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I am trying to calculate population growth over 5 years, and my rate is 1.41%. My starting point is 1,359,033. I should be ending up with a number of 1,486,521 after that 5 year period, but for some reason I keep getting 11 thousand something or 81. I am also in Algebra 1 and am in 8th grade. What would the formula for my population growth be, using the y=ab^x method?

GraciousBean
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  • So what was the expression you used to get your own answer? – Matti P. Feb 05 '20 at 14:06
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    Recognize, a growth of $1.41%$ corresponds to the next value being $1.0141$, i.e. $101.41%$ times that of the previous value. – JMoravitz Feb 05 '20 at 14:06
  • I agree with JMoravitz. That is really the core in understanding these calculations. – Matti P. Feb 05 '20 at 14:07
  • @MattiP. My expression was either y=(2.41)^5, or y=1,359,033(2.41)^5. I tried both of these multiple times, and continuously got the wrong answer. I also tried switching 2.41 out for 1.41. – GraciousBean Feb 05 '20 at 14:08
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    The calculation $1359033(2.41)^5$ would be for if you had an original amount of $1359033$ and a growth rate of $141%$, i.e. the population more than doubling each time period, a far greater growth rate than $1.41%$ where the population grows by only about a hundredth of the current amount each time period. – JMoravitz Feb 05 '20 at 14:10
  • Hint: the word "per cent" means that you have to divide that number by $100$ before using it in the equation. Therefore, you have to calculate $$ 1+ \frac{1.41}{100} $$ And not $$ 1+1.41 $$ – Matti P. Feb 05 '20 at 14:11
  • @MattiP. So would b be (1+0.0141) instead? – GraciousBean Feb 05 '20 at 14:12
  • Yes. If you have trouble remembering in general, at least try to remember a few common examples and check that the formula you are about to use works for those simple easy to remember examples... such as "growing by $100%$", i.e. doubling in size, or "growing by $0%$", i.e. staying the same size. – JMoravitz Feb 05 '20 at 14:14
  • @JMoravitz Okay, thank you! makes a lot more sense. – GraciousBean Feb 05 '20 at 14:17
  • So, using this information, my formula would be y=1359033(1.0141)^5 ? @JMoravitz – GraciousBean Feb 05 '20 at 14:19

3 Answers3

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For an initial population of $P_0$ and a $\color{blue}{\text{growth}}$ rate of $r\%$ per time period, the size of the population after $t$ time periods have passed will be

$$P(t) = P_0\cdot (1\color{blue}{+}\frac{r}{100})^t$$

Again, remembering that $r$ in the above was written as a percentage.

For your example, after five years we have $1359033\cdot (1+0.0141)^5 \approx 1457585$ as we expected.

Had it been a $\color{red}{\text{decay}}$ rate instead, we would be subtracting instead of adding.

Note also that some authors will prefer not to talk in terms of percentages, but will instead talk about raw rates. "Grows by a factor of $2$" for instance in which case the formula will need to be adjusted to accommodate.

In order to accommodate such changes in wording, I find it most helpful to spend the time to understand why the formula looks the way it does and not just memorize the formula itself. Honestly, half of the time I don't remember the formulas for annuities or growth rates and such exactly or doubt my memory on them and so just come up with the formulas again on the spot by recognizing what they are meant to represent. In your case of an exponential growth, we start with a value and after some period of time it has increased by some factor. After another period of time, the new amount increases again by that factor, and so on... leading to the general form of $a\cdot b^t$ for some appropriate choices of $a,b,t$

