Chabot Mathematics
§1.6 Exponent
Properties
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Review §
1.5
MTH 55
 Any QUESTIONS About
• §1.5 → (Word) Problem Solving
 Any QUESTIONS About HomeWork
• §1.5 → HW-01
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Exponent PRODUCT Rule
 For any number a and any positive
integers m and n,
Exponent
a a  a
m
n
m n
Base
 In other Words:
To MULTIPLY powers with the
same base, keep the base and
ADD the exponents
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Quick Test of Product Rule
a a  a
m
 Test
n
?
23
3 3 3
2
3
m n
3
5
3  3  9  27  243
2
3
3  3  3  3  3  3  3  3 3  3  3  9  27  243
5

Chabot College Mathematics
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Example  Product Rule
 Multiply and simplify each of the
following. (Here “simplify” means
express the product as one base to a
power whenever possible.)
a) x3  x5
c) (x + y)6(x + y)9
Chabot College Mathematics
5
b) 62  67  63
d) (w3z4)(w3z7)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Example  Product Rule
 Solution a) x3  x5 = x3+5
Base is x
= x8
Adding exponents
 Solution b) 62  67  63 = 62+7+3
Base is 6
= 612
 Solution c) (x + y)6(x + y)9 = (x + y)6+9
Base is (x + y)
= (x + y)15
 Solution d) (w3z4)(w3z7) = w3z4w3z7
3w 3z 4z 7
=
w
TWO Bases: w & z
= w6z11
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Exponent QUOTIENT Rule
 For any nonzero number a and any
positive integers m & m
mn
n for which m > n,
n
 In other Words:
To DIVIDE powers with the same
base, SUBTRACT the exponent of
the denominator from the
exponent of the numerator
a
a
a
Chabot College Mathematics
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Quick Test of Quotient Rule
m
a
mn

a
n
a
6 ?
 Test
5
6 4
2

5

5
4
5
64
5
555555
2


5

5

5

5
4
5
5555
6
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Example  Quotient Rule
 Divide and simplify each of the
following. (Here “simplify” means
express the product as one base to a
power whenever possible.)
x
• a) 3
x
87
b) 3
8
(6 y )14
• c)
6
(6 y )
7 9
6
r
t
d)
4r 3 t
9
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Example  Quotient Rule
9
x
 Solution a) 3  x93
 x6
x
Base is x
7
8
4
7 3
 Solution b)
8
8
3
8
Base is 8
(6 y )14
14  6
8
 (6 y )
 (6 y )
 Solution c)
6
(6 y )
Base is (6y)
7 9
7
9
6
r
t
6
r
t
 Solution d)
  3
3
4 r t
4r t
TWO Bases: r & t
6 7 3 91 3 4 8
  r t  r t
4
2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
The Exponent Zero
 For any number
a where a ≠ 0
a 1
0
 In other Words:
Any nonzero number raised
to the 0 power is 1
• Remember the base can be
ANY Number
–0.00073, 19.19, −86, 1000000, anything
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Example  The Exponent Zero


Simplify:
c) (4w)0
Solutions
a) 12450
d) (−1)80
b) (−3)0
e) −80
a) 12450 = 1
b) (−3)0 = 1
c) (4w)0 = 1, for any w  0.
d) (−1)80 = (−1)1 = −1
e) −80 is read “the opposite of 80” and is
equivalent to (−1)80: −80 = (−1)80
= (−1)1 = −1
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
The POWER Rule
 For any number a and any whole
numbers m and n
a 
n
m
a
mn
 In other Words:
To RAISE a POWER to a POWER,
MULTIPLY the exponents and
leave the base unchanged
Chabot College Mathematics
13
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Quick Test of Power Rule
a 
m n
a
7   7
?
3
2
23
mn
7
3
2 3
7   49  49 49 49
6
 7  7 7  7 7  7  7  7  7  7  7  7  7
 Test
7
Chabot College Mathematics
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23
6

Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Example  Power Rule
 Simplify:
a) (x3)4
 Solution a)
(x3)4 = x34
= x12
Base is x
 Solution b)
Base is 4
Chabot College Mathematics
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b) (42)8
(42)8 = 428
= 416
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Raising a Product to a Power
 For any numbers a and b and
any whole number n,
a  b
n
 a b
n
n
 In other Words:
To RAISE A PRODUCT to a
POWER, RAISE Each Factor to
that POWER
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Quick Test of Product to Power
n
n
n
a  b
 a b
2 11
3
 Test
?
 2 11
3
3
2 11  22  22 22 22  10648
3

3
2 11  8 1331  10648
Chabot College Mathematics
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3
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Example  Product to Power

