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Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
(For help, go to Lessons 1-2 and 1-6.)
Simplify each expression.
1
1. 23
2. 42
3. 42  22
4. (–3)3
5. –33
6. 62  12
Evaluate each expression for a = 2, b = –1, c = 0.5.
7.
a
2a
8. bc
c
9.
8-1
ab
bc
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
Solutions
1. 23 = 2 • 2 • 2 = 8
1 = 1 = 1
42
4 • 4 16
2
3. 42  22 = 42 = 4 • 4 = 16 = 4
2•2 4
2
2.
4. (–3)3 = (–3)(–3)(–3) = 9(–3) = –27
5. –33 = –(3 • 3 • 3) = –(9 • 3) = –27
6. 62  12 = 36  12 = 3
a for a = 2: 2 = 1
2•2 2
2a
8. bc for b = –1, c = 0.5: –1 • 0.5 = –1
c
0.5
9. ab for a = 2, b = –1, c = 0.5: 2 • (–1) = 2 = 4
bc
(–1) • 0.5 0.5
7.
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
Simplify.
a.
1
3–2 = 32
1
= 9
b. (–22.4)0 = 1
Use the definition of negative
exponent.
Simplify.
Use the definition of zero as an
exponent.
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
Simplify
a. 3ab –2 = 3a
1
b2
b.
Rewrite using a division symbol.
Use the definition of negative
exponent.
=
3a
b2
1
–3
=
1

x
–3
x
1
=1 3
x
Use the definition of negative
exponent.
Simplify.
= 1 • x3
1
Multiply by the reciprocal of x3 ,
which is x 3.
= x3
Identity Property of Multiplication
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
Evaluate 4x 2y –3 for x = 3 and y = –2.
Method 1: Write with positive exponents first.
4x 2
4x 2y –3 = y 3
Use the definition of negative exponent.
4(3)2
=
(–2)3
36
Substitute 3 for x and –2 for y.
1
= –8 = –4 2
Simplify.
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
(continued)
Method 2: Substitute first.
4x 2y –3 = 4(3)2(–2)–3
4(3)2
=
(–2)3
36
Substitute 3 for x and –2 for y.
Use the definition of negative exponent.
1
= –8 = –4 2
Simplify.
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
In the lab, the population of a certain bacteria doubles every
month. The expression 3000 • 2m models a population of 3000
bacteria after m months of growth. Evaluate the expression for m = 0
and m = –2. Describe what the value of the expression represents in
each situation.
a. Evaluate the expression for m = 0.
3000 • 2m = 3000 • 20
= 3000 • 1
Substitute 0 for m.
Simplify.
= 3000
When m = 0, the value of the expression is 3000. This represents the
initial population of the bacteria. This makes sense because when m = 0,
no time has passed.
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
(continued)
b. Evaluate the expression for m = –2.
3000 • 2m = 3000 • 2–2
Substitute –2 for m.
= 3000 • 1
4
= 750
Simplify.
When m = –2, the value of the expression is 750. This represents the 750
bacteria in the population 2 months before the present population of 3000
bacteria.
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
Simplify each expression.
1. 3–4
1
81
2. (–6)0
k
m–3
1
3. –2a0b–2 – 2
2
4.
5. 8000 • 40
6. 4500 • 3–2
b
8000
8-1
km3
500
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
(For help, go to Lesson 1-6.)
Rewrite each expression using exponents.
1.
t•t•t•t•t•t•t
2. (6 – m)(6 – m)(6 – m)
3.
(r + 5)(r + 5)(r + 5)(r + 5)(r + 5)
4. 5 • 5 • 5 • s • s • s
Simplify.
5. –54
6.
(–5)4
7. (–5)0
8.
(–5)–4
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Solutions
1. t • t • t • t • t • t • t = t7
2. (6 – m)(6 – m)(6 – m) = (6 – m)3
3. (r + 5)(r + 5)(r + 5)(r + 5)(r + 5) = (r + 5)5
4.
5.
6.
7.
8.
5 • 5 • 5 • s • s • s = 53 • s3 = 53s3
–54 = –(5 • 5 • 5 • 5) = –(25 • 25) = –625
(–5)4 = (–5)(–5)(–5)(–5) = (25)(25) = 625
(–5)0 = 1
(–5)–4 = (– 1 )4
5
= (– 1 )(– 1 )(– 1 )(– 1 )
5
5
= ( 1 )( 1 )
25 25
= 1
625
5
5
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Rewrite each expression using each base only once.
a. 73 • 72 = 73 + 2
= 75
b. 44 • 41 • 4–2 = 44 + 1 – 2
= 43
Add exponents of powers with the
same base.
Simplify the sum of the exponents.
Think of 4 + 1 – 2 as 4 + 1 + (–2) to
add the exponents.
