Question

The initial population of a town is 4100​, and it grows with a doubling time of 10 years. What will the population be in 8 ​years?

Answer 1

To estimate the population of the town in 8 years, we use the exponential growth formula P(8) = 4100 * 2^(8/10), yielding an estimated population of approximately 7134 residents.

Step-by-step explanation:

To calculate the future population of a town given its initial population and doubling time, we use the formula for exponential growth. Since the town has a doubling time of 10 years, and we want to find the population at 8 years, we will use the formula P(t) = P0 * 2^(t/D), where P(t) is the future population, P0 is the initial population, t is the time in years, and D is the doubling time.

Using this formula:

1. P0 = 4100 (initial population)
2. t = 8 (years)
3. D = 10 (doubling time in years)

The calculation is therefore:

P(8) = 4100 * 2^(8/10)

First, calculate 8/10, which is 0.8, and then raise 2 to the power of 0.8, and finally multiply this by 4100.

To find 2 raised to the power of 0.8, we can use a calculator. This gives us approximately 1.74. Now multiply 4100 by 1.74:

P(8) = 4100 * 1.74 = 7134

Therefore, the estimated population of the town in 8 years will be approximately 7134 residents.

Answer 2

if the population is doubling every 10 years, the rate of change is 100%, for the period of 10 years, so, whatever it happens to be at the time, it grows by 100%, namely it doubles.


`\bf \textit{Periodic Exponential Growth}\\\\ A=I(1 + r)^{(t)/(p)}\qquad \begin{cases} A=\textit{accumulated amount}\\ I=\textit{initial amount}\to &4100\\ r=rate\to 100\%\to (100)/(100)\to &1.00\\ t=\textit{elapsed time}\to &8\\ p=period\to &10 \end{cases} \\\\\\ A=4100(1 + 1)^{(8)/(10)}\implies A=4100(2)^{(4)/(5)}\implies A=4100\sqrt[5]{2^4}`