Example 1: The population of an island grows at an exponential rate. From 2015 to 2016, the population increases from 20,000 to 22,800. What is the population’s growth rate? 22,800 - 20,000 = 2,800 new people. 2,800 ÷ 20,000 = 0. 14, so the population is growing by 0. 14 per year. This is small enough that the estimate will be fairly accurate.
22,800 - 20,000 = 2,800 new people. 2,800 ÷ 20,000 = 0. 14, so the population is growing by 0. 14 per year. This is small enough that the estimate will be fairly accurate.
Example 1 (cont): The island had a growth rate of 0. 14, written as a decimal fraction. This represents 0. 141{\displaystyle {\frac {0. 14}{1}}}. Multiply the numerator and denominator by 100 to get 0. 141x100100=14100={\displaystyle {\frac {0. 14}{1}}x{\frac {100}{100}}={\frac {14}{100}}=} 14% per year.
Example 1 (cont): The growth rate was 14%, so the number of time intervals required is 7014=5{\displaystyle {\frac {70}{14}}=5}.
Example 1 (cont): In this case, since we measured the growth across one year, each time interval is one year. The island population doubles every 5 years. Example 2: The second, spider-infested island nearby is much less popular. It also grew from a population of 20,000 to 22,800, but took 20 years to do it. Assuming its growth is exponential, what is this population’s doubling time? This island has a 14% growth rate over 20 years. The “rule of 70” tells us it will also take 5 time intervals to double, but in this case each time interval is 20 years. (5 time intervals) x (20 years/time interval) = 100 years for the spider-infested island’s population to double.
This island has a 14% growth rate over 20 years. The “rule of 70” tells us it will also take 5 time intervals to double, but in this case each time interval is 20 years. (5 time intervals) x (20 years/time interval) = 100 years for the spider-infested island’s population to double.
To make sense of this formula, picture a $100 investment with a 0. 02 annual interest rate. Every time you calculate growth, you multiply the amount you have by 1. 02. After one year, that’s ($100)(1. 02), after two years that’s ($100)(1. 02)(1. 02), and so on. This simplifies to (1. 02)t{\displaystyle (1. 02)^{t}}, where t is the number of time periods. Note: If r and t do not use the same time unit, use the formula Af=A0(1+rn)nt{\displaystyle A_{f}=A_{0}(1+{\frac {r}{n}})^{nt}}, where n is the number of times growth is calculated per time period. For example, if r = 0. 05 per month and t = 4 years, use n = 12, since there are twelve months in a year.
This formula is often used to approximate population growth, and always when calculating continuously compounded interest. In situations where growth is calculated at regular intervals, such as annually compounded interest, the formula above is more accurate. You can derive this from the formula from the one above using calculus concepts.
2A0=A0(e)rt{\displaystyle 2A_{0}=A_{0}(e)^{rt}} Divide both sides by A0{\displaystyle A_{0}} 2=ert{\displaystyle 2=e^{rt}}
2=ert{\displaystyle 2=e^{rt}} ln(2)=ln(ert){\displaystyle ln(2)=ln(e^{rt})} ln(2)=rt{\displaystyle ln(2)=rt} ln(2)r=t{\displaystyle {\frac {ln(2)}{r}}=t}
Now that you know this formula, you can adjust it to solve similar problems. For example, find “tripling time” with the formula ttriple=ln(3)r{\displaystyle t_{triple}={\frac {ln(3)}{r}}}.