Simplify the sum of the exponents.
= 60
Add exponents of powers with the
same base.
Simplify the sum of the exponents.
=1
Use the definition of zero as an exponent.
c. 68 • 6–8 = 68 + (–8)
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Simplify each expression.
a.
p2 • p • p5 = p 2 + 1 + 5
= p8
Add exponents of powers with the same base.
Simplify.
b. 4x6 • 5x–4 = (4 • 5)(x 6 • x –4) Commutative Property of Multiplication
= 20(x 6+(–4))
Add exponents of powers with the
same base.
= 20x 2
Simplify.
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Simplify each expression.
a.
a 2 • b –4 • a 5 = a 2 • a 5 • b –4
Commutative Property of Multiplication
Add exponents of powers with the
same base.
= a 2 + 5 • b –4
a7
= 4
b
Simplify.
b. 2q • 3p3 • 4q4 = (2 • 3 • 4)(p 3)(q • q 4)
Commutative and Associative
Properties of Multiplication
= 24(p 3)(q 1 • q 4)
Multiply the coefficients. Write q as q 1.
= 24(p 3)(q 1 + 4)
Add exponents of powers with the
same base.
= 24p 3q 5
Simplify.
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Simplify (3  10–3)(7  10–5). Write the answer in scientific
notation.
(3 
10–3)(7

10–5) =
(3 • 7)(10–3 • 10–5)
Commutative and Associative
Properties of Multiplication
= 21  10–8
Simplify.
= 2.1  101 • 10–8
Write 21 in scientific notation.
= 2.1 
Add exponents of powers with the
same base.
101 + (– 8)
= 2.1  10–7
Simplify.
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
The speed of light is 3  108 m/s. If there are1  10–3 km in
1 m, and 3.6  103 s in 1 h, find the speed of light in km/h.
Speed of light =
meters
kilometers
seconds
•
•
seconds
meters
hour
m
km
s
= (3  108) s • (1  10–3)
• (3.6  103)
m
h
= (3 • 1 • 3.6)  (108 • 10–3 • 103)
= 10.8  (108 + (– 3) + 3)
Use dimensional analysis.
Substitute.
Commutative and Associative
Properties of Multiplication
Simplify.
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
(continued)
= 10.8  108
Add exponents.
= 1.08  101 • 108
Write 10.8 in scientific notation.
= 1.08  109
Add the exponents.
The speed of light is about 1.08  109 km/h.
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Simplify each expression.
1. 34 • 35
2. 4x5 • 3x–2
39
3. (3  104)(5  102) 1.5  107
12x3
4. (7  10–4)(1.5  105) 1.05  102
5. (–2w –2)(–3w2b–2)(–5b–3) – 305
b
6. What is 2 trillion times 3 billion written in scientific notation? 6  1021
8-3
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
(For help, go to Lesson 8-3.)
Rewrite each expression using each base only once.
1. 32 • 32 • 32
2. 23 • 23 • 23 • 23
3. 57 • 57 • 57 • 57
4. 7 • 7 • 7
Simplify.
5. x3 • x3
6. a2 • a2 • a2
7. y–2 • y–2 • y–2
8. n–3 • n–3
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Solutions
1. 32 • 32 • 32 = 3(2 + 2 + 2) = 36
2. 23 • 23 • 23 • 23 = 2(3 + 3 + 3 + 3) = 212
3. 57 • 57 • 57 • 57 = 5(7 + 7 + 7 + 7) = 528
4. 7 • 7 • 7 = 73
5. x3 • x3 = x(3 + 3) = x6
6. a2 • a2 • a2 = a(2 + 2 + 2) = a6
7. y–2 • y–2 • y–2 = y(–2 + (–2) + (–2)) = y–6 = 16
y
8. n–3 • n–3 = n(–3 + (–3)) = n–6 = 16
n
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Simplify (a3)4.
(a3)4 = a3 • 4
= a12
Multiply exponents when raising a
power to a power.
Simplify.
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Simplify b2(b3)–2.
b2(b3)–2 = b2 • b3 • (–2)
Multiply exponents in (b3)–2.
= b2 • b–6
Simplify.
= b2 + (–6)
Add exponents when multiplying
powers of the same base.
= b–4
Simplify.
1
= b4
Write using only positive exponents.
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Simplify (4x3)2.
(4x3)2 = 42(x3)2
Raise each factor to the second power.
= 42x6
Multiply exponents of a power raised
to a power.
= 16x6
Simplify.
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Simplify (4xy3)2(x3)–3.
(4xy3)2(x3)–3 = 42x2(y3)2 • (x3)–3
Raise the three factors to the second
power.
= 42 • x2 • y6 • x–9
Multiply exponents of a power raised
to a power.
= 42 • x2 • x–9 • y6
Use the Commutative Property of
Multiplication.
