Transcript Ch. 5.2 power point

Chapter 5
Section 2
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
5.2
Integer Exponents, and Quotient
Rule
1
Use 0 as an exponent.
2
Use negative numbers as exponents.
3
Use the quotient rule for exponents.
4
Use combinations of rules.
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Integer Exponents and the Quotient Rule
In all earlier work, exponents were positive integers. Now, to
develop a meaning for exponents that are not positive integers,
consider the following list.
24  16
23  8
22  4
Each time the exponent is reduced by 1, the value is divided by
2 (the bases). Using this pattern, the list can be continued to
smaller and smaller integers.
1
1
21 
22 
21  2
20  1
2
4
From the preceding list, it appears that we should define 20 as
1 and negative exponents as reciprocals.
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Slide 5.2 - 3
Objective 1
Use 0 as an exponent.
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Slide 5.2 - 4
Use 0 as an exponent.
The definitions of 0 and negative exponents must
satisfy the rule for exponents from Section 5.1. For
example if 60 = 1, then
0
2
2
2
0
2
0 2
2
6  6  1 6  6 and 6  6  6  6
so that the product rule is satisfied. Check that the
power rules are also valid for a 0 exponent. Thus we
define a 0 exponent as follows.
For any nonzero real number a,
Example: 170 = 1
a0 = 1.
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Slide 5.2 - 5
EXAMPLE 1
Using Zero Exponents
Evaluate.
Solution:
7
 7 
7
1
0
0
0
1
 1 7
0
 1
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Slide 5.2 - 6
Objective 2
Use negative numbers as
exponents.
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Slide 5.2 - 7
Use negative numbers as exponents.
Since 22  1 and 23  1 , we can deduce that 2−n should equal
8
4
1
. Is the product rule valid in such a case? For example, if we
n
2
multiply
2
6 6  6
2
2 2
6
0
The expression 6−2 behaves as if it were the reciprocal of 62: Their
1
product is 1. The reciprocal of 62 is also 2 , leading us to define
6
1
−2
6 as 2 . This is a particular case of the definition of negative
6
exponents.
For any nonzero real number a and any integer n, a
Example: a  n  1n .
n
1
 n.
a
a
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Slide 5.2 - 8
EXAMPLE 2
Simplify.
Solution:
1
1
2
 2 
4
4
16
1
 
4
3
 3
 
5
2
 43
 64
52
 2
3
25

9
Using Negative Exponents
15 1 2
    
25 5 2
5 2
 
10 10
7

10
1
1
3
 m  0
2 5
m
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1
 3
m
Slide 5.2 - 9
Use negative numbers as exponents.
Consider the following:
1
23 23
1 1
1 34 34

 3  4  3  3.
4
1 2 3
3
2 1 2
34
23 34
Therefore, 4  3 .
3
2
For any nonzero numbers a and b and any integers m and n,
-m
m
a m bn
a
b

=
and  
 
b n a m
b
 
a
Example:
35 2 4
 5
4
2
3
3
and
4
5
   
5
 4
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3
Slide 5.2 - 10
EXAMPLE 3
Changing from Negative to
Positive Exponents
Simplify by writing with positive exponents. Assume
that all variables represent nonzero real numbers.
Solution:
2
5
33
4h 5
2
m k
3
27

25
3
 2
5
2
4m
 5
hk
 x 
 3
 2y 
2
3

3
2y
 
x 
2
3
3
8y 9
 6
x
We cannot use this rule to change negative exponents to positive
exponents if the exponents occur in a sum or difference of terms. For
1
1
example,

2
52  31
5
3
7  23 would be written with positive exponents as 7  1 .
23
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Slide 5.2 - 11
Objective 3
Use the quotient rule for
exponents.
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Slide 5.2 - 12
Use the quotient rule for exponents.
We know that
65 6  6  6  6  6
2


6
.
3
6
666
Notice that the difference between the exponents, 5 − 3 = 2,
this is the exponent in the quotient. This example suggests the
quotient rule for exponents.
For any nonzero real number a and any integer m and n,
am
m n

a
.
n
a
(Keep the same base; subtract the exponents.)
58
8 4
4

5

5
Example:
54
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Slide 5.2 - 13
EXAMPLE 4
Using the Quotient Rule
Simplify by writing with positive exponents. Assume
that all variables represent nonzero real numbers.
7
4
45
5
Solution:
 475  42  16
4
1
1
5 7
2
4  2 
7 4
4
4
16
x 6
6( 12)
 x6
12  x
x
84 m9 n 3
45
910
32

8

m

n
85 m10 n 2
1
81 m 1n 5


5
8mn
1
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Slide 5.2 - 14
The product, quotient, and power rules are the same for positive and
negative exponents.
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Slide 5.2 - 15
Objective 4
Use combinations of rules.
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Slide 5.2 - 16
EXAMPLE 5
Using Combinations of Rules
Simplify. Assume that all variables represent nonzero
real numbers.
Solution:
2
2
 34 
33
2
2
2 4
36


6
5
y
38
5
y
83
5
 3  3  3  243    2  2 4  4
25y
5y
3
6
 6 
 4x   4x
2
2
 4x  4 x
 412 x22
 43 x 4
 64x 4
2
2
3  x y
9
2
2
2
33  x 4 y
393  x 4 y 2

x 4 y
36
 3
y
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729
 3
y
Slide 5.2 - 17