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Index Concepts

2  2  2  2  2. a  a  a  …  a. (5 times). ( n times). index/exponent. base. Index Concepts. 2 5. =. “ a to the power n ”. a n =. 記住 : a zero. a  a  …  a. a  a  …  a. . ( m times). ( n times). a  a  …  a. a m  a n. =. ( m x n times).

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Index Concepts

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  1. 2  2  2  2  2 a a a  … a (5 times) (n times) index/exponent base Index Concepts 25 = “a to the power n” an= 記住:azero

  2. aa…a aa…a  (m times) (n times) aa…a aman = (m x n times) aman =am+ n law 1 Class Activity 1 Law of positive integral indices 23 x 24 = (2 x 2 x 2) x (2 x 2 x 2 x 2) = 23+4 = 27 a2 x a3 = (a x a) x (a x a x a) = a2+3 = a5 b5 x b4 = (b x b x b x b x b) x (b x b x b x b) = b5+4 = b9 aman=

  3. Class Activity 2

  4. = , where m>n and a 0 , where m<n and a 0 Law 2

  5. EXAMPLE 1 –Simplify the following expressions (a) a7.a3 = a7+3 = a10 an.a3 = an+3 (b) a7 a9-2 = (c) (d)

  6. Class Activity (23)2 = (2  2  2)  (2  2  2) = 232= 26 (a  a  a  a) = (a  a  a  a)  (a  a  a  a)  a43= a12 (a4)3 = (bbb)= (bbb)  (bbb)  (bbb)  b35= a15 (bbb)  (b3)5= n times of am (aa…a) (aa…a)  (aa…a) ……  nm times of a n m a Law 3

  7. Example 2- Simplify the following expressions (a5)2= a52= a10 (a) (an)3= (b) an3= a3n (a4)m= a4m= (c) a4m Classwork- Simplify the following expressions x12 (x4)3= x43= (1) x6n (xn)6= xn6= (2) x5m (x5)m= x5m= (3)

  8. Class Activity (2  3) (2  3) (2  3) (2  3) (2  3)4= =(2  2  2  2) (3  3  3  3) =2434 (3  5) (3  5) (3  5) (3  5) (3  5) (3  5)5= (5  5  5  5  5) =(3  3  3  3  3) =3555 (a  b) (a  b) (a  b) (ab)3= =(a  a  a) (b  b  b) =a3b3

  9. (a  b) (a  b)…. (a  b) (ab)m= mtimes of ab (ab)m= (a  a  …  a)  (b  b  …  b) mtimes of a mtimes of b (ab)m= am bn Law 4

  10. 4 (ntimes of ) 4 = = = 4 5 5 = = = 5 3 3 = = = 3 Law 5 (where b  0)

  11. EXAMPLE 3: Simplify the following expressions 3 3 3 (2x)5= 25x5= 32x5 (1) (2) = = 3 2 5 2 5 5 (3) 2 = = = 2 2 2 2 2 (3a)3= 33a3= 27a3 (a) 4 4 4 = (b) = 4 3 3 2 2 3 2 6 (c) 3 = = = 3 3 3 Classwork 1.3 2

  12. EXAMPLE 4: Simplify the following expressions (a) (b) (c)

  13. Classwork 1.4 1. 2. 40 3.

  14. Exercise 1A Level 1 (1b) a2.a3.a4=a2+3+4=a9 (2b) (3b) (x7)3 =x73 =x21 (4b) (4a2)3 =43.(a2)3=64. a23 =64. a6

  15. Exercise 1A Level 2 (5b) 23.(b2)3. 42(b3)2 (6b) (2b2)3(4b3)2= =23.(b2)3. 42(b3)2 =8. b6. 16b6 =8. 16 b6. b6 =128 b6+6 =128 b12

  16. (7b) (8b)

  17. (9b) (10b)

  18. Mistakes students always make x5 x3+2= (x3)2= 23 54= (2  5) 7 (2  5) 7= 107 Wrong ! since these numbers has no common base. 3  46= 126 Wrong ! since these numbers has no common base. 3a.2a=5a

  19. ∵ (wherea 0) Thus, 1 Law 6

  20. ∴ Thus From Law 2 Any integer m and n Law 7

  21. Example 5- Evaluate the following expressions = = = 3 = -2 = = 2 = (1) 2 3 (a) (-2)-3 (b) 2 (3-3)2(20) (c) = = = 3 2

  22. Classwork 1.5 1. 2. 3. 4.

  23. Example 6 - Simplify the following expressions and express your answers in positive indices - 1 2 2 . p 3 3 p q - - - - - - = = 1 2 ( 1 ) ( 3 ) 1 3 3 = 2 p q 2 p q - - 1 3 p q 2 4 2 b c - 4 4 2 = a b c - - - - - = 2 2 2 2 1 2 - - - = ( a ) ( b ) ( c ) 2 2 1 2 a b c ) 4 a (a) (b) (c) (

