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# Compound Growth Help (page 2)

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By McGraw-Hill Professional
Updated on Oct 4, 2011

#### Example

• Find the compound amount of \$5000 after eight years, earning 12% annual interest, compounded continuously.

A = 5000 e 0.12(8) = 5000 e 0.96 ≈ 5000(2.611696) ≈ 13,058.48

The compound amount is \$13,058.48.

The compound growth formula for continuously compounded interest is used for other growth and decay problems. The general exponential growth model is n ( t ) = n 0 e rt , where n ( t ) replaces A and n 0 replaces P . Their meanings are the same— n ( t ) is still the compound growth, and n 0 is still the beginning amount. The variable t represents time in this formula; although, time will not always be measured in years. The growth rate and t need to have the same unit of measure. If the growth rate is in days, then t needs to be in days. If the growth rate is in hours, then t needs to be in hours, and so on. If the “population” is getting smaller, then the formula is n ( t ) = n 0 e −rt .

#### Examples

• The population of a city is estimated to be growing at the rate of 10% per year. In 2000, its population was 160,000. Estimate its population in the year 2005.
• The year 2000 corresponds to t = 0, so the year 2005 corresponds to t = 5; n 0 , the population in year t = 0, is 160,000. The population is growing at the rate of 10% per year, so r = 0.10. The formula n ( t ) = n 0 e rt becomes n ( t ) = 160,000 e 0.10 t . We want to find n ( t ) for t = 5.

n (5) = 160,000 e 0.10(5) ≈ 263,795

The city’s population is expected to be 264,000 in the year 2005 (estimates and projections are normally rounded off).

• A county is losing population at the rate of 0.7% per year. If the population in 2001 is 1,000,000, what is it expected to be in the year 2008?
• n 0 = 1,000,000, t = 0 is the year 2001, t = 7 is the year 2008, and r = 0.007. Because the county is losing population, we will use the decay model: n ( t ) = n 0 e −rt . The model for this county’s population is n(t) = 1,000,000e −.007 t . We want to find n ( t ) for t = 7.

n (7) = 1,000,000 e −.007(7) ≈ 952,181

The population is expected to be 952,000 in the year 2008.

• In an experiment, a culture of bacteria grew at the rate of 35% per hour. If 1000 bacteria were present at 10:00, how many were present at 10:45?

n 0 = 1000, r = 0.35, t is the number of hours after 10:00

The growth model becomes n(t) = 1000 e 0.35 t . We want to find n ( t ) for 45 minutes, or t = 0.75 hours.

n (0.75) = 1000 e 0.35(0.75) = 1000 e 0.2625 ≈ 1300

At 10:45, there were approximately 1300 bacteria present in the culture.

Find practice problems and solutions for these concepts at: Exponents and Logarithms Practice Test.

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