Question

The population of a southern city follows the exponential law. If the population doubled in size over 17 months and the current population is 40,000, what will the population be 4 years from now? I've keep trying to get this right and have failed over and over. Can anyone please help me?

Answer 1

Sure. Can you show me what you've tried?

Answer 2

The problem resets each time you get it incorrect with a similar problem, do you want to see my attempts on the old problem(s) or just this one?

Answer 3

well, you can tell me and I won't reset :-)

so, with the current problem (doubling every 17 months), how would you write the general equation? Say the current population is P0...

Answer 4

\[40,000e ^{K(48)} \] Where K is the rate?

Answer 5

Errr A(t) = that, that is.

Answer 6

okay, you can do it that way, but there's another way which might be easier. If we double every 17 months, then how about multiplying the population by \(2^{t/17}\) where \(t\) is in months? at t = 0 months, that will evaluate to \(2^0 = 1\). At \(t = 17\) months, that will evaluate to \(2^{17/17} = 2^1 = 2\). At \(t=34\) months, that will evaluate to \(2^{34/17} = 2^2 = 4\), etc.

Answer 7

Saves having to compute the decay/growth coefficient (which admittedly is just saving you dividing 0.693/17)

Answer 8

No, you don't use e at all in this form...

Answer 9

\[P(t) = P_0 2^{t/t_{dbl}}\]where \(t_{dbl}\) is the time to double the population.

Answer 10

or if you were doing exponential decay instead of growth, it would be the time to halve the population.

the advantage of doing it this way is if you have a different sort of growth (for example, say your population triples every 17 months), you just use the different base for the exponential. tripling would be \(P_0*3^{t/17}\), etc.

Answer 11

Okay, re-did it, got 283156.8

Answer 12

Oh nvm that is correct, thank you. ^^

Answer 13

Well, we can work it out both ways, and you can do whichever you like. I'll do it this way, and you do it with the decay/growth constant.

\[P(t) = P_0 2^{t/17}\]\[P(t) = 40000*2^{t/17}\]but if we want to take an input value of \(t\) in years, then we need to multiply \(t*12\) to get months in our calculation, maybe this is what is tripping you up...

\[P(t) = 40000*(2)^{12t/17}\]\[P(4) = 40000*(2)^{12*4/17} = 40000*2^{48/17} = 283157\]

Answer 14

Now if we went with the e-based version, the growth factor would be \[\lambda = 12*\frac{\ln 2}{17} = 0.48928\] (including the conversion from years to months)

\[P(t) = P_0 e^{\lambda t} = 40000*e^{0.48928t} \]\[P(4) = 40000*e^{0.48928*t} = 283157\]

Answer 15

I understand now, thank you. :)

Answer 16

Great!

The 2^t is a bit easier for initial understanding, I think, and the e^lambda t version is easier when you start doing differential equations...