2. Welcome to Year 11 Mathematics B Term 3 Topic 5: Exponential & Logarithmic Functions I

3. Index Laws Solutions of equations involving indicies Definitions of a x and log a x for a > 1 Logarithmic laws Graphs of, and the relationships between, y = a x, y = log a x for different values of a (done via an assignment) Applications of exponential and logarithmic functions

4. Index Laws An expression in the form a n is called a power. a is the base n is the index There are a number of rules involving powers

5. 1. a m  a n = a m+n Consider 100  10 000 = 1 000 000 10 2  10 4 = 10 6 +=

6. 2. a m  a n = a m-n Consider 10 5  10 3 = 10 2 - =

7. 3. (a m ) n = a mn Consider (10 2 ) 3 = 10 2  10 2  10 2 = 10 6

8. 4. (ab) m = a m.b m Consider (3t) 4 = 3t  3t  3t  3t = (3  3  3  3)  (t  t  t  t) = 3 4  t 4

9. Exercise 1.2, page 8, No. 5 & 8

10. 5. a 0 = 1 (a  0) 10 3 = 1 000 10 2 = 100 10 1 = 10 10 0 =  10 1

11. 6. 10 3 = 1 000 10 2 = 100 10 1 = 10 10 0 = 1 10 -1 = = 0.1

13. Exercise 1.2 Page 8 No. 1, 2, 5 & 8 Exercise 1.2 Page 8 No. 1, 2, 5 & 8

15. 7. Use your calculator to evaluate …

16. 7. Use your calculator to evaluate...

17. Fractional indices produce surds (roots)

18. 8. Use your calculator to evaluate these.

19. Consider Do you raise to the power of 3 first, or do you take the 4 th root first? Do it both ways and check for yourself

20. Therefore, it can be done both ways!

21. Exercise 1.2 Page 8 No. 3, 4, 6 & 7 Exercise 1.2 Page 8 No. 3, 4, 6 & 7

22. Summary of Index Laws For all real values of a, b, m and n a m  a n = a m+n a m  a n = a m-n (a m ) n = a mn (ab) m = a m.b m a 0 = 1 (a  0) a -n =

23. 1. Fractional Indicies worksheet 2. Additional Exercises 1.2 3. Worksheet 7.1 Numbers 1 - 5

24. Equations With Indices  If a m = a n, then we have powers with the same base being equal, therefore their indices must also be equal.  Solving the most simple index equations involves this concept, e.g. Solve for x if 2 x = 32

25. 2 x = 32 2 x = 2 5  x = 5 Try these 3 x = 81 5 x = 78 125 49 x = 7

26. Index Equations If the bases are not the same, we must attempt to express both sides of the equation as powers of the same base. Consider this. 9 x = 243 We see that 243 is not an integer power of 9, so we must find a common base that we can express both 9 and 243 in.

27. 9 = 3 2 243 = 3 5  If 9 x = 243 Then (3 2 ) x = 3 5  3 2x = 3 5  2x = 5  x = 2.5

28. Now try these. 2 x = 4 3 9 x = 2 187 8 x = 4 7.5

29.  Exercise 10.2 Page 346 No. 1

30. Additional Exercises 10.2 Number 1

31. Quadratic Form 2 2x – 5(2 x ) + 4 = 0 This equation has three features that make it a quadratic equation. 1. The bases are the same (base 2). 2. One index is twice the other. 3. There are three terms

32. A simple piece of substitution will enable us to factorise this as a normal quadratic. 2 2x – 5(2 x ) + 4 = 0

33. A simple piece of substitution will enable us to factorise this as a normal quadratic. 2 2x – 5(2 x ) + 4 = 0 1. Let A = 2 x

34. A simple piece of substitution will enable us to factorise this as a normal quadratic. 2 2x – 5(2 x ) + 4 = 0 1. Let A = 2 x  2 2x = (2 x ) 2 = A 2

35. A simple piece of substitution will enable us to factorise this as a normal quadratic. 2 2x – 5(2 x ) + 4 = 0 1. Let A = 2x2x 2.  Equation becomes A2 A2 – 5A + 4 = 0 3. Factorise:(A – 4) (A – 1) = 0 4. Solve: A = 4orA = 1 2x 2x = 42x 2x = 1 x = 2 x = 0  2 2x = (2 x ) 2 = A 2

