Introduction to Finding the Growth Rate
We can find the growth rate of a population if we have reason to believe that it is growing exponentially and if we know the population level at two different times. We will use the first population level as n _{0} . Because we will know another population level, we have a value for n ( t ) and for t . This means that the equation n ( t ) = n _{0} e ^{rt} will have only one unknown, r . We can find r using natural logarithms in the same way we found t in the problems above.
Examples
 The population of a country is growing exponentially. In the year 2000, it was 10 million and in 2005, it was 12 million. What is the growth rate? In the year t = 0 (2000), the population was 10 million, so n _{0} = 10. The growth formula becomes n ( t ) = 10 e ^{rt} . When t = 5 (the year 2005), the population is 12 million, so n ( t ) = 12. We will solve the equation 12= 10 e ^{5 r} for r .
The country’s population is growing at the rate of 3.6% per year.
 Suppose a bacteria culture contains 2500 bacteria at 1:00 and at 1:30 there are 6000. What is the hourly growth rate?

Because we are asked to find the hourly growth rate, t must be measured in hours and not minutes. Initially, at t = 0, the population is 2500, so n _{0} = 2500. Half an hour later, the population is 6000, so t = 0.5 and n ( t ) = 6000. We will solve for r in the equation 6000 = 2500 e ^{0.5 r} .
The bacteria are increasing at the rate of 175% per hour.
 A certain species of fish is introduced in a large lake. Wildlife biologists expect the fish’s population to double every four months for the first few years. What is the annual growth rate?

If n _{0} represents the fish’s population when first put in the lake, then it will double to 2 n _{0} after . The growth formula becomes . This equation has two unknowns, n _{0} and r , not one. But after we divide both sides of the equation by n _{0} , r becomes the only unknown.
The fish population is expected to grow at the rate of 208% per year.
Finding the Growth Rate Practice Problems
Practice
 The population of school children in a city grew from 125,000 to 200,000 in five years. Assuming exponential growth, find the annual growth rate for the number of school children.
 A corporation that owns a chain of retail stores operated 500 stores in 2000 and 700 stores in 2003. Assuming that the number of stores is growing exponentially, what is its annual growth rate?
 At 10:30, 1500 bacteria are present in a culture. At 11:00, 3500 are present. What is the hourly growth rate?
Solutions
The population of school children grew at the rate of 9.4% per year.
The number of stores is growing at the rate of 11.2% per year.
The bacteria are increasing at the rate of 169% per hour.
Find practice problems and solutions for these concepts at: Exponents and Logarithms Practice Test.
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