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"The greater part of our happiness or misery depends on our dispositions, and not on our circumstances." Martha Dandridge Custis Washington 1731 – 1802.

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Presentation on theme: ""The greater part of our happiness or misery depends on our dispositions, and not on our circumstances." Martha Dandridge Custis Washington 1731 – 1802."— Presentation transcript:

1 "The greater part of our happiness or misery depends on our dispositions, and not on our circumstances." Martha Dandridge Custis Washington 1731 – 1802

2 Chapter 3 Exponential and Logarithmic Functions

3 Day III Properties of Logarithms (3.3)

4 Logarithmic functions are often used to model scientific observations like human memory.

5 GOAL I. To rewrite logarithmic functions with a different base

6 I. Change of Base

7 Calculators only have two types of log keys. The common log and the natural log. The bases are 10 and e, respectively.

8 Change-of-Base Formula

9 Let a, b, and x be positive real numbers such that a  1 and b  1. Then log a x can be converted to a different base as follows:

10 Base b log a x = log b x log b a

11 Base 10 log a x = log x log a

12 Base e log a x = ln x ln a

13 Example 1. Changing Bases Using Common Logarithms

14 Using a calculator and the common log setting, evaluate the expression to 1/10000. log 7 4 = 0.7124 log 4/log 7 =

15 Your Turn

16 Using a calculator and the common log setting, evaluate the expression to 1/10000. 1.log 1/4 5 = 2. log 20 0.125 = -1.1610 -0.6941

17 Example 2. Changing Bases Using Natural Logarithms

18 Using a calculator and the natural log setting, evaluate the expression to 1/10000. log 7 4 = 0.7124 ln 4/ln 7 =

19 Your Turn

20 Using a calculator and the natural log setting, evaluate the expression to 1/10000. 1.log 1/4 5 = 2. log 20 0.125 = -1.1610 -0.6941

21 GOAL II. To use properties of logarithms to evaluate or rewrite logarithmic expressions

22 II. Properties of Logarithms

23 Summative Math Algebra 2 Standard 14.0.1 Students understand the properties of logarithms (log laws).

24 Let a be a positive number such that a  1, and let n be a real number. If u and v are positive real numbers, the following properties are true.

25 1. log a (uv) = log a u + log a v log (uv) = log u + log v ln (uv) = ln u + ln v

26 2. log a = log a u – log a v log = log u – log v ln = ln u – ln v uvuv uvuv uvuv

27 3. log a u n = n log a u log u n = n log u ln u n = n ln u

28 Example 3. Using Properties of Logarithms

29 Use the properties of logarithms and the given values to find the logarithm indicated. NO CALCULATORS!!! log 7  0.8 log 8  0.9 log 12  1.1

30 1.log = 7878 log 7 – log 8  0.8 – 0.9  – 0.1

31 2.log 64 = log 8 2  2 log 8  2(0.9)  1.8

32 3.log 96 = = log 8 + log 12  0.9 + 1.1  2.0 log 8  12

33 Your Turn

34 1.log = 7 12 log 7 – log 12  0.8 – 1.1  – 0.3

35 2.log 49 = log 7 2  2 log 7  2(0.8)  1.6

36 3.log 1008 = log 7 + log 12 2  0.8 + 2(1.1)  3.0 = log 7 + 2log 12  0.8 + 2.2

37 Example 4. Using Properties of Logarithms

38 Use the properties of logarithms and the given values to find the logarithm indicated. NO CALCULATORS!!!

39 1.log 9 = log 9 4 -2 = -2B = -2log 9 4 1 16 log 9 7 = A log 9 4 = B log 9 10 = C

40 2.log 7 392 = =log 7 8 + log 7 7 2 = S + 2 = log 7 8 + 2log 7 7 log 7 6 = R log 7 8 = S log 7 10 = T log 8 8  7 2 = S + 2(1)

41 3.log 8 = log 8 = log 8 12 + log 8 8 – log 8 9 2 = log 8 32 27 log 8 12 = P log 8 5 = Q log 8 9 = R 96 81 12  8 9 2 3333 = P + 1 – 2log 8 9 = P + 1 – 2R

