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

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.

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