= 42 • x–7 • y6
Add exponents of powers with the
same base.
16y6
= x7
Simplify.
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
An object has a mass of 102 kg. The expression
102 • (3  108)2 describes the amount of resting energy in joules the
object contains. Simplify the expression.
102 • (3  108)2 = 102 • 32 • (108)2
Raise each factor within parentheses
to the second power.
= 102 • 32 • 1016
Simplify (108)2.
= 32 • 102 • 1016
Use the Commutative Property of
Multiplication.
= 32 • 102 + 16
Add exponents of powers with the
same base.
= 9  1018
Simplify.
Write in scientific notation.
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Simplify each expression.
1.
(x4)5
3. (5x4)3
5.
x(x5y–2)3
x20
2.
125x12
4. (1.5  105)2
(2w–2)4(3w2b–2)3
432
b6w2
x16
y6
2.25  1010
6. (3  10–5)(4  104)2
8-4
4.8  103
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
(For help, go to Skills Handbook page 724.)
Write each fraction in simplest form.
5
1. 20
5.
6
15
9. 5xy
15x
125
2. 25
6.
8
30
10. 6y2
60
124
4
3. 100
4.
7. 10
8. 18
11. 3ac
12. 24m
12a
6mn2
35
3x
8-5
63
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
Solutions
1.
5
= 5•1 = 1
20
5•4
4
2.
125
25 • 5
=
=5
25
25 • 1
3.
60
20 • 3
3
=
=
100
20 • 5
5
4.
124
4 • 31
=
= 31
4
4•1
5.
6
3•2 2
=
=
15
3•5 5
6.
8
2•4
4
=
=
30 2 • 15
15
7.
10
5•2
2
=
=
35
5•7
7
8.
18
9•2
2
=
=
63
9•7
7
9.
5xy
5•x•y
y
=
=
15x 5 • 3 • x 3
10. 6y2 = 3 • 2 • y2 = 2y2
3x
3•x
x
11. 3ac = 3 • a • c = c
12a
3•4•a 4
12.
8-5
24m
6•4•m
4
=
=
6mn2
6 • m • n2
n2
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
Simplify each expression.
a.
Subtract exponents when dividing
powers with the same base.
x4
4–9
x
=
9
x
= x–5
Simplify the exponents.
1
Rewrite using positive exponents.
= x5
b.
p3 j –4
p3 – (–3)j
=
–3
6
p j
= p6 j –10
p6
= j10
–4 – 6
Subtract exponents when dividing
powers with the same base.
Simplify.
Rewrite using positive exponents.
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
A small dog’s heart beats about 64 million beats in a year. If
there are about 530 thousand minutes in a year, what is its average
heart rate in beats per minute?
6.4  107 beats
64 million beats
=
530 thousand min
5.3  105 min
Write in scientific notation.
6.4
 107–5
Subtract exponents when dividing
powers with the same base.
6.4
 102
Simplify the exponent.
= 5.3
= 5.3
1.21  102
= 121
Divide. Round to the nearest hundredth.
Write in standard notation.
The dog’s average heart rate is about 121 beats per minute.
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
3
y3
Simplify
3
y3
4
4
.
34
= 34
(y )
Raise the numerator and the
denominator to the fourth power.
34
= y 12
Multiply the exponent in the denominator.
81
= y 12
Simplify.
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
a. Simplify
2
3
–3
=
–3
2
3
3
2
.
3
2
Rewrite using the reciprocal of 3 .
Raise the numerator and the
denominator to the third power.
33
= 23
27
3
= 8 or 3 8
Simplify.
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
(continued)
4b
b. Simplify – c .
– 4b
c
–2
=
– c
4b
2
=
– c
4b
2
Rewrite using the reciprocal of – 4b .
c
Write the fraction with a negative
numerator.
(–c)2
= (4b)2
Raise the numerator and denominator
to the second power.
c2
= 16b2
Simplify.
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
Simplify each expression.
1.
a8
a–2
4.
1.6  103
2.
a10
4  10–2
4  104
5.
w3
w7
24
5
8-5
1
w4
2
256
6
or
10
25
25
4
–2
3. (3a) (2a )
6a2
6.
4x 3
3x 2
27
–3
27
64x3
Geometric Sequences
ALGEBRA 1 LESSON 8-6
(For help, go to Lesson 5-6.)
Find the common difference of each sequence.
1. 1, 3, 5, 7, ...
2. 19, 17, 15, 13, ...
3. 1.3, 0.1, –1.1, –2.3, ...
4. 18, 21.5, 25, 28.5, ...
Use inductive reasoning to find the next two numbers in
each pattern.
5. 2, 4, 8, 16, ...
6. 4, 12, 36, ...
7. 0.2, 0.4, 0.8, 1.6, ...
