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Applications of Exponential Functions

Applications of Exponential Functions. Population Growth. Population growth can be modeled by the basic form of the exponential function y = ab x . Growth: b > 1 b = “ growth factor ” a = “ initial amount ” x = time y = ending amount. Population Growth.

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Applications of Exponential Functions

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  1. Applications of Exponential Functions

  2. Population Growth • Population growth can be modeled by the basic form of the exponential function y = abx. • Growth: b > 1 • b = “growth factor” • a = “initial amount” • x = time • y = ending amount

  3. Population Growth • In 2003, the population of the popular town of Smithville was estimated to be 35,000 people with an annual rate of increase (growth) of about 2.4%. • What is the growth factor? • After one year: 35,000 + (0.024)(35,000) • Factor out 35,000 • 35,000(1 + 0.024) = 35,000(1.024) • So, the growth factor is 1.024

  4. Population Growth • 2. Write an equation to model future population growth in Smithville. • y = abx • y = a(1.024)x • So, y = 35,000(1.024)x, where x is the number of years since 2003.

  5. Population Growth • 3. Use the equation that you’ve written to estimate the population of Smithville in 2007 to the nearest one hundred people. • y = 35,000(1.024)4 = 38,482.91 = 38,500

  6. Compound Interest • What is interest? • Compound Interest: Interest that is earned on both the principal and any interest that has been earned previously. • Balance: The sum of the Principal and the Interest

  7. Compound Interest • Formula: • A: the ending amount • P: the beginning amount (or "principal”) • r: the interest rate (expressed as a decimal) • n: the number of compoundings in a year • t: the total number of years

  8. Compound Interest • Jackie deposits $325 in an account that pays 4.1% interest compounded annually. How much money will Jackie have in her account after 3 years? • A = 325(1 + 0.041)1(3) 1 • A = 325(1.041)3 • A = $366.64 Jackie will have $366.64 in her account after 3 years.

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