Help With Exponential Growth Please and Thank You

Question

A population of 280 flying squirrels in a town in Honduras is growing at a rate of nine

percent every six months.

Part A: Write a function that represents the exponential growth of the flying squirrel

population.

Part B: Fuera Bichos is a local full-service pest and wildlife animal control company in a

Honduras. They are developing a social and broadcast media ad to spread the

word on the services that they provide to control the flying squirrel population

growth. The ad is centered on the amount of flying squirrels in the town (if it is not

controlled) in five years. What should that number be in order to represent the

most accurate amount of flying squirrels in five years?

Answer 1

First lets see what the growth in number of squirrels looks like

Now we have 280

In six months we will have an additional 9% or 0.09*280.

The total number of squirrels will be 280 + 280*0.09 = 280*(1 + 0.09) =280*(1.09) = 305 Squirrels (Equation 1)

In the second six months this repeats. We have a 9% increase from the 305. So the increase is 305*(0.09)

And the total would be 305+305*(0.09)=305*(1+0.09)=305*1.09 (Equation 2)

From Equation 1 above we see that the 305 came from 280*(1.09). So, if we plug this into Equation 2 for 305 we get

280*1.09*1.09=280*1.09^2

If we repeat this we will see that at the end of 18 months the number of squirrels will be 280*1.09^3. (Equation 3)

Also we can see that the exponent increases by 1 every 6 months

For t=6 months the exponent is 1.

For t=12 months the exponent is 2.

For t =18 months the exponent is 3.

So the exponent is the number of time periods and the general equation could be written as

S = 280*1.09^n (Equation 4)

So the exponent can be expressed as n = c*t. Where n is the number of time periods, c is a constant and t is time in months

To find c we can use the exponent from any of the time periods.

For example from Equation 2, the exponent is 2 and time is 12 months.

So c*12=2. (Equation 5)

Dividing both sides of Equation 4 by 12 we get

c=2/12 or 1/6

So the exponent for any time can be expressed as n = t/6

Using t/6 as the exponent in Equation 1, Equation 2, or Equation 3 we get the general equation

S=280*1.09^(t/6). Where S = the number of squirrels and t= the number of months. (Equation 5)

Applying equation 5 at 5 years (60 months) we get

S=280*1.09^(60/6) = 280*1.09^10 = 663 squirrels