Simplify: a) (3x)4
b) (−2x3)2
c) (a2b3)7(a4b5)
 Solutions
a) (3x)4 = 34x4 = 81x4
b) (−2x3)2 = (−2)2(x3)2 = (−1)2(2)2(x3)2 = 4x6
c) (a2b3)7(a4b5) = (a2)7(b3)7a4b5
= a14b21a4b5
Multiplying exponents
= a18b26
Adding exponents
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Raising a Quotient to a Power
n
 For any real
numbers a and b,
b ≠ 0, and any
whole number n
 In other Words:
To Raise a Quotient to a power,
raise BOTH the numerator &
denominator to the power
a
a
   n
b
b
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
n
Quick Test of Quotient to Power
n
a
a
   n
b
b
3
n
 Test
3
5 5
   3
7 7
?
3
5 5
   3
7 7
3

 5   5   5   5  125 5  5  5 5

 3
       
 7   7   7   7  343 7  7  7 7
3
Chabot College Mathematics
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3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Example  Quotient to a Power
 Simplify: a)
w
 
4
3
 3 
b)  5 
b 
4
 2a 
c)  4 
 b 
5
3
 Solution a)
 Solution b)
 Solution c)
Chabot College Mathematics
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3
3
w
w
w
 
   3 
64
4
4
4
4
3
3
 
 5  5 4
(b )
b 
2
 2a 5 
(2a 5 ) 2
 4   4 2
(b )
 b 
81
81
 54  20
b
b
22 (a5 )2 4a10

 8
4 2
b
b
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
2
Negative Exponents
 Integers as Negative Exponents
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Negative Exponents
 For any real number a that is
nonzero and any integer n
a
n
1
 n
a
 The numbers a−n and an are thus
RECIPROCALS of each other
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Example  Negative Exponents
 Express using POSITIVE exponents,
and, if possible, simplify.
a) m–5 b) 5–2
c) (−4)−2 d) xy–1
 SOLUTION
1
–5
a) m = 5
m
b) 5–2
Chabot College Mathematics
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1
1
= 52  25
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Example  Negative Exponents
 Express using POSITIVE exponents,
and, if possible, simplify.
a) m–5 b) 5–2
c) (−4)−2 d) xy−1
 SOLUTION
1
1
1
c) (−4)−2 = (4) 2  (4)(4)  16
 1
1 x
d)
= x 1   x  
 y y
y 
• Remember PEMDAS
xy–1
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
More Examples
 Simplify. Do NOT use NEGATIVE
exponents in the answer.
5
3
a) w  w
b) (x4)3
c) (3a2b4)3
5
a
d) 6
a
1
e) b 9
w 7
f)
6
z
 Solution
a) w5  w3  w5 ( 3)  w2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
More Examples

Solution
b) (x−4)−3 = x(−4)(−3) = x12
c) (3a2b−4)3 = 33(a2)3(b−4)3
6
27a
= 27 a6b−12 = 12
b
a 5
5 ( 6 )
1

a

a
a
d) a 6
1
 ( 9 )
9

b

b
e) 9
b
6
w7
1
1
z
7
6

w



z
 7
f)
6
6
7
z
z
w
w
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Factors & Negative Exponents
 For any nonzero real numbers a
and b and any integers m and n
n
m
a
b

m
n
b
a
 A factor can be moved to the other
side of the fraction bar if the sign of
the exponent is changed
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Examples  Flippers
 Simplify
20 x 6
3 4
4y z
 SOLUTION
 We can move the negative factors to
the other side of the fraction bar if we
change the sign of each exponent.
20 x 6 5 z 4
 3 6
3 4
4y z
y xz
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Reciprocals & Negative Exponents
 For any nonzero real numbers a
and b and any integer n
a
 
b
n
b
 
a
n
 Any base to a power is equal to the
reciprocal of the base raised to the
opposite power
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Examples  Flippers
 Simplify  a 
 
 3b 
4
2
2
 SOLUTION  a    3b 
 
 4
3
b
a 
 
4
2
(3b) 2
 4 2
(a )
2
2
2
3 b
9b
 8  8
a
a
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Summary – Exponent Properties
a1 = a
0 as an exponent
a0 = 1
Negative
Exponents
(flippers)
an
1
 n,
a
a n bm  a 
 n,  
m
b
a
b
The Product Rule
a m  a n  a mn .
The Quotient Rule
am
 a mn .
n
a
The Power Rule
(am)n = amn
The Product to
a Power Rule
(ab)n = anbn
The Quotient to
a Power Rule
Chabot College Mathematics
32
n
n
a a
   n.
b b
n
b
 
a
n
This summary assumes that no
denominators are 0 and that 00 is not
considered. For any integers m and n
1 as an exponent
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
WhiteBoard Work
 Problems From §1.6 Exercise Set
• 14, 24, 52, 70, 84, 92, 112, 130

Base &
Exponent →
Which is Which?
Chabot College Mathematics
33
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
All Done for Today
Astronomical
Unit
(AU)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
35
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt