3.5 exponents and order of operations; complex fractions · pdf file3.5 exponents and order of...
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200
3.5 Exponents and Order of Operations; Complex Fractions Since exponents represent repeated multiplication, we are in a position to compute exponents for
fractions, mixed numbers, and decimals. For example, to compute 113
!"#
$%&2
, we first convert to
fractions, compute the multiplication, then convert back to mixed form. The steps are:
11
3
!"#
$%&
2
=4
3
!"#
$%&
2
converting to fraction form
=4
3•
4
3writing out the exponent
=16
9multiplying fractions
= 17
9converting to mixed number
Example 1 Compute the following exponents.
a. !3
4
"#$
%&'3
b. !21
4
"#$
%&'2
c. 0.5( )3
d. !1.4( )2
Solution a. Computing the exponent:
!
3
4
"#$
%&'
3
= !3
4
"#$
%&'
• !3
4
"#$
%&'
• !3
4
"#$
%&'
writing out the exponent
= !27
64multiplying fractions
201
b. Computing the exponent:
!21
4
"#$
%&'
2
= !9
4
"#$
%&'
2
converting to fraction form
= !9
4
"#$
%&'
• !9
4
"#$
%&'
writing out the exponent
=81
16multiplying fractions
= 51
16converting to mixed number
c. Computing the exponent:
0.5( )3= 0.5( ) • 0.5( ) • 0.5( ) writing out the exponent
= 0.125 multiplying decimals
d. Computing the exponent:
!1.4( )2= !1.4( ) • !1.4( ) writing out the exponent
= 1.96 multiplying decimals
Another type of exponent problem involved with fractions is that of repeated exponents. Consider the problem:
2
3
!"#
$%&3
3
4
!"#
$%&2
We could apply the order of operations to this, computing the exponents first, and then the multiplication. However, consider the approach of writing out the exponents. That is:
2
3
!"#
$%&3
3
4
!"#
$%&2
=2
3•2
3•2
3•3
4•3
4
202
If we now factor to primes and cancel, we have:
2
3
!"#
$%&3
3
4
!"#
$%&2
=2
3•2
3•2
3•3
4•3
4
=2
3•2
3•2
3•3
2 • 2•3
2 • 2
=/2 • /2 • /2 • /3 • /3
/2 • /2 • /2 • 2 • 3 • /3 • /3
=1
6
The only multiplication to perform is in the final step, after all simplifying has been done. Example 2 Simplify the following exponents.
a. 3
5
!"#
$%&3
5
6
!"#
$%&2
b. 8
9
!"#
$%&2
3
4
!"#
$%&4
c. 5
12
!"#
$%&2
8
15
!"#
$%&3
d. 5x
6y
!"#
$%&
2
8x
15y
!"#
$%&
2
203
Solution a. Writing out the exponents and simplifying:
3
5
!"#
$%&
35
6
!"#
$%&
2
=3
5•
3
5•
3
5•
5
6•
5
6writing out the exponents
=3
5•
3
5•
3
5•
5
2 • 3•
5
2 • 3prime factorizations
=3 • /3 • /3 • /5 • /5
2 • 2 • /3 • /3 • 5 • /5 • /5cancelling factors
=3
20simplifying
b. Writing out the exponents and simplifying:
8
9
!"#
$%&
23
4
!"#
$%&
4
=8
9•
8
9•
3
4•
3
4•
3
4•
3
4writing out the exponents
=2 • 2 • 2
3 • 3•
2 • 2 • 2
3 • 3•
3
2 • 2•
3
2 • 2•
3
2 • 2•
3
2 • 2prime factorizations
=/2 • /2 • /2 • /2 • /2 • /2 • /3 • /3 • /3 • /3
/2 • /2 • /2 • /2 • /2 • /2 • 2 • 2 • /3 • /3 • /3 • /3cancelling factors
=1
4simplifying
c. Writing out the exponents and simplifying:
5
12
!"#
$%&
28
15
!"#
$%&
3
=5
12•
5
12•
8
15•
8
15•
8
15writing out the exponents
=5
2 • 2 • 3•
5
2 • 2 • 3•
2 • 2 • 2
3 • 5•
2 • 2 • 2
3 • 5•
2 • 2 • 2
3 • 5prime factorizations
=/2 • /2 • /2 • /2 • 2 • 2 • 2 • 2 • 2 • /5 • /5
/2 • /2 • /2 • /2 • 3 • 3 • 3 • 3 • 3 • 5 • /5 • /5cancelling factors
=32
1215simplifying
204
d. Writing out the exponents and simplifying:
5x
6y
!"#
$%&
2
8x
15y
!"#
$%&
2
=5x
6y•
5x
6y•
8x
15y•
8x
15ywriting out the exponents
=5 • x
2 • 3 • y•
5 • x
2 • 3 • y•
2 • 2 • 2 • x
3 • 5 • y•
2 • 2 • 2 • x
3 • 5 • yprime factorizations
=/2 • /2 • 2 • 2 • 2 • 2 • /5 • /5 • x • x • x • x
/2 • /2 • 3 • 3 • 3 • 3 • /5 • /5 • y • y • y • ycancelling factors
=2 • 2 • 2 • 2 • x • x • x • x
3 • 3 • 3 • 3 • y • y • y • yremaining factors
=16x4
81y4simplifying
We now turn our attention to order of operations problems involving fractions. Recall the order of operations agreement from Chapter 1: Order of Operations 1. First compute all parentheses. 2. Compute all exponents next. 3. Compute all multiplications and divisions (working left to right). 4. Compute all additions and subtractions (working left to right). Except for the fact that the steps in each problem may take a little longer, the problems are generally of the same form.
205
Example 3 Compute the following expressions.
a. 1
4!1
2•2
3
b. 1
4!1
2
"#$
%&'•2
3
c. 5
6!2
3
"#$
%&'2
d. 1
12!1
6!1
2
"#$
%&'2
Solution a. First compute the multiplication:
1
2•2
3=1• /2
/2 • 3=1
3
The problem then becomes a subtraction:
1
4!1
2•2
3=1
4!1
3
=1
4•3
3!1
3•4
4
=3
12!4
12
= !1
12
b. First compute the subtraction within the parentheses:
1
4!1
2=1
4!1
2•2
2
=1
4!2
4
= !1
4
206
The problem then becomes a multiplication:
1
4!1
2
"#$
%&'•2
3= !
1
4•2
3
= !1
2 • 2•2
3
= !1• /2
/2 • 2 • 3
= !1
2 • 3
= !1
6
c. First compute the exponent:
2
3
!"#
$%&2
=2
3•2
3=4
9
The problem then becomes a subtraction:
5
6!2
3
"#$
%&'2
=5
6!4
9
=5
6•3
3!4
9•2
2
=15
18!8
18
=7
18
d. First compute the subtraction within the parentheses:
1
6!1
2=1
6!1
2•3
3
=1
6!3
6
= !2
6
= !1• /2
/2 • 3
= !1
3
207
Thus the problem becomes:
1
12!1
6!1
2
"#$
%&'2
=1
12! !
1
3
"#$
%&'2
Now computing the exponent:
!1
3
"#$
%&'2
= !1
3
"#$
%&'• !
1
3
"#$
%&'=1
9
Now computing the subtraction:
1
12! !
1
3
"#$
%&'2
=1
12!1
9
=1
12•3
3!1
9•4
4
=3
36!4
36
= !1
36
Order of operations also applies to mixed numbers, as the next example illustrates. Example 4 Compute the following expressions. Answer using mixed numbers.
