3. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN
4. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent base
5. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 base
6. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 base
7. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2 base
8. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) base
9. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 base
10. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 base
11. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) base
12. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base
13. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base
14. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base Rules of Exponents
15. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base Rules of Exponents Multiplication Rule: ANAK =AN+K
16. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base Rules of Exponents Multiplication Rule: ANAK =AN+K Example B. a. 5354
17. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base Rules of Exponents Multiplication Rule: ANAK =AN+K Example B. a. 5354 = (5*5*5)(5*5*5*5)
18. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base Rules of Exponents Multiplication Rule: ANAK =AN+K Example B. a. 5354 = (5*5*5)(5*5*5*5) = 53+4 = 57 b. x5y7x4y6
19. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base Rules of Exponents Multiplication Rule: ANAK =AN+K Example B. a. 5354 = (5*5*5)(5*5*5*5) = 53+4 = 57 b. x5y7x4y6 = x5x4y7y6
20. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base Rules of Exponents Multiplication Rule: ANAK =AN+K Example B. a. 5354 = (5*5*5)(5*5*5*5) = 53+4 = 57 b. x5y7x4y6 = x5x4y7y6 = x9y13
21. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base Rules of Exponents Multiplication Rule: ANAK =AN+K Example B. a. 5354 = (5*5*5)(5*5*5*5) = 53+4 = 57 b. x5y7x4y6 = x5x4y7y6 = x9y13 AN = AN β K Division Rule: AK
22. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base Rules of Exponents Multiplication Rule: ANAK =AN+K Example B. a. 5354 = (5*5*5)(5*5*5*5) = 53+4 = 57 b. x5y7x4y6 = x5x4y7y6 = x9y13 AN = AN β K Division Rule: AK 56 Example C. 52
23. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base Rules of Exponents Multiplication Rule: ANAK =AN+K Example B. a. 5354 = (5*5*5)(5*5*5*5) = 53+4 = 57 b. x5y7x4y6 = x5x4y7y6 = x9y13 AN = AN β K Division Rule: AK 56 (5)(5)(5)(5)(5)(5) Example C. = 52 (5)(5)
24. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base Rules of Exponents Multiplication Rule: ANAK =AN+K Example B. a. 5354 = (5*5*5)(5*5*5*5) = 53+4 = 57 b. x5y7x4y6 = x5x4y7y6 = x9y13 AN = AN β K Division Rule: AK 56 (5)(5)(5)(5)(5)(5) Example C. = 52 (5)(5)
25. Exponents We write the quantity A multiplied to itself N times as AN, i.e. A x A x A β¦.x A = AN exponent Example A. 43 = (4)(4)(4) = 64 (xy)2= (xy)(xy) = x2y2 xy2 = (x)(yy) βx2 = β(xx) base Rules of Exponents Multiplication Rule: ANAK =AN+K Example B. a. 5354 = (5*5*5)(5*5*5*5) = 53+4 = 57 b. x5y7x4y6 = x5x4y7y6 = x9y13 AN = AN β K Division Rule: AK 56 (5)(5)(5)(5)(5)(5) = 56 β 2 = 54 Example C. = 52 (5)(5)
29. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4
30. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320
31. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 A1
32. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 A1
33. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0 A1
34. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1
35. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0
36. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = AK AK
37. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K AK AK
38. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK
39. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK 1 Negative-Power Rule: AβK = , A = 0 AK
40. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK 1 Negative-Power Rule: AβK = , A = 0 AK Example D. Simplify a. 30
41. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK 1 Negative-Power Rule: AβK = , A = 0 AK Example D. Simplify a. 30 = 1
42. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK 1 Negative-Power Rule: AβK = , A = 0 AK Example D. Simplify a. 30 = 1 b. 3β2
43. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK 1 Negative-Power Rule: AβK = , A = 0 AK Example D. Simplify a. 30 = 1 1 b. 3β2 = 32
44. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK 1 Negative-Power Rule: AβK = , A = 0 AK Example D. Simplify a. 30 = 1 1 1 b. 3β2 = = 9 32
45. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK 1 Negative-Power Rule: AβK = , A = 0 AK Example D. Simplify a. 30 = 1 1 1 b. 3β2 = = 9 32 2 c. ( )β1 5
46. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK 1 Negative-Power Rule: AβK = , A = 0 AK Example D. Simplify a. 30 = 1 1 1 b. 3β2 = = 9 32 2 1 c. ( )β1 = = 5 2/5
47. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK 1 Negative-Power Rule: AβK = , A = 0 AK Example D. Simplify a. 30 = 1 1 1 b. 3β2 = = 9 32 2 1 5 5 c. ( )β1 1* = = = 5 2/5 2 2
48. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK 1 Negative-Power Rule: AβK = , A = 0 AK Example D. Simplify a. 30 = 1 1 1 b. 3β2 = = 9 32 2 1 5 5 c. ( )β1 1* = = = 5 2/5 2 2 a b In general ( )βK = ( )K b a
49. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK 1 Negative-Power Rule: AβK = , A = 0 AK Example D. Simplify a. 30 = 1 1 1 b. 3β2 = = 9 32 2 1 5 5 c. ( )β1 1* = = = 5 2/5 2 2 a b In general ( )βK = ( )K b a 2 d. ( )β2 5
50. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK 1 Negative-Power Rule: AβK = , A = 0 AK Example D. Simplify a. 30 = 1 1 1 b. 3β2 = = 9 32 2 1 5 5 c. ( )β1 1* = = = 5 2/5 2 2 a b In general ( )βK = ( )K b a 2 5 d. ( )β2 = ( )2 5 2
51. Exponents Power Rule: (AN)K = ANK Example D. (34)5 = (34)(34)(34)(34)(34)= 34+4+4+4+4 = 34*5 = 320 A1 Since = 1 = A1 β 1 = A0, we obtain the 0-power Rule. A1 0-Power Rule: A0 = 1, A = 0 1 A0 Since = = A0 β K = AβK, we get the negative-power Rule. AK AK 1 Negative-Power Rule: AβK = , A = 0 AK Example D. Simplify a. 30 = 1 1 1 b. 3β2 = = 9 32 2 1 5 5 c. ( )β1 1* = = = 5 2/5 2 2 a b In general ( )βK = ( )K b a 2 25 5 d. ( )β2 = ( )2 = 5 4 2
57. Exponents 1 1 1 1 1 β 1* β = e. 3β1 β 40 * 2β2 = = 3 22 3 4 12 Although the negative power means to reciprocate, for problems of consolidating exponents, we do not reciprocate the negative exponents.
58. Exponents 1 1 1 1 1 β 1* β = e. 3β1 β 40 * 2β2 = = 3 22 3 4 12 Although the negative power means to reciprocate, for problems of consolidating exponents, we do not reciprocate the negative exponents. Instead we add or subtract them using the multiplication and division rules first.
