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Say the population of a city is increasing at a constant rate of 11.5% per year. If the population is currently 2000, estimate how long it will take for the population to reach 3000.

How can this be solved using the formula below. I know how to solve when we input x number of years but don't know how to solve when the number of years is the unknown.

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jaykirby
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  • Yes I understand the formula and have tried but I get stuck. I know how to use the formula to predict what the population would be in x number of years, but don't know what the do when the number of years is the unknown.. – jaykirby Jul 03 '13 at 10:25
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    Do you know about logarithms? – Raskolnikov Jul 03 '13 at 10:26
  • No I'm not up to logarithims yet, this problem is meant to be solved without using them. Can you help? – jaykirby Jul 03 '13 at 10:27
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    Yes, I can help. Just wanted to know what you expect me to use in the answer. – Raskolnikov Jul 03 '13 at 10:28

1 Answers1

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So, we have that after $x$ years, there are 3000 people where we started from 2000. Thus

$$3000=2000 e^{rx}$$

which can be rewritten as

$$e^{rx}=1.5$$

We also know that after 1 year, there is a 11.5% increase. Thus

$$e^r=1.115$$

From the conjugation of those two formulas, we have that

$$(1.115)^x=1.5$$

So all we have to do is see how many times we have to multiply 1.115 by itself to obtain 1.5. Since this will likely involve a non-integer number, we just take the integer solution such that we exceed 1.5. Which is 4. So somewhere in the fourth year, population exceeds 3000.

Raskolnikov
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  • I follow your work up to the point where you write that e^r=1.115. How did you get that value? Doesn't r=0.115 – jaykirby Jul 03 '13 at 10:38
  • @jaykirby You have been given the values $P(0) = 2000$ and $P(1) = P(0) + \frac{11.5}{100}\cdot P(0) = 1.115\cdot P(0)$. On the other hand, $P(1) = P(0)\cdot e^{r\cdot 1}$. – Daniel Fischer Jul 03 '13 at 10:47
  • Doesn't 11.5/100 = 0.115? – jaykirby Jul 03 '13 at 10:48
  • @raskolnikov: could you please clarify? – jaykirby Jul 03 '13 at 10:51
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    @jaykirby: You are right, 11.5/100=0.115. But you forget something. That part is just what is coming extra. There's also the hundred percent, that is the 2000 people you already had. That's 100/100=1 and you have to take those into account too, which is why $e^r=1+0.115=1.115$. – Raskolnikov Jul 03 '13 at 12:24