8. 200, 100, 50, 25, ...
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Solutions
1. 1, 3, 5, 7, ...
7 – 5 = 2, 5 – 3 = 2, 3 – 1 = 2
Common difference: 2
2.
19, 17, 15, 13, ...
13 – 15 = –2, 15 – 17 = –2,
17 – 19 = –2
Common difference: –2
3. 1.3, 0.1, –1.1, –2.3, ...
–2.3 – (–1.1) = –1.2, –1.1
– 0.1 = –1.2, 0.1 – 1.3 = –1.2
Common difference: –1.2
4.
18, 21.5, 25, 28.5, ...
28.5 – 25 = 3.5, 25 – 21.5 = 3.5,
21.5 – 18 = 3.5
Common difference: 3.5
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Solutions (continued)
5. 2, 4, 8, 16, ...
2(2) = 4, 4(2) = 8, 8(2) = 16,
16(2) = 32, 32(2) = 64
Next two numbers: 32, 64
6. 4, 12, 36, ...
4(3) = 12, 12(3) = 36,
36(3) = 108, 108(3) = 324
Next two numbers: 108, 324
7. 0.2, 0.4, 0.8, 1.6, ...
(0.2)2 = 0.4, 0.4(2) = 0.8, 0.8(2) =
1.6, 1.6(2) = 3.2, 3.2(2) = 6.4
Next two numbers: 3.2, 6.4
8. 200, 100, 50, 25, ...
200  2 = 100, 100  2 = 50,
50  2 = 25, 25  2 = 12.5,
12.5  2 = 6.25
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Find the common ratio of each sequence.
a. 3, –15, 75, –375, . . .
–15
3
(–5)
–375
75
(–5)
(–5)
The common ratio is –5.
3
3
3
b. 3, 2 , 4 , 8 , ...
3
2
3

1
2
3
4

3
8
1
2

1
2
1
The common ratio is 2 .
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Find the next three terms of the sequence 5, –10, 20, –40, . . .
5
–10
(–2)
20
(–2)
–40
(–2)
The common ratio is –2.
The next three terms are –40(–2) = 80, 80(–2) = –160, and –160(–2) = 320.
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Determine whether each sequence is arithmetic
or geometric.
a. 162, 54, 18, 6, . . .
62
54
1
3
18
1
3
6
1
3
The sequence has a common ratio.
The sequence is geometric.
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
(continued)
b. 98, 101, 104, 107, . . .
98
101
+3
104
+3
107
+3
The sequence has a common difference.
The sequence is arithmetic.
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Find the first, fifth, and tenth terms of the sequence that has
the rule A(n) = –3(2)n – 1.
first term: A(1) = –3(2)1 – 1 = –3(2)0 = –3(1) = –3
fifth term: A(5) = –3(2)5 – 1 = –3(2)4 = –3(16) = –48
tenth term: A(10) = –3(2)10 – 1 = –3(2)9 = –3(512) = –1536
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Suppose you drop a tennis ball from a height of 2 meters.
On each bounce, the ball reaches a height that is 75% of its previous
height. Write a rule for the height the ball reaches on each bounce. In
centimeters, what height will the ball reach on its third bounce?
The first term is 2 meters, which is 200 cm.
Draw a diagram to help understand the problem.
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
(continued)
The ball drops from an initial height, for which there is no bounce. The
initial height is 200 cm, when n = 1. The third bounce is n = 4. The
common ratio is 75%, or 0.75.
A rule for the sequence is A(n) = 200 • 0.75n – 1.
A(n) = 200 • 0.75n – 1
Use the sequence to find the height of
the third bounce.
A(4) = 200 • 0.754 – 1
Substitute 4 for n to find the height of
the third bounce.
= 200 • 0.753
Simplify exponents.
= 200 • 0.421875
Evaluate powers.
= 84.375
Simplify.
The height of the third bounce is 84.375 cm.
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
1. Find the common ratio of the geometric sequence –3, 6, –12, 24, . . .
–2
2. Find the next three terms of the sequence 243, 81, 27, 9, . . .
1
3, 1,
3
3. Determine whether each sequence is arithmetic or geometric.
a. 37, 34, 31, 28, . . . arithmetic
b. 8, –4, 2, –1, . . .
geometric
4. Find the first, fifth, and ninth terms of the sequence that has the rule
A(n) = 4(5)n–1. 4, 2500, 1,562,500
5. Suppose you enlarge a photograph that is 4 in. wide and 6 in. long so
that its dimensions are 20% larger than its original size. Write a rule for
the length of the copies. What will be the length if you enlarge the
photograph five times? (Hint: The common ratio is not just 0.2. You
must add 20% to 100%.)
A(n) = 6(1.2)n-1; about 14.9 in.
8-6