a. 21
3!11
2• 31
3
b. 21
3!11
2
"#$
%&'• 31
3
c. 31
4!12
3
"#$
%&'• 5
1
3! 2
1
2
"#$
%&'
d. 23
4
!"#
$%&2
' 31
2
!"#
$%&2
208
Solution a. First compute the multiplication (remember to convert to fractions first):
11
2• 31
3=3
2•10
3
=3
2•2 • 5
3
=/2 • /3 • 5
/2 • /3
= 5
The problem then becomes a subtraction:
21
3!11
2• 31
3= 2
1
3! 5
=7
3!5
1•3
3
=7
3!15
3
= !8
3
= !22
3
b. First compute the subtraction within the parentheses:
21
3!11
2=7
3!3
2
=7
3•2
2!3
2•3
3
=14
6!9
6
=5
6
209
The problem then becomes a multiplication:
21
3!11
2
"#$
%&'• 31
3=5
6• 31
3
=5
6•10
3
=5
2 • 3•2 • 5
3
=/2 • 5 • 5
/2 • 3 • 3
=25
9
= 27
9
c. First compute the subtractions within each parentheses:
31
4!12
3=13
4!5
3
=13
4•3
3!5
3•4
4
=39
12!20
12
=19
12
51
3! 2
1
2=16
3!5
2
=16
3•2
2!5
2•3
3
=32
6!15
6
=17
6
The problem then becomes a multiplication:
31
4!12
3
"#$
%&'• 5
1
3! 2
1
2
"#$
%&'=19
12•17
6
=323
72
= 435
72
210
d. First compute the exponents:
23
4
!"#
$%&2
=11
4
!"#
$%&2
=11
4•11
4=121
16
31
2
!"#
$%&2
=7
2
!"#
$%&2
=7
2•7
2=49
4
The problem then becomes a subtraction:
23
4
!"#
$%&2
' 31
2
!"#
$%&2
=121
16'49
4
=121
16'49
4•4
4
=121
16'196
16
= '75
16
= '411
16
A particular type of fraction is called a complex fraction. A complex fraction is a “fraction
within a fraction”, such as
3
4
2
3
or
1
3+ 3
5
6
. Let’s start with the fraction
3
4
2
3
. One approach in
simplifying this fraction is to interpret it as division, then convert to a multiplication problem. The steps are:
3
42
3
=3
4÷
2
3writing as division
=3
4•
3
2converting to multiplication
=9
8 or 1
1
8multiplying fractions
211
There is a second approach to use for this problem. The LCM for 3 and 4 is 12, so we multiply
the numerator and denominator by the form of 1 which is 1212
(just as if we were building the
fraction). The steps are:
3
42
3
=
3
4•12
2
3•12
multiplying by 12
12
=3 • 3
2 • 4simplifying (cancelling)
=9
8simplifying fractions
This second approach is generally easier to apply, especially with more complicated fractions. Applying this same approach with the second fraction:
1
3+ 3
5
6
=
1
3+ 3
!"#
$%&
• 6
5
6• 6
multiplying by 6
6
=
1
3• 6 + 3 • 6
30
6
distributive property
=2 +18
5computing
=20
5simplifying
= 4 simplifying
We will use this second approach for the next example.
212
Example 5 Simplify the following complex fractions.
a.
5
6
7
12
b. 4 +
2
3
5
12
c.
3
4+5
6
2 !2
3
d. 11
2+ 2
1
3
34
5!11
6
Solution a. Multiplying by 1212
and simplifying the resulting fraction:
5
67
12
=
5
6•12
7
12•12
multiplying by 12
12
=5 • 2
7 •1simplifying (cancelling)
=10
7simplifying fractions
213
b. Multiplying by 1212
and simplifying the resulting fraction:
4 +2
35
12
=
4 +2
3
!"#
$%&
•12
5
12•12
multiplying by 12
12
=
4 •12 +2
3•12
5 •1distributive property
=48 + 8
5simplifying fractions
=56
5simplifying
c. Multiplying by 1212
and simplifying the resulting fraction:
3
4+
5
6
2 !2
3
=
3
4+
5
6
"#$
%&'
•12
2 !2
3
"#$
%&'
•12
multiplying by 12
12
=
3
4•12 +
5
6•12
2 •12 !2
3•12
distributive property
=3 • 3+ 5 • 2
24 ! 2 • 4simplifying fractions
=9 +10
24 ! 8multiplying
=19
16simplifying
214
d. Multiplying by 3030
and simplifying the resulting fraction (first convert
mixed numbers to fractions):
11
2+ 2
1
3
34
5!1
1
6
=
3
2+
7
319
5!
7
6
converting to fractions
=
3
2+
7
3
"#$
%&'
• 30
19
5!
7
6
"#$
%&'
• 30
multiplying by 30
30
=
3
2• 30 +
7
3• 30
19
5• 30 !
7
6• 30
distributive property
=3 •15 + 7 •10
19 • 6 ! 7 • 5simplifying fractions
=45 + 70
114 ! 35multiplying
=115
79simplifying
215
Example 6 Find the average of 2 13
, 312
, and4 56
.