  24. Classwork 1.6 (1) - - - - 1 3 1 3 3 x . x 3 x . x1 - - - + - = = (2) 1 2 3 1 2 3 x 2 2 2 ( 3 x ) 3 x 1 1 - - = = = 3 4 3 x 3 4 4 3 x 27 x (3) (4)

  25. Example 7: Simplify the following expressions and express your answers in positive indices - - - - - = 2 3 2 1 2 2 4 6 2 4 ( a b ) ( a b ) ( a b )( a b ) - - = 4 2 6 4 ) ( a a )( b b 6 a + - - - = = = 4 2 6 4 6 10 a b a b 10 b - [ ] [ ] 1 y - 1 - - - - - = = 2 3 6 3 1 6 3 = 2 ( x y ) 2 x y 2 x y 6 2 x (a) 3 (b)

  26. Classwork 1.7 - - - - - 3 6 4 6 3 3 2 2 2 1 6 6 4 3 - = x y x y = ( x y ) ( x y ) x x y y 12 x + - + - = = = 6 6 4 3 12 x y x y y - - - - - - - - = 2 1 1 3 2 1 1 2 2 2 6 ( 2 xy ) ( 3 x y ) 2 x y . 3 x y - - - - = 1 2 1 2 2 6 2 3 x x y y + - -1 2 2 6 x y = 2 2 . 3 x - 4 xy = = 4 18 y 18 (1) 1 (2)

  27. 3 [ ] [ ] 12 x - - 2 2 - - - = = = 3 2 6 2 12 4 ( x y ) x y x y 4 y - + 2 0 2 5 2 3 a b a b - - - - - - - = = = 3 0 1 2 5 1 0 2 2 2 1 5 ( b ) ( 2 a b ) ( 8 b ) b 2 a b 8 b 2 2 . 8 32 4

  28. Additional example + 4 4 4 4 8 x . x x x - - - - - = = = = 1 2 4 2 1 2 4 8 4 2 ( x y ) ( x y ) ( x y )( x y ) + 8 2 8 2 10 y . y y y - - - - = 0 2 1 3 2 2 2 2 2 3 4 [ 3 ( 5 x y ) ( 2 y ) ] 1 ( 5 x y ) ( 2 y ) 4 2 4 2 4 4 25 x y 25 x y 25 x 25 x = = = = - 3 4 12 12 2 10 ( 2 y ) 16 y y y

  29. Exercise 1B (1c) (1b) (3b) (4b) 8 a 4 (5a) - - + - + - = = . . = 6 2 2 5 2 ( 5 ) 6 ( 2 ) 3 - 4 ( 4 a b )( 2 a b ) 4 2 b a 8 b a b 3 (6a) - 2 4 4 2 1 ( 4 ) x 16 x . 2 x - - - = = 2 0 2 2 2 1 ( 4 x y ) ( x y ) 2 2 y 2 y + 4 2 6 x 32 x 32 x 2 2 = = 2 2 y y

  30. 1.4 SIMPLE EXPONENTIAL EQUATIONS Equation involves unknowns in the exponents (a0, a1, a-1) a2x=ax 2x a = 1 x a = 0 2x x - a a 2x-x=0 ∴ x=0

  31. Example 8: Solve the following exponential equations (4) 2.13x=2 (3) 4x=256 (1) 11x=1 3x=81 13x=1 4x=44 11x=110 3x=34 ∴ ∴ ∴ x=4 13x=130 x=0 x=4 x=0 ∴ (a) 6x=1 (b) 2x=8 2x=23 6x=60 ∴ x=3 ∴ x=0 Classwork 1.8 (2)

  32. Example 9: Solve the following exponential equations (b) 42x+1=8x-2 52x=625 (a) (22)2x+1=(23 )x-2 22(2x+1)=23(x-2) 52x=54 244x+2=23x-6 2x=4 4x+2=3x-6 ∴ x=2 4x-3x=-6-2 ∴ x=-8 Classwork 1.9 (4) 94x-1=272x+4 (3) 82x=16x+4 (32)4x-1=(33) 2x+4 32x=81 42x=256 (1) (2) (23)2x=(24) x+4 32(4x-1)=33(2x+4) 32x=34 42x=44 26x=24(x+4) 38x-2=26x+12 2x=4 26x=24x+16 2x=4 8x-2=6x+12 6x=4x+16 x=2 6x-4x=16 8x-6x=12+2 ∴ x=2 ∴ 2x=16 2x=14 x=8 ∴ ∴ x=7

