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Chapter 5

Chapter 5. Exponential and Logarithmic Functions. Section 5.9 Building Exponential, Logarithmic , and Logistic Models from Data. Objectives. Build an Exponential Model from Data Build a Logarithmic Model from Data Build a Logistic Model from Data.

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Chapter 5

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  1. Chapter 5 Exponential and Logarithmic Functions

  2. Section 5.9Building Exponential, Logarithmic, and Logistic Models from Data

  3. Objectives Build an Exponential Model from Data Build a Logarithmic Model from Data Build a Logistic Model from Data

  4. Example: Fitting an Exponential Function to Data Mariah deposited $20,000 in a well-diversified mutual fund 6 years ago. The data in the table represents the value of the account at the beginning of each year for the last 7 years.

  5. Example: Continued (a) Using a graphing utility, draw a scatter diagram with year as the independent variable. (b) Using a graphing utility, build an exponential model from the data. (c) Express the function found in part (b) in the form A = A0ekt. (d) Graph the exponential function found in part (b) or (c) on the scatter diagram. (e) Using the solution to part (b) or (c), predict the value of the account after 10 years. (f) Interpret the value of k found in part (c).

  6. Example: Continued Solution: • Enter the data into the graphing utility and draw the scatter diagram.

  7. Example: Continued (b) A graphing utility fits the data in the table to an exponential model of the form y = abxusing the EXPonentialREGression option. This shows y= abx= 19,820.43(1.085568)x. Notice that |r|= 0.999, which is close to 1, indicating a good fit.

  8. Example: Continued (c) To express y = abxin the form A = A0ekt, where x = t and y = A, proceed as follows: If x = t = 0, then a = A0. This leads to

  9. Example: Continued (c) Because y = abx= 19,820.43(1.085568)x, this means that a = 19,820.43 and b = 1.085568. a = A0 = 19,820.43 and b = ek = 1.085568 To find k, rewrite ek = 1.085568 as a logarithm to obtain k = ln(1.085568) ≈ 0.08210 As a result A = A0ekt= 19,820.43e0.08210t.

  10. Example: Continued (d) (e) Let t = 10 in the function found in part (c). The predicted value of the account after 10 years is A = A0ekt= 19,820.43e0.08210(10) ≈ $45,047 (f) The value of k = 0.08210 = 8.210%, represents the annual growth rate of the account. It represents the rate of interest earned, assuming the account is growing continuously.

  11. Example: Fitting a Logarithmic Function to Data Jodi, a meteorologist, is interested in finding a function that explains the relation between the height of a weather balloon (in kilometers) and the atmospheric pressure (measured in millimeters of mercury) on the balloon. She collects the data shown.

  12. Example: Continued (a) Using a graphing utility, draw a scatter diagram of the data with atmospheric pressure as the independent variable. (b) It is known that the relation between atmospheric pressure and height follows a logarithmic model. Using a graphing utility, build a logarithmic model from the data. (c) Draw the logarithmic function found in part (b) on the scatter diagram. (d) Use the function found in part (b) to predict the height of the weather balloon if the atmospheric pressure is 560 millimeters of mercury.

  13. Example: Continued Solution: • Enter the data into the graphing utility and draw the scatter diagram.

  14. Example: Continued (b) A graphing utility fits the data in the table to a logarithmic function of the form y = a + b ln x by using the LOGarithmREGression option. The logarithmic model from the data is h(p) = 45.7863 – 6.9025 ln p where h is the height of the weather balloon and p is the atmospheric pressure. Notice that |r| is close to 1, indicating a good fit.

  15. Example: Continued (c) The graph of h(p) = 45.7863 – 6.9025 ln p on the scatter diagram. (d) Using the function found in part (b), Jodi predicts the height of the weather balloon when the atmospheric pressure is 560 to be

  16. Example: Fitting a Logistic Function to Data The data represents the amount of yeast biomass in a culture after thours.

  17. Example: Continued (a) Using a graphing utility, draw a scatter diagram of the data with time as the independent variable. (b) Using a graphing utility, build a logistic model from the data. (c) Using a graphing utility, graph the function found in part (b) on the scatter diagram. (d) What is the predicted carrying capacity of the culture? (e) Use the function found in part (b) to predict the population of the culture at t= 19 hours.

  18. Example: Continued Solution: (a)

  19. Example: Continued (b) A graphing utility fits the data to a logistic growth model of the form by using the LOGISTIC regression option. The logistic model from the data is where y is the amount of yeast biomass in the culture and x is the time.

  20. Example: Continued (c) (d) Based on the logistic growth model found in part (b), the carrying capacity of the culture is 663. (e) Using the logistic growth model found in part (b), the predicted amount of yeast biomass at t = 19 hours is

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