JMoravitz
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  • Thanks! Can you let me know what P stands for in this formula, or is it just the same thing as y=ab^x but with different variables? Also, what does it mean when at exponent is at the bottom of a number (0) instead of the top? – GraciousBean Feb 05 '20 at 14:34
  • @GraciousBean $P$ is the name given to "Population" by several authors. The subscript is also incredibly commonly used to indicate elements in a sequence... and in particular a subscript of $0$ indicates "the initial or starting value" of the sequence. So here, $P_0$ just reads "the starting population" while $P(t)$ or $P_t$ indicates the population after $t$ units of time. – JMoravitz Feb 05 '20 at 14:50
  • @GraciousBean Yes, I just wrote it with different names than "$a$,$b$,$x$" but a rose by any other name would smell as sweet so it doesn't really matter what you call them in the long run. That being said, having variables whose names correspond to what they represent (P for Population, r for Rate, t for Time) generally help to remember the physical meanings of the different terms in the expressions. – JMoravitz Feb 05 '20 at 14:52
  • Okay, thank you so much. I have been seeing this formula everywhere and I had no idea what it meant. Now that I do, I should be able to finish answering my question! – GraciousBean Feb 05 '20 at 14:54
  • So, my ending formula would be P^n = 1359033(1.0141)^5, correct? – GraciousBean Feb 05 '20 at 14:59
  • As alluded to in the other comment I just posted, I would have written it with ascii instead as P_n rather than P^n, as the ^ which commonly denotes superscript might have been interpreted as an exponent rather than a sequential index. Otherwise, yes, P_n = 1359033(1.0141)^5 is correct – JMoravitz Feb 05 '20 at 15:01
  • Great! I can answer my question now. Thank you so much. – GraciousBean Feb 05 '20 at 15:04
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If the starting population is $P_0$ and the annual growth rate, in percent, is $r$, then the appropriate formula for the population after $n$ years is

$$P_n=P_0\left(1+{r\over100}\right)^n$$

However, something is a bit funky here, because $1{,}359{,}033(1.0141)^5\approx1{,}457{,}585$, not $1{,}486{,}521$, as asserted in the OP.

Barry Cipra
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  • The reason it is a bit off is because I found my result on a website that tells me what the population was in 2018 (5 years after 2013, my starting point.) It was just something that would help me make sure my answer was around the right area. If you don't mind me asking, What does the P variable represent? – GraciousBean Feb 05 '20 at 14:43
  • The variable $P$ here represents Population. It's often helpful, when modeling "real world" problems, to let the symbols come from the words of the problem, e.g., $A$ for Area, $V$ for volume, etc. More precisely $P_0$ stands for the starting population (in "year $0$") and $P_n$ for the population $n$ years later. – Barry Cipra Feb 05 '20 at 14:49
  • So, my ending formula would be P^n = 1359033(1.0141)^5? – GraciousBean Feb 05 '20 at 14:57
  • The right hand side is OK, but the left hand side should be $P_n$, not $P^n$. Superscripts usually indicate exponentiation, whereas subscripts usually indicate terms in a sequence. Incidentally, the choice of the symbol $r$ came from the first letter of rate. – Barry Cipra Feb 05 '20 at 14:57
  • Thanks so much, this all makes much more sense now! – GraciousBean Feb 05 '20 at 14:59
  • @GraciousBean as a nitpick, although not impossible, it is incredibly uncommon for a superscript to be used to indicate different terms in a sequence and in ascii plaintext, a ^ character generally indicates superscript. It is far more common to see it written P_n instead or P^n – JMoravitz Feb 05 '20 at 15:00
  • Alright, I didn't know that. Thanks for clarifying – GraciousBean Feb 05 '20 at 15:01
  • Incidentally, to get from $1359033$ to $1486521$ in $5$ years, you need an annual growth rate close to $1.81$%. – Barry Cipra Feb 05 '20 at 15:05
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Let $P_{\alpha,\gamma}^\beta$ the population from $\gamma$ over $\alpha$ years and with rate $\beta$. I have: $P_{\alpha,\gamma}^\beta=\gamma\cdot(1+ \alpha/100)^\beta$. With your inputs: $$P_{5,13590033}^{0.41}=1359033\cdot(1+0.41/100)^5=1486521$$

Matteo
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