  33. Example 10: Solve the following exponential equations ∴

  34. Classwork 1.10 - Solve the following exponential equations (1) (1) ∴ ∴

  35. Scientific Notation科學記數法 In scientific world, we often study objects which are very large or very small e.g. Mass of an element of hydrogen is 0.000 000 000 000 000 167 724 567 kg. Velocity of light = 300 000 000 m/s Difficult to read and write!!! To overcome, we use a method called scientific notation to express these numbers approximately 0.000 000 000 000 000 167 724 567 = 1.68  10-16corr. to 3 sig fig 300 000 000 = 3  108kg 25870945830231 = 2.59  10 13corr. to 3 sig fig

  36. (a) 0.000 001 = 1 0.000 001 = 1 10-6 (b) 18 000 = 1.8 10 000 = 1 104 (c) 2 000 000 = 2 1 000 000 = 2 106 (d) 0.000 216 = 2.16 0. 000 1 = 2.16 10-4 Example 11- express the following numbers in scientific notation Positive number N can be expressed as N = d 10n, where n is an integer and 1  d < 10. (e) 345104 = 3.45  100  104 = 3.45  102 104 = 3.45106 (f) 0.050 210-3 = 5.02  0.01  10-3 = 5.02  10-2 10-3 = 5.0210-5

  37. Classwork 1.1- express the following numbers in scientific notation (1) 3 740 000 = 3.74 100 000 = 1 106 (2) 0.000 0007 89 = 7.89 0.000 0001 = 1 10-7 (3) 6 311011 = 6.311001011 = 6.31  1021011 = 6.311013 (4) 0.1 203710-5=1.20370.110-5=1.20 3710-110-5 =1.20 3710-5 (5)–0.0004 508103=-4.5080.0001104=-4.50810-4104=-4.50810-8 (6) 0.7 00810-4 =7.0080.110-4=7.00810-110-3=7.00810-4

  38. Example 12- Evaluate the following,express your answers in scientific notation and correct to 3 significant figures - -  9 9 8 . 16 10 8 . 16 10 -  = 6 = - =   1 . 41913 10 6 1 . 42 10  - - 3 3 5 . 75 10 5 . 75 10 Corr. to 3 sig. Fig. =(3.527.91 )(1081012) (3.52108)(7.911012) (a) =27.84321020 =2.784321011020 =2.784321021 =2.781021 Corr. to 3 sig. Fig. (b)

  39. Classwork 1.12 ,express your answers in scientific notation and correct to 3 significant figures (1) =(2.3058.2)(107105) =18.9011012 (2.305107)(8.21012) =1.891011012 =1.891013 (2) =1.41011011 =141011 (210-12)(71023) =(27)(10-121023) = 1.41012 =3.0410110-21 (3) (3.810-10)(810-11) =(3.88)(10-1010-11) =30.410-21 = 3.0410-20 (4)  2 3 10     - - - - = = = 1 1 1 2 0 . 46875 10 4 . 69 10 10 4 . 69 10  3 6 . 4 10  6 (5) 8 . 009 10  = = 10 10 2 . 2882 10 2 . 29 10  - 4  3 . 5 10 (6) - 5 4 . 67  10  - - = = 1 1 3 . 3357 10 3 . 34 10  - 4  1 . 4 10

  40. (a) 2105=2 100 000 =200 000 (b) (2) (1) 4.1210-5=4.120.0000 1 = 0.0000412 3.1410-4=3.14 0.000 1 = 0.000314 6.36106=6.36 1000000 = 6.360000 (c) (2.510-3)(3.610-4)=2.53.610(-3)+(-4) =910-7 =90.0000001 =0.0000009 (3) =7.41100 (1.310-5)(5.7107)= 1.35.710-5+7 =7.41102 =741 -  2 9 . 75 10 (4) =  =  = 7 1 . 3 10 1 . 3 10000000 13000000 -  9 7 . 5 10 Example 13 –Express the values of the following as integers or decimal numbers Classwork 1.13

  41. =134217728 =282475249 =10000000 =43046721 7 =2.82108 10 10 =1107 Xy = =1.34108 =4.30107 9 8 8 9 7 Xy = Xy = Xy = Example 14- express the following numbers in scientific notation and arrange them in ascending order Use calculator 89, 98, 710, 107 710 107 98 89 107< 98 < 89 <710 Compare powers of 10. The greater the power of 10, the greater the number. If powers of 10 are equal, compare values of d of the numbers

