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August 1st, 2017, 05:52 AM   #1
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Puzzled by population growth problem

Hi guys,

Could someone point me in the right direction with this question. I've made a go of it but not convinced I'm on the right track.

"The population of a country grows according to the formula $\displaystyle N=N_{0}e^{kt}$ where $\displaystyle N$ is the population (in millions) at time $\displaystyle t$ (in years). $\displaystyle N_{0}$ is the initial population (in millions) and $\displaystyle k$ is a constant."

Question 1:
If the initial population was 40 million and after 120 years it is 360 million, find an expression for $\displaystyle k$ in terms of $\displaystyle \ln$.

Answer:
I'm hoping I got this part down:

$\displaystyle 360,000,000 = 40,000,000e^{120k}$

$\displaystyle 9 = e^{120k}$

$\displaystyle \ln9 = \ln e^{120k}$

$\displaystyle \ln9 = 120k$

$\displaystyle k = \frac {\ln9}{120}$

$\displaystyle k = \frac {\ln3^2}{120}$

$\displaystyle k = \frac {2\ln3}{120}$

$\displaystyle k = \frac {\ln3}{60}$

Question 2:
Use this expression for $\displaystyle k$ and the other information given above, to find how long it takes for the population to double.

Answer:
This one is worrying me because I'm not getting to something that looks like the correct answer, which makes me think I've got the first one wrong.

$\displaystyle 2N_{0} = N_{0}e^{\frac {\ln3}{60}t}$

$\displaystyle 2 = \frac {\ln3}{60}t$

$\displaystyle 120 = (\ln3)t$

$\displaystyle \frac {120}{\ln3} = t$

How much of a mess have I made here?

Last edited by skipjack; August 1st, 2017 at 08:01 AM.
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August 1st, 2017, 06:18 AM   #2
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Originally Posted by PuzzleBot View Post

Question 2:
Use this expression for $\displaystyle k$ and the other information given above, to find how long it takes for the population to double.

Answer:
This one is worrying me because I'm not getting to something that looks like the correct answer, which makes me think I've got the first one wrong.

$\displaystyle 2N_{0} = N_{0}e^{\frac {\ln3}{60}t}$

$\displaystyle 2 = \frac {\ln3}{60}t$

$\displaystyle 120 = (\ln3)t$

$\displaystyle \frac {120}{\ln3} = t$

How much of a mess have I made here?
I think question 1 is ok.

$\displaystyle e^{\frac {\ln 3}{60}t} \ne \frac {\ln3}{60}$

Rather,

$\displaystyle e^{\frac {\ln 3}{60}t} = e^{\ln ( 3^{\frac{1}{60}}) t} = 3^{\frac{1}{60}}$

Last edited by skipjack; August 1st, 2017 at 08:03 AM.
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