37. Graphic Calculator Solutions What can we do when the numbers in the equation cannot be written as powers with a common base? Consider 3 x = 12

38. Method 1. A Graphical Approach 1. Enter y = 3 x into Y 1 in the Y= editor 2. Enter y = 12 into Y 2 in the Y= editor 3. Set window range, x = -1 to 4, y = 0 to 14 4. Graph 5. Choose intersect from the Calculate menu and follow the prompts

40. Method 2. A Calculator Approach 1. Choose the Solver option from the Math menu. 2. Enter the equation, be sure to have it rearranged to equal zero and hit enter. 3. Alpha, solve to evaluate x.

41. Attempt the following Use both methods 1. 2 x = 9 2. 2 x = 15 3. 1 – 5 x = 25 4. 3 x + 7 = 21 5. 10 x = 25 6. 7 – 10 x = -49 Additional Exercises 10.2 Numbers 2 – 10 Worksheet 7.1 Numbers 6 - 10

42. Logarithms A logarithm is an index. Number 1101001 000

43. Logarithms A logarithm is an index. Number 1101001 000 Power of 10 10 0 10 1 10 2 10 3

44. Logarithms A logarithm is an index. Number 1101001 000 Power of 10 10 0 10 1 10 2 10 3 Log (Base 10) 0123

45. Earthquakes, Sound & Logs Earthquake magnitude is measured by the amount of ground movement. This can vary from fractions of a millimetre to metres. Numbers with such a wide range are difficult to fit on a number line when graphing.

46. To overcome this, the Richter scale is based on exponents, base 10.

47. deci-Bells

48. Using a linear scale, the difference successive values on a number line is the same. On a logarithmic scale, the ratio between successive values is the same. One unit greater 10 time greater

49. If a x = N then x = log a N Index

50. If a x = N then x = log a N Base

51. If a x = N then x = log a N Number

52. If a x = N then x = log a N i.e. a x = N  x = log a N (N > 0, a > 0, a  1) Why not???

54. Without use of a calculator, evaluate Log 10 100 000 Either… Let x = log 10 100 000 x = log 10 10 5 x = 5 Or… Let x = log 10 100 000 10 x = 100 000 10 x = 10 5 x = 5

55. Logs (to base 10), are referred to as common logs, and the base 10 can be left out. i.e. log 10 100 000  log 100 000 Whereas log 5 200 must include the base.

56. Evaluate log 3 243 Either… Let x = log 3 243 x = log 3 3 5 x = 5 Or… Let x = log 3 243  3 x = 243 3 x = 3 5  x = 5

57. log 0.01 Let x = log 0.01 10 x = 0.01 10 x = 10 -2 x = -2

58. Additional Exercises 10.3

59. Properties of Common Logarithms

60. 1. log a MN = log a M + log a N L.H.S. = log 10 (100  1000) = log 10 (10 2  10 3 ) = log 10 (10 5 ) = 5 R.H.S. = log 10 100 + log 10 1000 = log 10 10 2 + log 10 10 3 = 2 + 3 = 5 = L.H.S. log 10 (100  1000) = log 10 100 + log 10 1000

61. Evaluate log 3 (9  81) = log 3 9 + log 3 81 = log 3 3 2 + log 3 3 4 = 2 + 4 = 6 i.e. 9  81 = 3 6

62. Express as a single logarithm log 3 5 + log 3 7 = log 3 (5  7) = log 3 35

63. 2. log a = log a M – log a N

64. Express as a single logarithm log 2 9 – log 2 5 = log 2 (9  5) = log 2 1.8

65. Try These a) log 3 5 + log 3 7 b) log 2 6 + log 2 9 c) log 8 6 – log 8 3 d) log 3 6 – log 3 5 e) log 4 3 – log 4 2 f) log 6 50 + log 6 2 g) log 2 6 + log 2 3 – log 2 5 h) log 5 6 + log 5 10 – log 5 4

66. 3. log a N p = p.log a N L.H.S. = log 10 100 2 = log 10 (10 2 ) 2 = log 10 10 4 = 4 R.H.S. = 2.log 10 100 = 2.log 10 10 2 = 2  2 = 4 = L.H.S. Consider log 10 100 2 = 2.log 10 100

67. Evaluate log 3 81 3 = 3.log 3 81 = 3.log 3 3 4 = 3  4 = 12

68. Evaluate log 10 1 = log 10 10 0 = 0 Evaluate log 10 10 = log 10 10 1 = 1  log a 1 = 0  log a a = 1

69. Summary of Logarithm Laws 1. If N = a x then log a N (N > 0, a>0, a  1) 2. log a MN = log a M + log a N 3. 4. log a a = 1 5. log a 1 = 0 6. log a a x = x 7. log a N p = p.log a N 8.

