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August 1st, 2017, 06:52 AM  #1 
Newbie Joined: Jun 2017 From: Cape Town Posts: 6 Thanks: 0  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 09:01 AM. 
August 1st, 2017, 07:18 AM  #2  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,519 Thanks: 506 Math Focus: Yet to find out.  Quote:
$\displaystyle e^{\frac {\ln 3}{60}t} \ne \frac {\ln3}{60}$ 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 09:03 AM.  

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