln(2)}{5}=r}\qquad therefore\qquad \boxed{A=e^{(ln(2))/(5)\cdot t}} \\\\\\ \textit{how long to tripling?}\quad \begin{cases} A=3\\ I=1 \end{cases}\implies 3=1\cdot e^{(ln(2))/(5)\cdot t}`


`\bf 3=e^{(ln(2))/(5)\cdot t}\implies ln(3)=ln\left( e^{(ln(2))/(5)\cdot t} \right)\implies ln(3)=\cfrac{ln(2)}{5} t \\\\\\ \cfrac{5ln(3)}{ln(2)}=t\implies 7.9\approx t`

b)

A = 10,000, t = 3


`\bf \begin{cases} A=10000\\ t=3 \end{cases}\implies 10000=Ie^{(ln(2))/(5)\cdot 3}\implies \cfrac{10000}{e^{(3ln(2))/(5)}}=I \\\\\\ 6597.53955 \approx I`