In January of 2010, the population of a small town called Riley was 1,250. Its population has been decreasing at a steady rate of 8% per year. According to this model, what was Riley's population on January 2000?

Question

sloppycanada · Answer

I know right away that I use the following formula - 

A = P * e^(r)(t))

this is the formula for continuous growth or decay. In our case it's decay.

sloppycanada · Answer

From there, I'm not quite sure what I want to do. Do I set up the equation as the following -

A = 1250 * e^(-.08)(-10)

sloppycanada · Answer

Am I correct on my thinking or...?

jim_thompson5910 · Answer

```
I know right away that I use the following formula - 

A = P * e^(r)(t))

this is the formula for continuous growth or decay
```

sorry that's not correct. The population isn't decreasing at a continuous rate. It's only decreasing on an annual basis

jim_thompson5910 · Answer

You actually use this formula

$$\Large A = P(1+r)^t$$

where 

A = final amount after t years
P = initial amount
r = growth or decay rate
if r > 0 then we have growth, if r < 0 then we have decay
t = time in years

mathstudent55 · Answer

I would have thought that you use a yearly compounding formula.
Also, the initial value in 2010 is unknown.
The future value is 1250 in 2010.

jim_thompson5910 · Answer

`In January of 2010, the population of a small town called Riley was 1,250. Its population has been decreasing at a steady rate of 8% per year. According to this model, what was Riley's population on January 2000? `

in this case,

A = 1250
P = unknown
r = 0.08
t = 10

use these values and the formula to solve for P

jim_thompson5910 · Answer

sorry I meant to make r negative. It should be r = -0.08

sloppycanada · Answer

But as a hint, it says to use the continuously compounding formula?

jim_thompson5910 · Answer

You're right, the key phrase `at a steady rate of ` is why you use the continuously compounding formula. My bad

jim_thompson5910 · Answer

somehow I missed that too

sloppycanada · Answer

Wouldn't t  = 10? Since it's the time between 2000 and 2010?

jim_thompson5910 · Answer

correct

sloppycanada · Answer

Alright, so using the continuous growth or decay formula, I'm looking at 

1250/(e^(-.08)(10) = P?

sloppycanada · Answer

But if I solve things like that, I strangely get a number greater than 1250. Which makes me think that I'm using the wrong values for P and A.

mathstudent55 · Answer

If the population decreased steadily for 10 years and ended up at 1250 in 2010, then it must have been greater than 1250 to start with in the year 2000. You are looking for a number greater than 1250.

sloppycanada · Answer

Okay, good. Right, forgot about that. So in 2000, the population was about 2782

mathstudent55 · Answer

Sounds reasonable.

jim_thompson5910 · Answer

As you can see, the graph slowly goes down hill following along a curve of sorts. See attached

So yes, as said earlier, the initial amount should be larger

mathstudent55 · Answer

$\large A = Pe^{rt} $

$\large A = 2782 e^{-0.08 \times 10} $

$\large A = 1250.033178$

nincompoop · Answer

You should get a larger population in 2010 than in 2010 if the rate decreases every year even for whatever model it is.  It should automatically prompt you that your slope is going to be negative so that means the population is more in the past than in the future in this case.

sloppycanada · Answer

Thanks for the help!

mathstudent55 · Answer

@nincompoop 
You wrote above:
"You should get a larger population in 2010 than in 2010"
???

nincompoop · Answer

2000 than in 2010

nincompoop · Answer

my badness typoness