42 Your Turn

43 1.log 5 = log 5 12 -1 = -R = -1log 5 12 1 12 log 5 12 = R log 5 9 = S log 5 11 = T

44 2.log 8 729 = =3log 8 9 = 3B log 8 6 = A log 8 9 = B log 8 10 = C log 8 9 3

45 3.log 7 = log 7 = log 7 3 + log 7 10 – log 7 8 2 = log 7 15 32 log 7 3 = X log 7 8 = Y log 7 10 = Z 30 64 3  10 8 2 2222 = X + Z – 2log 7 8 = X + Z – 2Y

46 What do you get when you cross a fawn and a hornet? Bambee

47 GOAL III. To use properties of logarithms to expand or condensed logarithmic expressions

48 III. Rewriting Logarithmic Expressions

49 Summative Math Algebra 2 Standard 14.0.3 Students use the properties of logarithms to identify their approximate values (expanding).

50 Example 5. Expanding Logarithmic Expressions

51 Use properties of logarithms to expand the expression as a sum, difference, and/or constant multiple of logarithms. Assume all variables are positive.

52 1.ln, x > 1 x 2 – 1 x 3 = ln (x + 1)(x – 1) x 3 = ln(x + 1) + ln (x – 1) – ln x 3 = ln(x + 1) + ln (x – 1) – 3ln x

53 2.ln  x 2 (x + 2) = ln x 2 (x + 2) = ½ [ln x 2 + ln (x + 2)] = ½ [2ln x + ln (x + 2)] ½ = ln x + ½ ln (x + 2)

54 Your Turn

55 Use properties of logarithms to expand the expression as a sum, difference, and/or constant multiple of logarithms. Assume all variables are positive.

56 1.ln x  x 2 + 1 = ln x (x 2 + 1) 1/2 = ln x – ln (x 2 + 1) ½ = ln x – ½ln (x 2 + 1)

57 2.ln x2y3x2y3 = ln x2x3x2x3 = ½[ln x 2 – ln y 3 ] = ½ [2ln x – 3ln y] ½ = ln x – ln y 3232

58 Summative Math Algebra 2 Standard 14.0.2 Students use the properties of logarithms to simplify logarithmic numeric expressions (condensing).

59 Example 6. Condensing Logarithmic Expressions

60 Condense the expression to the logarithmic of a single quantity.

61 1. 4[lnz + ln(z + 2)] – 2ln(z – 5) = ln [z(z + 2)] 4 (z – 5) 2 = ln z 4 (z + 2) 4 (z – 5) 2

62 Your Turn

63 1.2ln 8 + 5ln z = ln 8 2 + ln z 5 = ln 64z 5

64 2. 2[lnx – ln(x + 1) – ln(x – 1)] = ln x 2 (x + 1)(x – 1) = ln x 2 x 2 – 1 = 2 lnx – [ln(x + 1) + ln(x – 1)]

65 GOAL IV. To use logarithmic functions to model and solve real-life applications

66 IV. Applications

67 Logarithmic functions are often used to model scientific observations like human memory.

68 Example 7. Finding a Mathematical Model

69 Students participating in a psychological experiment attended several lectures and were given an exam.

70 Every month for a year after the exam, the students were retested to see how much of the material they remembered.

71 The average score of the group can be modeled by the memory model f(t) = 90 – 15 log (t + 1), 0  t  12 where t is the time in months.

72 1. What was the average score on the original exam (t = 0)? f(0) = 90 – 15 log (t + 1) 0 f(0) = 90 – 15 log 1 0 90 points

73 2. What was the average score after six months? f(t) = 90 – 15 log (t + 1) 6 6 f(6) = 90 – 15 log 7 f(6)  77 points

74 Your Turn

75 3. What was the average score after 12 months? f(12) = 90 – 15 log (12 + 1) f(12) = 90 – 15 log 13 f(12)  73 points

76 4. When will the average score decrease to 75? f(t) = 90 – 15 log (t + 1) 75 -15 = – 15 log (t + 1) 1 = log (t + 1) 10 1 = t + 1 9 months

77 Acupuncture is a jab well done. Pun for the Day


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