5.2 Integer Exponents and The Quotient Rule
5.2 Integer Exponents and The Quotient Rule
Feb 24, 2016
5.2 Integer Exponents and The Quotient Rule. For Example:. 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. . - PowerPoint PPT Presentation
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5.2
Integer Exponents and The Quotient
Rule
4
3
2
2 16
2 8
2 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.
12 2 02 1 1 122
2 124
From the preceding list, it appears that we should define 20 as 1 and negative exponents as reciprocals.
For any nonzero real number a, a0 = 1.Example: 170 = 1
For Example:
EXAMPLE 1Evaluate.
07
Solution:
1
01 7
1
1
07
07
Since and , we can deducethat 2−n should equal
2 124
3 128
12n
2 2 2 2 06 6 6 6
For any nonzero real number a and any integer n, 1 .n
naa
EXAMPLE 2Simplify.
24
Solution:
2
14
34
5 25
12 21
5
3
1m
5 210 10
710
2
2
53
314
235
1 12 5
3 0m m
116
64
259
Consider the following:
For any nonzero numbers a and b and any integers m and n,
and
Therefore,
3 4 43
4 3 4 3 3
4
12 1 1 1 3 32
13 2 3 2 1 23
.
3 4
4 33.2 3
2
m n
n m
a bb a
-m ma b=b a
Example: and5 4
4 5
3 22 3
3 34 5
5 4
EXAMPLE 3Simplify by writing with positive exponents. Assume that all variables represent nonzero real numbers.Solution:
3
2
35
2725
2
5
4mh k
3
3
3
2
2y
x
2
3
53
5
2
4hm k
32
32xy
9
6
8yx
€
600
−600
1x−4
a−2b3d−3
Use the quotient rule for exponents.We know that
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.
52
3
6 6 6 6 6 6 66 6 6 6
.
For any nonzero real number a and any integer m and n,
(Keep the same base; subtract the exponents.)
Example:
.m
m nn
a aa
88 4 4
4
5 5 55
Simplify by writing with positive exponents. Assume that all variables represent nonzero real numbers.
6
12
xx
Solution:7
5
44
5
7
44
7 54
4 9 3
5 10 2
88
m nm n
6 ( 12)x 6x24
2
14
5 74 241
16
1 1 581
m n
4 5 9 10 3 28 m n
16
5
18mn
Simplify. Assume that all variables represent nonzero real numbers.
Solution:225
6y
24
3
3
3
224 4x x 29 2
3 4
3
3
x y
x y
8
3
33
8 33 53
2 2 24 4x x 1 2 2 24 x 3 44 x
464x
2 4
2
56
y
2
2 4
65 y
4
3625y
9 3 4 2
4
3 x yx y
6
3
3y
3
729y
243
Homework5.1: 1 – 87 EOO5.2: 1 – 77 ODD
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