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YearBudget in Billions of Dollars 1997355 1999390 2001442 2002485 2004554 m0246 T12.028.085.443.66 For questions 1 – 3, do NOT use exponential regression.

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Presentation on theme: "YearBudget in Billions of Dollars 1997355 1999390 2001442 2002485 2004554 m0246 T12.028.085.443.66 For questions 1 – 3, do NOT use exponential regression."— Presentation transcript:

1 YearBudget in Billions of Dollars 1997355 1999390 2001442 2002485 2004554 m0246 T12.028.085.443.66 For questions 1 – 3, do NOT use exponential regression. 1.The table below shows the temperature T of a certain liquid after it has been cooled in a freezer for m minutes. Show that the data can be modeled with an exponential function. 2. What is the decay factor per minute? Briefly show work. 3.Write an exponential model for the data. 4.The defense budget for the United States in various years is show in the table at the right. Use exponential regression to find a model for this data using years since 1997. Round all values to the nearest thousandth. T = 12.02(.82 m ) B = 348.454(1.067 t ) where B = budget in billions of dollars and t = number of years since 1997.67 1/2 .82 The decay factor per minute is.82 0245702457

2 Remember to download from D2L and print a copy of the Final Group Project. DateSection October 304.4 November 4 4.4 Continued November 6Review for test 3 November 11Test 3 November 135.2 November 185.4 November 205.5 November 25No Class (Fall Break) November 27No Class (Fall Break) December 25.6 December 4Last Day of Class Final Group Project due Dec 9 – 15 Final exam week – specific dates and times will be announced in future

3 Uranium 239 (U 239 ) is an unstable isotope of uranium that decays rapidly. In order to determine the rate of decay, 10 grams were placed in a container, and the amount remaining was measured at 5-minute intervals and recorded in the table below. Time in minutes Grams remaining 010 58.63 107.45 156.43 205.55 1.Show that an exponential model would be appropriate for this data. 2.What is the decay factor per minute (nearest thousandth)? 3.Find an exponential model for the data. 4.What is the decay rate each minute? What does this number mean in practical terms? 5.Use functional notation to express the amount remaining after 13 minutes and then calculate that value. 6.The half-life of a radioactive element is the time it takes for the mass to decay by half. Use the graphing calculator to find the half-life of U 239 to the nearest hundredth of a minute. U Periodic Table of Elements G(t) = 10

4 Uranium 239 (U 239 ) is an unstable isotope of uranium that decays rapidly. In order to determine the rate of decay, 10 grams were placed in a container, and the amount remaining was measured at 5-minute intervals and recorded in the table below. Time in minutes Grams remaining 010 58.63 107.45 156.43 205.55 6. The half-life of a radioactive element is the time it takes for the mass to decay by half. Use the graphing calculator to find the half-life of U 239 to the nearest hundredth of a minute. U Periodic Table of Elements 23.55 minutes G(t) = 10

5 Our goal for the rest of this class period is to find an algebraic way of solving the equation (and others like it) x = 23.55 5 = 10(.971 x )

6 Answers to even-numbered HW problems Section 4.3 S-2 Exponential S-10 y = 5.592(.814 x ) Ex 6 tNTN 021.34204.4 137.55359.7 266.06633.1 395.3 Suspect data point

7 Our goal for the rest of this class period is to find an algebraic way of solving the equation (and others like it) x = 23.55.5 =.971 x 10 Since the variable is part of the exponent, the only way to isolate the variable is to use logarithms. 5 = 10(.971 x )

8 10 x = 10

9 10 1 = 10

10 10 x = 100

11 10 1 = 10 10 2 = 100

12 10 1 = 10 10 2 = 100 10 x = 40 What is the value of x?

13 10 1 = 10 10 2 = 100 10 x = 40 The question can be rephrased: If 10 is raised to a power and the result is 40, what is the exponent? What is the value of x?