  42. Place holder Place values Distinguish values between 365 and 3065 ten basic numerals: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (less than 10) The place value of each digit is ten times the place value of the digit on its right-hand side. Numeral Systems x is an integer more than 1. a, b, c and d are all non-negative integers less than x. In the algebraic expressionax3+ bx2 + cx1 + dx0 Denary System (十進制)- we use everyday When x=10, abcd is a denary number. abcd = a  103 + b 102 + c 101 + d 100 e.g. 3065(10)=3  103+ 0 102 + 6 101+ 5 100 3065(10)=3  1000+ 0 100+ 6 10+ 5 1

  43. Example 15 (a) Write down the place values of each digit of 50943 (b) Express 50943 in the expanded form with base 10 50943(10)=510000 + 01000+ 9100 + 410 + 3 1 =5104+0103 + 9102 + 4101 + 3 100

  44. Classwork 1.15 (1) Write down the place value of each digit in the following numbers (a) 24071 Digit 2 4 0 7 1 Place Value 10000 1000 100 10 1 (b) 9145 Digit 9 1 4 5 Place Value 1000 100 10 1 (2) Express the following as denary numbers (a) 2541 = 2103 + 5102 + 4101 + 1 100 (b) 205041 = 2105 + 0104 + 5103 + 0 102 + 4 101 + 1 100

  45. What are the expansions of 12.302? Challenge 12.302=1 101 + 2 100 + 3 10-1 + 0 10-2 + 2 10-3 Example 16 Express the following as denary numbers. (a) 2 102 + 3 101 + 7 100 =200 +30 + 7 =237 (b) 9 105 + 5 103 + 6 101 + 7 100 =9 105 + 0 104 + 5 103 + 0 102 + 6 101 + 7 100 =905067

  46. Classwork 1.16 Express the following as denary numbers. (1) 6103 + 3101 =6103+ 0102 + 3101 + 0100 = 6030 410000+ 210+ 9 (2) =4 104 + 0 103 +0 102 + 2 101 + 9 100 =40029 5  10000+ 4 100+ 3  10 (3) =5 104 +0 103 + 4 102 + 3 101 + 0 100 =50430

  47. (a) 1232 100, 1 (b) 123423 10000, 10 (a) 119= 1 102 + 1 101 + 9 100 (b) 2318= 2 103 + 3 102 + 1 101 + 8 100 (a) 6 10 + 8 1 = 6 101 + 8 100 =68 (c) 8 100000 + 4100 + 1 = 8105 + 0104 + 0103 + 4102 + 0101+ 1100 =800401 (e) 11000 + 4100 + 9 1 = 1103 + 4102 + 0101+ 9100 =1409 Greatest 642, smallest 246 Greatest 5310, smallest 1035 10000, 1 10000/1=10000 10000/10=1000 10000, 10 (1) Write down the place value of each digit 2 in the following denary numbers (2) Express the following denary numbers in the expanded form with base 10 Exercise 1E (3) Express the following as denary numbers (4) Find the greatest and smallest denary numbers (a) 2, 6, 4 (b) 0,1,3,5 (5) (a) 40014 (b) 241243

  48. Denary System $100 $10 $1 0 to1 (<2) 213(10) = 2 100+ 1 10+ 3 1 0 to 9 (<10) 213(10) = 2 102+ 1 101+ 3 100 Binary System 23(10)= ++  4+ 2+1 1 0 1 1 1 =10111(2) 10111(2)= 1 24+0  23+ 1  22+ 1  21+ 1  20

  49. Denary System $100 $10 $2 $4 $16 $8 $1 $1 0 to1 (<2) 213(10) = 2 100+ 1 10+ 3 1 0 to 9 (<10) 213(10) = 2 102+ 1 101+ 3 100 Binary System 1 0 1 1 1 23(10)= ++ + +  =10111(2) 10111(2)= 1 24+0  23+ 1  22+ 1  21+ 1  20

  50. Denary System (十進制) Place holder Place values Distinguish values between 111 and 1011 2 basic numerals: 0 to1 (< 2) The place value of each digit is two times the place value of the digit on its right-hand side. e.g. 3065(10)=3  103+ 0 102 + 6 101+ 5 100 0 to 9 (<10) Binary System (二進制)- only computer understand When x=2, abcd is a binary number. abcd(2)= a  23 + b 22 + c 21 + d 20 e.g. 1011(2)=1  23+ 0 22 + 1 21+ 1 20 1011(2)=1  8+ 0 4+ 1 2+ 1 1

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