doubling time question

Question

The initial population of a town is 104,243 and it grows with the doubling time of 88 years. What will the population be in 21 years

Answer 1

The initial population of a town is 104,243 and it grows with the doubling time of 88 years. What will the population be in 21 years

To start out I like to write down my given info and what the quesiton is asking

Given:

Initial population is 104,243

Doubles in 88 years

Find:

Population in 21 years.

This is and exponential growth/decay problem. Since the population is growing, it's a growth problem.

Formula for exponential growth y(t) = a e^kt

Where

a is the initial condition

k is the rate of growth

t is the time

In our case: a = 104,243

Now they tell us the the population doubles in 88 years. That means that y(t) = double initial conditions at 88 years. We can use that info to find the value of k

y(t) = 2a at 88 years. so

2a = a e^{k 88}

2 = e^{k 88}

ln 2 = 88k

k = ln2 / 88 ≈ 0.007877

Now we need to find what the population would be at 21 years.

y(21) = 104,243 e ^{(0.007877)(21)}

y(21) = 122,994

To do a sanity check. The population at 21 years should be more then the initial conditions, but less than doubling. Our answer fits.