Solution Recall that the average of a group of numbers is the sum of those numbers divided by the amount of numbers. To find the sum of these three mixed numbers, first add their fractional portions:
1
3+1
2+5
6=1
3•2
2+1
2•3
3+5
6
=2
6+3
6+5
6
=10
6
=/2 • 5
/2 • 3
=5
3
= 12
3
Therefore the sum is given by:
21
3+ 31
2+ 4
5
6= 9 +1
2
3= 10
2
3
Thus, the average of the three numbers is given by the complex fraction:
21
3+ 3
1
2+ 4
5
6
3=
102
3
3computing the sum
=
32
3
3writing as a fraction
=
32
3• 3
3 • 3multiplying by
3
3
=32
9simplifying
= 35
9converting to mixed number
216
We conclude this section with order of operations with decimals. The rules are identical to those with fractions and mixed numbers. Example 7 Compute the following expressions. a. 13.2 ! (5.4)(3.7) b. 4.06 ! (1.4)
2 c. (4.06 !1.4)
2 d. (3.5 ! 8)(4 ! 8.6) Solution a. Computing using the order of operations:
13.2 ! (5.4)(3.7) = 13.2 !19.98 multiplying decimals
= 13.2 + (!19.98) writing subtraction as addition
= !6.78 adding decimals
b. Computing using the order of operations:
4.06 ! (1.4)2= 4.06 !1.96 computing the exponent
= 2.1 subtracting decimals
c. Computing using the order of operations:
(4.06 !1.4)2= (2.66)2 subtracting decimals
= 7.0756 computing the exponent
d. Computing using the order of operations:
(3.5 ! 8)(4 ! 8.6) = 3.5 + (!8)( ) 4 + (!8.6)( ) writing as addition
= (!4.5)(!4.6) adding decimals
= 20.7 multiplying decimals
Terminology order of operations complex fractions
217
Exercise Set 3.5 Compute the following exponents.
1. !2
5
"#$
%&'2
2. !4
7
"#$
%&'2
3. !3
5
"#$
%&'3
4. !5
8
"#$
%&'3
5. !4
5
"#$
%&'2
6. !5
6
"#$
%&'2
7. !32
3
"#$
%&'2
8. !51
4
"#$
%&'2
9. !11
3
"#$
%&'3
10. !21
2
"#$
%&'3
11. (0.2)3 12. (!0.3)3
13. (!1.3)2 14. (!1.1)2
15. !0.82 16. !1.2
2 Simplify the following exponents.
17. 3
4
!"#
$%&2
2
3
!"#
$%&3
18. 3
4
!"#
$%&3
4
5
!"#
$%&2
19. 9
14
!"#
$%&3
7
9
!"#
$%&2
20. 5
6
!"#
$%&3
4
5
!"#
$%&2
21. 4
9
!"#
$%&2
3
8
!"#
$%&3
22. 5
12
!"#
$%&3
8
15
!"#
$%&2
23. x
2y
!"#
$%&
3
4y
x
!"#
$%&2
24. 3x
2y
!"#
$%&
2
4y
9x
!"#
$%&2
25. 2b
3a
!"#
$%&2
9a
8b
!"#
$%&2
26. 4b
5a
!"#
$%&3
15a
16b
!"#
$%&2
218
Compute the following expressions. Express your answers as fractions.
27. 3
8!1
8•2
3 28. 3
5!1
6•3
4
29. 3
7!1
12•6
7 30. 3
4!5
8•6
11
31. 1
8!1
3
"#$
%&'•2
5 32. 3
4!7
8
"#$
%&'•4
13
33. 1
5!7
10
"#$
%&'• !
8
15
"#$
%&'
34. 1
7!5
14
"#$
%&'• !
7
12
"#$
%&'
35. 3
8!1
4
"#$
%&'2
36. 7
12!2
3
"#$
%&'2
37. 8
15!3
5
"#$
%&'2
38. 5
6!1
2
"#$
%&'3
39. 1
4!2
3
"#$
%&'•1
2!1
5
"#$
%&'
40. 1
8!3
4
"#$
%&'•1
3!5
6
"#$
%&'
41. 1
12!3
8
"#$
%&'•1
6!7
10
"#$
%&'
42. 1
15!9
10
"#$
%&'•5
8!7
12
"#$
%&'
43. 5
12!1
8!1
2
"#$
%&'2
44. 2
3!4
5!7
10
"#$
%&'2
45. 5
16!1
2!2
3
"#$
%&'2
46. 3
8!1
3!5
12
"#$
%&'2
219
Compute the following expressions. Answer using mixed numbers.
