Question

The population of a town grows at a rate proportional to the population present at time t. The initial population of 500 increases by 25% in 10 years. What will be the population in 50 years? (Round your answer to the nearest person.)

How fast is the population growing at
t = 50?
(Round your answer to two decimal places.)

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}\\\\ -------------------------------\\\\ \textit{now, it was 500, in increased by 25\% so }500+\stackrel{\textit{25\% of 500}}{125}\implies 625`


`\bf \textit{in 10 years}\implies \begin{cases} A=625\\ I=500\\ t=10 \end{cases}\implies 600=500e^(r10)\implies \cfrac{600}{500}=e^(10r) \\\\\\ \cfrac{625}{500}=e^(10r)\implies ln\left((625)/(500) \right)=ln(e^(10r))\implies ln\left((625)/(500) \right)=10r \\\\\\ \cfrac{ln\left((625)/(500) \right)}{10}=r\implies 0.022314\approx r`


`\bf -------------------------------\\\\ A=500e^(0.0.022314t)\qquad \qquad \stackrel{t=50}{A}=500e^(0.022314\cdot 50)\implies A\approx 1526`

how fast is the population growing? well, I'd think is "r", or the rate, which is 0.022314 * 100 or 2.2314%.