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"A population of an organism grows such that after t hours the number of organisms is N thousand, where N is given by the equation N = A - $8e^{-kt}$

Initially there are 3000 organisms and this number doubles after 5 hours."

Find the value of : i) A ii) k

This is all I need help for currently, and I have attempted many ways. Firstly, I did a table, so that when t is 1(as it is initial), N is 3000 and then when t is 6 (5 hours later), N is 6000(doubled).

What I did next was to ln the equation, getting ln N = lnA - 8kt. Using simultaneous equations, I have:

ln 3000 = ln A - 8k and ln 6000 = ln A - 48k and this is where I'm stuck, send help D;

Jia Xuan Ng
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  • t=0 is the initial time. So A should be easier to calculate. Then you need to solve 2N = A - 8$e^{-5k}$, which should also be doable. – Joel Pereira Jan 01 '19 at 14:23

1 Answers1

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First note that N is how many thousand of organisms there are, not how many organisms there are.

At time $t=0$, $N=3$. Plugging this into your equation gives $$3 = A - 8e^{-0k}$$ $$3 = A - 8$$ $$A = 11$$.

At time $t=5$, $N=6$. Plugging this into your equation gives $$6 = 11 - 8e^{-5k}$$ $$-5 = -8e^{-5k}$$ $$e^{-5k} = \frac58$$ $$-5k = \ln\frac58$$ $$k = -\frac{1}{5}\ln\frac58$$

Erik Parkinson
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