47. 114! 2
1
3• 31
2 48. 3
1
3! 4
1
2• 31
4
49. 11
4! 2
1
3
"#$
%&'• 31
2 50. 3
1
3! 4
1
2
"#$
%&'• 31
4
51. 51
4+ 2
1
3• 31
5 52. 4
1
5+11
2• 23
4
53. 51
4+ 2
1
3
!"#
$%&• 31
5 54. 4
1
5+11
2
!"#
$%&• 23
4
55. 61
2•11
4!11
3• 21
2 56. 4
1
5• 21
4! 2
1
4• 31
3
57. 31
2• 21
4! 5
1
3• 31
3 58. 2
1
3• 21
4! 4
1
2• 53
4
59. 31
2!13
4
"#$
%&'• 4
1
3! 2
1
2
"#$
%&'
60. 21
3! 31
2
"#$
%&'• 3
1
2! 5
3
4
"#$
%&'
61. 21
4! 5
5
8
"#$
%&'• 1
1
3! 4
5
6
"#$
%&'
62. 21
3! 4
3
4
"#$
%&'• 1
1
5! 2
3
10
"#$
%&'
63. 31
4
!"#
$%&2
' 21
2
!"#
$%&2
64. 23
4
!"#
$%&2
' 11
3
!"#
$%&2
65. 21
3
!"#
$%&2
' 51
2
!"#
$%&2
66. 31
4
!"#
$%&2
' 51
2
!"#
$%&2
220
Simplify the following complex fractions.
67.
3
4
5
6
68.
5
8
5
12
69.
15
16
5
6
70.
16
21
4
7
71. 3+
1
4
2
3
72. 4 +
1
2
3
5
73. 5 !
3
4
5
6
74. 4 !
3
5
7
10
75.
2
3+3
4
4 !1
3
76.
5
6+2
3
3!1
2
77.
3
4!1
3
5
6+1
2
78.
2
3+1
6
3
4!5
8
79. 11
2! 2
2
3
31
4+11
6
80. 11
3! 4
1
2
21
4+ 31
8
81. 21
6+ 4
2
3
21
3! 5
3
4
82. 11
2+ 2
2
3
11
2! 2
2
3
221
Find the average of the following sets of numbers.
83. 112,23
4,51
4 84. 3
1
3,41
2,51
6
85. 63
10,31
2,53
5 86. 5
3
4,47
8,95
6
87. !22
3,11
4,!3
1
2 88. !6
3
5,!2
1
4,41
2
89. !42
3,!3
1
4,21
2,61
6 90. !4
5
6,!2
3
4,31
3,!5
1
2
91. 5.46, 3.85, 2.7, 8.5 92. 4.9, 8.69, 5.04, 12.9 93. –4.5, 1.08, –5.59, 3.2 94. –6.8, 2.59, 1.8, –13.5 Compute the following expressions. 95. 14.96 ! (3.1)(2.8) 96. 18 ! (3.5)(2.7) 97. 6 ! (4.6)(2.9) 98. 4.62 ! (5.7)(3.5) 99. (6.7)(4.5) ! (2.1)(3.9) 100. (5.9)(4.2) ! (3.2)(3.6) 101. 5.05 ! (1.8)2 102. 12 ! (2.6)2 103. 4.16 ! (5.2)2 104. 8.83! (4.9)2 105. (3.1!1.9)2 106. (2.4 ! 4.1)2 107. (2.6 ! 8)(5 ! 3.7) 108. (3.1! 7.5)(6 ! 9.4) 109. (4.6 ! 8)(5.2 !10) 110. (3.7 ! 8.3)(4.5 ! 6.2) 111. (5.2)(!4) ! (3! 6.1)2 112. (3.5)(!6) ! (4.2 ! 7)2 113. (3.1! 6.3)2 ! (5 ! 7.3)2 114. (6 !10.2)2 ! (8.5 ! 4.9)2