1. Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
(For help, go to Lessons 1-2 and 1-6.)
Simplify each expression. 1 42
 22
4. (–3)3 5. –  12
Evaluate each expression for a = 2, b = –1, c = 0.5. a 2a bc c ab bc
8-1

2. Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
1. 23 = 2 • 2 • 2 = 8
= =
3. 42  22 = = = = 4
4. (–3)3 = (–3)(–3)(–3) = 9(–3) = –27
5. –33 = –(3 • 3 • 3) = –(9 • 3) = –27
6. 62  12 = 36  12 = 3
for a = 2: =
for b = –1, c = 0.5: = –1
for a = 2, b = –1, c = 0.5: = = 4
Solutions 1 42 1
4 • 4 1 16 42 22
4 • 4
2 • 2 16 4 a 2a 2
2 • 2 1 2 bc c
–1 • 0.5
0.5 ab bc
2 • (–1)
(–1) • 0.5 2
0.5
8-1

3. Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
Simplify. =
Use the definition of negative exponent. 1 32 a.
3–2
Simplify. 1 9 = b.
(–22.4)0
Use the definition of zero as an exponent.
= 1
8-1

4. Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
Simplify 1
x –3 a.
3ab –2 1 b2
Use the definition of negative exponent.
= 3a b.
Rewrite using a division symbol.
= 1  x –3
= 1  1
x 3
Use the definition of negative exponent.
Simplify. 3a
b 2 =
= 1 • x 3
Multiply by the reciprocal of , which is x 3. 1 x3
= x 3
Identity Property of Multiplication
8-1

5. 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 2y –3 =
Use the definition of negative exponent.
4x 2
y 3
Substitute 3 for x and –2 for y.
4(3)2
(–2)3 = 36 –8 –4 1 2 =
Simplify.
8-1

6. Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
(continued)
Method 2: Substitute first.
4x 2y –3 = 4(3)2(–2)–3
Substitute 3 for x and –2 for y.
4(3)2
(–2)3 =
Use the definition of negative exponent. 36 –8 –4 1 2 =
Simplify.
8-1

7. 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   Substitute 0 for m.
= 3000 • 1 Simplify.
= 3000
When m = 0, the value of the expression is This represents the initial population of the bacteria. This makes sense because when m = 0, no time has passed.
8-1

8. 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 • Simplify.
= 750 1 4
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

9. Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
Simplify each expression.
1. 3–4 2. (–6)0
3. –2a0b–2 4.
• • 3–2
1 81 1 2 b2 – k
m–3
km3
8000
500
8-1

10. 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

11. 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 • 5 • 5 • s • s • s = 53 • s3 = 53s3
5. –54 = –(5 • 5 • 5 • 5) = –(25 • 25) = –625
6. (–5)4 = (–5)(–5)(–5)(–5) = (25)(25) = 625
7. (–5)0 = 1
8. (–5)–4 = (– )4
= (– )(– )(– )(– )
= ( )( ) = 1 5 1 5 1 5 1 5 1 5 1 25 1 25 1
625
8-3

12. Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Rewrite each expression using each base only once.
Add exponents of powers with the
same base.
73 + 2 = a.
73 • 72
= 75
Simplify the sum of the exponents.
Think of – 2 as (–2) to add the exponents.
– 2 = b.
44 • 41 • 4–2
= 43
Simplify the sum of the exponents.
Add exponents of powers with the
same base.
68 + (–8) = c.
68 • 6–8
= 60
Simplify the sum of the exponents.
Use the definition of zero as an exponent.
= 1
8-3

13. Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Simplify each expression. a.
p2 • p • p5
Add exponents of powers with the same base. p =
= p 8
Simplify.
4x6 • 5x–4 b.
Commutative Property of Multiplication
(4 • 5)(x 6 • x –4) =
Add exponents of powers with the same base.
= 20(x 6+(–4))
Simplify.
= 20x 2
8-3

14. Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Simplify each expression. a.
a 2 • b –4 • a 5
Commutative Property of Multiplication
a 2 • a 5 • b –4 =
= a • b –4
Add exponents of powers with the same base.
Simplify.
a 7
b 4 =
Commutative and Associative Properties of Multiplication
(2 • 3 • 4)(p 3)(q • q 4) = b.
2q • 3p3 • 4q4
= 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

15. 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) =
Commutative and Associative Properties of Multiplication
(3 • 7)(10–3 • 10–5)
= 21  10–8
Simplify.
= 2.1  101 • 10–8
Write 21 in scientific notation.
= 2.1  (– 8)
Add exponents of powers with the same base.
= 2.1  10–7
Simplify.
8-3

16. 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
seconds
kilometers
hour •
Use dimensional analysis.
= (3  108) • (1  10–3) • (3.6  103) m s km h
Substitute.
= (3 • 1 • 3.6)  (108 • 10–3 • 103)
Commutative and Associative Properties of Multiplication
= 10.8  (108 + (– 3) + 3)
Simplify.
8-3

