Question

A town's population is currently 2,183. If the population doubles every 28 years, what will the population be 140 years from now?

Answer 1

=69805

Explanation:

This is an exponential growth problem. If the population doubles periodically, it follows a law like this:

P(t) = P(0)e^kt

where P(0) is the initial population at time t=0, and k is a constant with units of years-1.

To find k, let t=0. Then P(t) = P(0) = initial population = 2183.

Since the population doubles every 28 years, we can write

P(t+28) = 2P(t)

P(0)ek(t+28) = 2[P(0)e^kt]

Simplifying,

e²⁸k = 2

k = ln(2) / 28 = 0.02475 years-¹

Finally,

P(t) = 2183e⁰.⁰²⁴⁷⁵t, t in years

P(t)= 2183e^0.02475t, t in years

Then at t=140 years from now,

P(140) = 2183e⁰.⁰²⁴⁷⁵×¹⁴⁰

P(140) = 2183e^(0.02475 × 140) = 69804.61168

=69805

Answer 2

Answer: 69,856

Step-by-step explanation: First, find out how many times the population will double. Divide the number of years by how long it takes for the population to double.

140÷28=5

The population will double 5 times.

Now figure out what the population will be after it doubles 5 times. Multiply the population by 2 a total of 5 times.

2,18322222=69,856

That calculation could also be written with exponents:

2,18325=69,856

After 140 years, the population will be 69,856 people.