59. Exponents 1 1 1 1 1 β 1* β = e. 3β1 β 40 * 2β2 = = 3 22 3 4 12 Although the negative power means to reciprocate, for problems of consolidating exponents, we do not reciprocate the negative exponents. Instead we add or subtract them using the multiplication and division rules first. Example E. Simplify 3β2 x4 yβ6 xβ8 y 23
60. Exponents 1 1 1 1 1 β 1* β = e. 3β1 β 40 * 2β2 = = 3 22 3 4 12 Although the negative power means to reciprocate, for problems of consolidating exponents, we do not reciprocate the negative exponents. Instead we add or subtract them using the multiplication and division rules first. Example E. Simplify 3β2 x4 yβ6 xβ8 y 23 3β2 x4yβ6xβ8y23
61. Exponents 1 1 1 1 1 β 1* β = e. 3β1 β 40 * 2β2 = = 3 22 3 4 12 Although the negative power means to reciprocate, for problems of consolidating exponents, we do not reciprocate the negative exponents. Instead we add or subtract them using the multiplication and division rules first. Example E. Simplify 3β2 x4 yβ6 xβ8 y 23 3β2 x4yβ6xβ8y23 = 3β2 x4 xβ8 yβ6y23
62. Exponents 1 1 1 1 1 β 1* β = e. 3β1 β 40 * 2β2 = = 3 22 3 4 12 Although the negative power means to reciprocate, for problems of consolidating exponents, we do not reciprocate the negative exponents. Instead we add or subtract them using the multiplication and division rules first. Example E. Simplify 3β2 x4 yβ6 xβ8 y 23 3β2 x4yβ6xβ8y23 = 3β2 x4 xβ8 yβ6y23 1 = x4 β 8yβ6+23 9
63. Exponents 1 1 1 1 1 β 1* β = e. 3β1 β 40 * 2β2 = = 3 22 3 4 12 Although the negative power means to reciprocate, for problems of consolidating exponents, we do not reciprocate the negative exponents. Instead we add or subtract them using the multiplication and division rules first. Example E. Simplify 3β2 x4 yβ6 xβ8 y 23 3β2 x4yβ6xβ8y23 = 3β2 x4 xβ8 yβ6y23 1 = x4 β 8yβ6+23 = xβ4y17 9 1 9
64. Exponents 1 1 1 1 1 β 1* β = e. 3β1 β 40 * 2β2 = = 3 22 3 4 12 Although the negative power means to reciprocate, for problems of consolidating exponents, we do not reciprocate the negative exponents. Instead we add or subtract them using the multiplication and division rules first. Example E. Simplify 3β2 x4 yβ6 xβ8 y 23 3β2 x4yβ6xβ8y23 = 3β2 x4 xβ8 yβ6y23 1 = x4 β 8yβ6+23 = xβ4y17 = y17 9 1 9 1 9x4
65. Exponents 1 1 1 1 1 β 1* β = e. 3β1 β 40 * 2β2 = = 3 22 3 4 12 Although the negative power means to reciprocate, for problems of consolidating exponents, we do not reciprocate the negative exponents. Instead we add or subtract them using the multiplication and division rules first. Example E. Simplify 3β2 x4 yβ6 xβ8 y 23 3β2 x4yβ6xβ8y23 = 3β2 x4 xβ8 yβ6y23 1 = x4 β 8yβ6+23 = xβ4y17 = y17 = 9 1 9 1 9x4 y17 9x4
66. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3
67. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3 23xβ8 26xβ3
68. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3 23xβ8 = 23 β 6xβ8β(β3 ) 26xβ3
69. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3 23xβ8 = 23 β 6xβ8β(β3 ) 26xβ3 = 2β3xβ5
70. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3 23xβ8 = 23 β 6xβ8β(β3 ) 26xβ3 = 2β3xβ5 1 1 1 = = 23 x5 * 8x5
71. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3 23xβ8 = 23 β 6xβ8β(β3 ) 26xβ3 = 2β3xβ5 1 1 1 = = 23 x5 * 8x5 (3xβ2y3)β2 x2 Example G. Simplify 3β5xβ3(yβ1x2)3
72. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3 23xβ8 = 23 β 6xβ8β(β3 ) 26xβ3 = 2β3xβ5 1 1 1 = = 23 x5 * 8x5 (3xβ2y3)β2 x2 Example G. Simplify 3β5xβ3(yβ1x2)3 (3xβ2y3)β2 x2 3β5xβ3(yβ1x2)3
73. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3 23xβ8 = 23 β 6xβ8β(β3 ) 26xβ3 = 2β3xβ5 1 1 1 = = 23 x5 * 8x5 (3xβ2y3)β2 x2 Example G. Simplify 3β5xβ3(yβ1x2)3 (3xβ2y3)β2 x2 3β2x4yβ6x2 = 3β5xβ3yβ3 x6 3β5xβ3(yβ1x2)3
74. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3 23xβ8 = 23 β 6xβ8β(β3 ) 26xβ3 = 2β3xβ5 1 1 1 = = 23 x5 * 8x5 (3xβ2y3)β2 x2 Example G. Simplify 3β5xβ3(yβ1x2)3 (3xβ2y3)β2 x2 3β2x4yβ6x2 3β2x4x2yβ6 = = 3β5xβ3yβ3 x6 3β5xβ3(yβ1x2)3 3β5xβ3x6yβ3
75. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3 23xβ8 = 23 β 6xβ8β(β3 ) 26xβ3 = 2β3xβ5 1 1 1 = = 23 x5 * 8x5 (3xβ2y3)β2 x2 Example G. Simplify 3β5xβ3(yβ1x2)3 (3xβ2y3)β2 x2 3β2x4yβ6x2 3β2x4x2yβ6 = = 3β5xβ3yβ3 x6 3β5xβ3(yβ1x2)3 3β5xβ3x6yβ3 3β2x6yβ6 = 3β5x3yβ3
76. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3 23xβ8 = 23 β 6xβ8β(β3 ) 26xβ3 = 2β3xβ5 1 1 1 = = 23 x5 * 8x5 (3xβ2y3)β2 x2 Example G. Simplify 3β5xβ3(yβ1x2)3 (3xβ2y3)β2 x2 3β2x4yβ6x2 3β2x4x2yβ6 = = 3β5xβ3yβ3 x6 3β5xβ3(yβ1x2)3 3β5xβ3x6yβ3 3β2x6yβ6 = = 3β2 β (β5) x6 β 3 yβ6 β (β3) 3β5x3yβ3
77. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3 23xβ8 = 23 β 6xβ8β(β3 ) 26xβ3 = 2β3xβ5 1 1 1 = = 23 x5 * 8x5 (3xβ2y3)β2 x2 Example G. Simplify 3β5xβ3(yβ1x2)3 (3xβ2y3)β2 x2 3β2x4yβ6x2 3β2x4x2yβ6 = = 3β5xβ3yβ3 x6 3β5xβ3(yβ1x2)3 3β5xβ3x6yβ3 3β2x6yβ6 = = 3β2 β (β5) x6 β 3 yβ6 β (β3) 3β5x3yβ3 = 33 x3 yβ3 =
78. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3 23xβ8 = 23 β 6xβ8β(β3 ) 26xβ3 = 2β3xβ5 1 1 1 = = 23 x5 * 8x5 (3xβ2y3)β2 x2 Example G. Simplify 3β5xβ3(yβ1x2)3 (3xβ2y3)β2 x2 3β2x4yβ6x2 3β2x4x2yβ6 = = 3β5xβ3yβ3 x6 3β5xβ3(yβ1x2)3 3β5xβ3x6yβ3 3β2x6yβ6 = = 3β2 β (β5) x6 β 3 yβ6 β (β3) 3β5x3yβ3 27x3 = 33 x3 yβ3 = y3
79. Exponents 23xβ8 Example F. Simplify using the rules for exponents. Leave the answer in positive exponents only. 26 xβ3 23xβ8 = 23 β 6xβ8β(β3 ) 26xβ3 = 2β3xβ5 1 1 1 = = 23 x5 * 8x5 (3xβ2y3)β2 x2 Example G. Simplify 3β5xβ3(yβ1x2)3 (3xβ2y3)β2 x2 3β2x4yβ6x2 3β2x4x2yβ6 = = 3β5xβ3yβ3 x6 3β5xβ3(yβ1x2)3 3β5xβ3x6yβ3 3β2x6yβ6 = = 3β2 β (β5) x6 β 3yβ6 β (β3) 3β5x3yβ3 27x3 = 33 x3 yβ3 = y3