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My daughter has this question as a homework:

If a population is known to double every 12 years and we know that the population in 2000 was of 100,000 individuals: a) What is the analytic expression of this growth according to the number of years? b) What was the population in 1995?

The only formula she has so far is this one:

P(n)=P(i)(1+rate)^n

So, I guess the analytic expression would be:

Population in 12 years= 100,000*(1+rate)^12

This would lead me to a rate of 0.594631.

That part, I'm not sure about but still, it makes sense.

But then, how to get back in time until 1995? Trying and guessing, I evaluated the same expression with the same rate for 1988 (12 years earlier), as the population was supposed to be about 50,000 individuals (half of the 2000 population).

If my analytic expression and rate here above are right, I should then have this:

50,000= 100,000*(1+0.594631)^-1.5

That exponent is beyond me as I don't see any relationship between -1.5 and 12 years in the past.

So this doesn't give me any clue on how to find, based on the initial expression, how many people were accounted in 1995...

Something tells me that I'm completely wrong from the beginning.

I searched a lot on internet but when it concerns growth of population, I just see expressions regarding population that doubles or predictions for the future, nothing related to prediction of the past (that includes the above expression that is).

I guess that going back in time involves the use of logs but still, I have the base (rate) but I'm still unsure about how to feed the 5 years back in time as an input to get an answer.

Any help would be appreciated.

Thanks.

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  • $\begingroup$ Just plug in $n = -5$. Negative numbers as exponents are just fine. $\endgroup$ – fleablood Dec 11 '17 at 21:27
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    $\begingroup$ The rate is $0.05946 = 2^{\frac{1}{12}}-1$ $\endgroup$ – Satish Ramanathan Dec 11 '17 at 21:29
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Let's start with the given formula. $$P(\text{final year}) = P(\text{initial year}) *(1+r)^n$$ Where $n= \text{final year}-\text{initial year}$. You are correct in that next you should use the information that population doubles every 12 years $$P(2012)=P(2000)*(1+r)^{12}$$ $$100,000=50,000*(1+r)^{12}$$ $$2=(1+r)^{12}$$ Therefore $r = \sqrt[12]{2}-1=0.\mathbf{0}594631$. You are mostly fine up to here but for some reason you started to guess and check. Just use your formula $$n = 1995 - 2000 = -5$$ $$P(1995) = P(2000)*1.0594631^{-5} = 100,000 * 1.0594631^{-5} \approx 74,915 $$ If you are uncomfortable with negative exponents or the fact that the initial year is after the final year make 1995 the initial year and 2000 the final year then $$P(2000) = 100,000 = P(1995) *1.0594631^{5}$$ $$P(1995) = \frac{100,000}{1.0594631^{5}}\approx 74,915$$

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  • $\begingroup$ Thank you. Of course, that's the first thing that came to my mind (-5 as an exponent) but if I went somewhere else, it is because I couldn't get a logical answer when I tried. I had in mind that the answer should be close to 75000 but when I entered the data in my calculator, I got something way below 50000. So, I probably misused the calculator. $\endgroup$ – Bachir Messaouri Dec 11 '17 at 22:15
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    $\begingroup$ You just accidentally lost a zero, happens to everyone ;) $\endgroup$ – Daniel Gendin Dec 12 '17 at 0:49
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$P_{1995} = \frac{P_{2000}}{(1+r)^5} = \frac{100000}{(1.0594)^5}=74915$

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  • $\begingroup$ Thank you very much. I'm sorry but I don't have enough reputation to declare your answer as useful but of course, it is. $\endgroup$ – Bachir Messaouri Dec 11 '17 at 22:16
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Let $P_i = 100,000$ equal population in $2000$.

Then $P(n) = P_i* (1 + rate)^n$ where $n$ is the number of years since $2000$.

So $P(12) = P_i*(1+rate)^{12} = 2*P_i$ and $(1+rate)^{12} = 2$ and so $1 + rate = \sqrt[12]{2}$ and $rate = \sqrt[12]{2} -1 = 0.0594631.$ (you are off by a factor of $10$).

So $P(n) = 100,000 * (1.0594631)^n$.

So since $1995$ is $5$ years before $2000$ plug in $n=-5$

$P(-5) = 100,000 * (1.0594631)^{-5} \approx 74,915$.

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Alternatively

$(1 + rate)^{12} = 2$. What is $(1 + rate)^5$?$

$(1+rate)^{12} = 2 \implies rate = .0594631$ so

$(1+rate)^5 = 1.594631^5 = 1.3348398541700343648308318811845$

So from $1995$ to $2000$ the population grew by a factor of $1.3348398541700343648308318811845$.

Or $x*1.3348398541700343648308318811845 = 100,000$ so

$x= \frac {100,000}{1.3348398541700343648308318811845} = 74,915$

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  • $\begingroup$ Thank you very much. I'm sorry but I don't have enough reputation to declare your answer as useful but of course, it is. $\endgroup$ – Bachir Messaouri Dec 11 '17 at 22:16

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