14 10 1 = 10 10 2 = 100 10 x = 40 The question can be rephrased: If 10 is raised to a power and the result is 40, what is the exponent? The word logarithm is a synonym for exponent.

15 10 1 = 10 10 2 = 100 10 x = 40 The question can be rephrased: If 10 is raised to a power and the result is 40, what is the exponent? This question can be rephrased: If 10 is the base, what is the logarithm that will give a value of 40?

16 10 1 = 10 10 2 = 100 10 x = 40 The question can be rephrased: If 10 is raised to a power and the result is 40, what is the exponent? This question can be rephrased: What is log 10 40 ? x = log 10 40 exponentbase 10result  1.6

17 The common logarithm of a, written log 10 a, is defined as the power of 10 that gives a. The word logarithm is a synonym for exponent. The common logarithm of 40 is 1.6 because using base 10, the exponent needed to get a value of 40 is 1.6. log 10 40 = 1.6 Can we get a more accurate value for log 10 40 ? log 10 40  1.602059991

18 Is it true that 2 5 = 32 ? log 2 32 = 5 Write an equivalent statement using logarithms. Identify the exponent, the base, and the result. 5 2 32

19 Is it true that 10 2.68 = 478.63 (approximately)? log 10 478.63 = 2.68 Write an equivalent statement using logarithms. Identify the exponent, the base, and the result. log 478.63 = 2.68 When no base is indicated, it is understood to be 10. 2.68 10 478.63

20 When the base of a logarithm is 10, we call it a common logarithm, and we do not indicate the base. log 10 40 is the same as log 40.

21 Find each of the following common logarithms on a calculator. Round to four decimal places. a) log 723,456 b) log 0.0000245 c) log (  4) When the base of a logarithm is 10, we call it a common logarithm, and we do not indicate the base. log 10 40 is the same as log 40. Write an equivalent exponential equation for each. ERR: nonreal ans 10  4.6108 =.0000245 =  4.6108 10 5.8594 = 723,456 = 5.8594

22 An important property of logarithms: log (a b ) = b (log a) Illustration: Does log (2 7 ) = 7(log 2)?

23 An important property of logarithms: log (a b ) = b (log a) Illustration: Does log (2 7 ) = 7(log 2)?

24 Take the logarithm on both sides.5 =.971 x log(.5) = log(.971 x ) log(.5) = x log (.971) -.3010 x = 23.55 Divide both sides by 10 10 5 = 10(.971 x ) Now let’s solve the equation x (-.0128) = Apply the property x = = 23.52 -.3010 -.0128 Compute the values Solve for x

25 Take the logarithm on both sides.5 =.971 x log(.5) = log(.971 x ) log(.5) = x log (.971) x = 23.55 Divide both sides by 10 10 5 = 10(.971 x ) Now let’s solve the equation = x Apply the property Solve for x 23.55 = x

26 Solve algebraically to the nearest thousandth

27 Solve algebraically to the nearest thousandth 48 48 15.3 = log(15.3) = log log(15.3) = x log (3.7) log(15.3) log (3.7) = x 2.085 = x

28 Section 4.4 (Do not read) Handout – Common Logs (correction to question 18. It should read “Solve question 17 algebraically.”)

29 The table below shows the average weekly amount of electricity used in five Michigan cities in 2011. Population of city in thousands Amount of electricity in 1000’s of kilowatts 73.5365 85.4395 97.2442 108.0503 120.9584 Additional Practice 1. Make a scatter plot of the data. Based on the graph, would an exponential model be appropriate? 2.Write an equation for an exponential model for the weekly amount of electricity used versus the population. Round coefficients to three decimal places (nearest thousandth). 3. Graph the exponential model. 4. What is the growth rate in weekly electricity use per 1,000 people? 5. Use your model to estimate the weekly amount of electricity used in 2011 in a Michigan city with a population of 90,000 people. 6. The weekly amount of electricity used by the people of another Michigan city in 2011 was 660 thousand kilowatts. Use your graph to estimate the population of the city that year. 7. Solve question 6 algebraically.


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