70. Find the value of 2 log 3 6 – log 3 4 = log 3 6 2 – log 3 4 = log 3 36 – log 3 4 = log 3 (36  4) = log 3 9 = log 3 3 2 = 2

71. Given that log 5.24  0.7193 Find the value of log 5 240 5 240 = 5.24  1000  log (5 240) = log (5.24  1000) = log 5.24 + log 1000 = 0.7193 + 3 = 3.7193

72. Use common logarithms to solve 1.8 x = 15 Take logs of both sides log 1.8 x = log 15 x.log 1.8 = log 15 x = log 15  log 1.8 x ≈ 1.176  0.2553 x ≈ 4.607 Worksheet 7.2

73.  Exercise 10.4 Page 350 No. 1 to 10  Exercise 10.4 Page 350 No. 1 to 10

74. Shape of Log Graphs y = log 10 xNo. 1

75. y = log 5 x No. 2

76. y = log 2 x No. 3

77. y = log (x – 2)No. 4

78. y = log 2 (x – 1)No. 5

79. y = log 4 (x + 3) No. 6

80. y = log (x) + 1 No. 7

81. y = log 3 (x + 1) + 2No. 8

83. Application The level of a leaking oil-storage tank drops by 1% of the height above the leak every hour. The leak starts at 2am and is at a height of 4.5m. The tank is a large cylinder of height 10m and diameter 20m. It was only ¾ full when the leak began.

84. a) Find an expression for the amount of oil that has leaked after time t hours. b) How much oil is spilt by 2.10 am? c) A spillage of more than 5000L must be reported to the environmental Protection agency. By what time must the leak be fixed to avoid notification? The level of a leaking oil-storage tank drops by 1% of the height above the leak every hour. The leak starts at 2 am and is at a height of 4.5m. The tank is a large cylinder of height 10m and diameter 20m. It was only ¾ full when the leak began. 10m

85. a)Find an expression for the amount of oil that has leaked after time t hours. b) How much oil is spilt by 2.10 am? c) A spillage of more than 5000L must be reported to the environmental Protection agency. By what time must the leak be fixed to avoid notification? 10m To calculate the volume of oil spilt, we need the height of the oil above the leak. Since it starts at 3m above the leak and each hour the level is 99% of the level of the previous hour  Height of oil above leak H 1 = 3  0.99 t  Height oil has dropped H 2 = 3 - 3  0.99 t

86. a)Find an expression for the amount of oil that has leaked after time t hours. b) How much oil is spilt by 2.10 am? c) A spillage of more than 5000L must be reported to the environmental Protection agency. By what time must the leak be fixed to avoid notification? To calculate the volume of oil spilt, we need the height of the oil above the leak. Since it starts at 3m above the leak and each hour the level is 99% of the level of the previous hour  Height of oil above leak H 1 = 3  0.99 t  Height oil has dropped H 2 = 3 - 3  0.99 t a) V = πr 2 = π  10 2  (3 - 3  0.99 t ) m 3 = 100 π  (3 - 3  0.99 t )  1000L  after t hours the amount of oil leaked = 300 000 π  (1 - 0.99 t ) L

87. a) V = πr 2 = π  10 2  (3 - 3  0.99 t ) m 3 = 100 π  (3 - 3  0.99 t )  1000L  after t hours the amount of oil leaked = 300 000 π  (1 - 0.99 t ) L b) Oil spilt by 2.10 am. V = 300 000π  (1 - 0.99 t ) L = 300 000π  (1 - 0.99 1/6 ) L ≈ 1577L c) Time at which 5000L has spilt 5000 = 300 000π  (1 - 0.99 t )L = 1 - 0.99 t 0.99 t = 1 – log 0.99 t = log (1 – ) t  log 0.99 = log (1 – ) t ≈  The leak should be fixed before about 2.32 am

88.  Exercise 10.5 Page 353