Question

How can I calculate the exponential decay?
http://prntscr.com/d3mndp

Answer 1

Thanks in advance @mathmate .

Answer 2

So you need to put together the function for population in terms of time P(t).
I would call the time t, years since 1992, since that is the first data point and will correspond to t=0...

Answer 3

So they give you two points on the function curve, (t , P)
(t , P) = (0 , 28)
(t , P) = (7 , 30)

right? Good on that stuff?

Answer 4

We have the form
\[\huge P(t)=Ae^{kt}\]
where t is \(years\) or \(years~difference\)??

Answer 5

Yes. That is a general form for an exponential curve. time t we can call years after 1992. and P(t) is population in millions.

Answer 6

You see how those two points are from the given info?
(t , P) = (0 , 28)
(t , P) = (7 , 30)

Answer 7

right

Answer 8

According to that, we are going to find out the two constants \(A\) and \(k\) by plugging the two points in our equation..

I would get \(A=28~and~k=9.856×10^{−3}\)

So the final form would be
\[\huge P(t)=28e^{(9.856*10^{-3})t}\]

Answer 9

But what I am not satisfied with is

"I would call the time t, years since 1992, since that is the first data point and will correspond to t=0..."

Why that? and when we substitute with t=1992 as it is?

Answer 10

you can if you want, it is messier. The graph is just shifted to the right to 1992. and the constants A and k will be different.

28=A*e^(1992*k)
30=A*e^(1999*k)

Answer 11

If I did it as I think (sub with 1992), would I reach the same result if I want the population at 2008?

Answer 12

What is your equation using those points instead (1992,28) and (1999,30) ??
that is not easy to solve i dont think

Answer 13

I think like that because of the fact:
"at the start (t=0), there was any population P(t)=0, not there was 28 million!!!!"

That confuses me!

Answer 14

You can use, if you want to be able to use the year number for t instead.
\[\huge P(t)=28*e^{.009856*(t-1992)}\]

Answer 15

You can solve for the rate by the 3rd formula in the attached graphic:
rate = 10^[log(end amt/bgn amt)/time] -1
rate = 10^[log(30/28)/time]-1
rate = 10^[log(1.0714285714)/7]-1
rate = 10^(0.0299632234/7) -1
rate = (10^0.0042804605) -1
rate = 1.0099048561 -1
rate = 0.0099048561

Now look at the second formula
2008 Population = 28,000,000 * (1 + 0.0099048561) ^ 16

2008 Population = 32,782,751

(and this checks out wuth the calculator here: http://www.1728.org/expgrwth.htm

Answer 16

Oh and 3mar, in this problem you are calculating exponential growth and not exponential decay.

Answer 17

@wolf1728 @3mar do any of you know spanish?

Answer 18

@wolf1728
where did you get this formulas??

I want the basic to build on it!

By the wat I am satisfied of what @DanJS has said and explained and it makes sense - and thanks to him of course - but I need to know when I sub. with x=1992 not just x=7 or even x=16.

And @DanJS, I have solved for A and k if I plugged in these points (1992,28) and (1999,30) and I got:
\[\huge A=6.0621×10^{-85}\]
\[\huge k=0.099021\]

So it would be at this form
\[\huge P(t)=6.0621×10^{-85}e^{0.099021}x\]

But what made me sad is the result I have gotten when I put x=2008 was 136.53, which is not like what we got from the first formula \(32.783~ million\)