17. 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

18. Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Simplify each expression.
1. 34 • x5 • 3x–2
3. (3  104)(5  102) 4. (7  10–4)(1.5  105)
5. (–2w –2)(–3w2b–2)(–5b–3)
6. What is 2 trillion times 3 billion written in scientific notation? 39
12x3
1.5  107
1.05  102 30 b5 –
6  1021
8-3

19. 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 • • 23 • 23 • 23
3. 57 • 57 • 57 • • 7 • 7
Simplify.
5. x3 • x3 6. a2 • a2 • a2
7. y–2 • y–2 • y–2 8. n–3 • n–3
8-4

20. More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Solutions
• 32 • 32 = 3( ) = 36
• 23 • 23 • 23 = 2( ) = 212
• 57 • 57 • 57 = 5( ) = 528
4. 7 • 7 • 7 = 73
5. x3 • x3 = x(3 + 3) = x6
6. a2 • a2 • a2 = a( ) = a6
7. y–2 • y–2 • y–2 = y(–2 + (–2) + (–2)) = y–6 =
8. n–3 • n–3 = n(–3 + (–3)) = n–6 = 1
y 6 1
n 6
8-4

21. More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Simplify (a3)4.
Multiply exponents when raising a power to a power.
(a3)4 = a3 • 4
Simplify.
= a12
8-4

22. 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.
Simplify.
= b–4 1 b4 =
Write using only positive exponents.
8-4

23. 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

24. 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

25. 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 •
Add exponents of powers with the same base.
= 9  1018
Simplify.
Write in scientific notation.
8-4

26. More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Simplify each expression.
1. (x4)5 2. x(x5y–2)3
3. (5x4)3 4. (1.5  105)2
5. (2w–2)4(3w2b–2)3 6. (3  10–5)(4  104)2
x16 y6
x20
2.25  1010
125x12
432 b6w2
4.8  103
8-4

27. Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
(For help, go to Skills Handbook page 724.)
Write each fraction in simplest form. 5 20
125 25 60
100
124 4 6 15 8 30 10 35 18 63
5xy
15x
6y2 3x
3ac
12a
24m
6mn2
8-5

28. Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
1. 2.
3. 4.
5. 6.
7. 8.
= = 5 20
5 • 1
5 • 4 1 4
125 25
25 • 5
25 • 1
= = 5 60
100
20 • 3
20 • 5 3
= =
124
= = 31
4 • 31
4 • 1 6 15 2
3 • 2
3 • 5
= = 8 30
2 • 4
2 • 15 10 35 7
5 • 2
5 • 7
= = 18 63
9 • 2
9 • 7
5xy
15x y
5 • x • y
5 • 3 • x
= =
6y2 3x
2y2 x
3 • 2 • y2
3 • x
= =
3ac
12a
3 • a • c
3 • 4 • a c
24m
6mn2
6 • 4 • m
6 • m • n2 n2
= =
Solutions
8-5

29. Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
Simplify each expression. x4 x9 =
Subtract exponents when dividing powers with the same base.
x4 – 9 a.
Simplify the exponents.
= x–5
Rewrite using positive exponents. 1 x5 =
p3 j –4
p–3 j 6 =
Subtract exponents when dividing powers with the same base.
p3 – (–3)j –4 – 6 b.
= p6 j –10
Simplify.
Rewrite using positive exponents. p6
j10 =
8-5

30. 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?
64 million beats
530 thousand min
6.4  107 beats
5.3  105 min =
Write in scientific notation.
6.4
5.3
 107–5 =
Subtract exponents when dividing powers with the same base.
6.4
5.3
 102 =
Simplify the exponent.
1.21  102
Divide. Round to the nearest hundredth.
= 121
Write in standard notation.
The dog’s average heart rate is about 121 beats per minute.
8-5

31. Division Properties of Exponents
ALGEBRA 1 LESSON 8-5 3
y 3 4
Simplify . 3
y 3 4 34
(y 3)4 =
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

32. Division Properties of Exponents
ALGEBRA 1 LESSON 8-5 2 3 –3
a. Simplify . 2 3 –3 =
Rewrite using the reciprocal of . 33 23 =
Raise the numerator and the denominator to the third power.
Simplify. 27 8 3 or =
8-5

33. Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
(continued) 4b c –
b. Simplify 4b c – –2 2 =
Rewrite using the reciprocal of .
Write the fraction with a negative numerator. c 4b – 2 =
Raise the numerator and denominator to the second power.
(–c)2
(4b)2 =
Simplify. c2
16b2 =
8-5

34. Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
Simplify each expression.
a8 a–2
w3 w7 1 w4
(3a)4(2a–2) 6a2
a10 27
1.6   10–2 24 5 2 –3
4  104
256 25
or 10 6
4x 3
3x 2 27
64x3
8-5

35. Find the common difference of each sequence.
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, , 17, 15, 13, ...
3. 1.3, 0.1, –1.1, –2.3, , 21.5, 25, 28.5, ...
Use inductive reasoning to find the next two numbers in
each pattern.
5. 2, 4, 8, 16, , 12, 36, ...
7. 0.2, 0.4, 0.8, 1.6, , 100, 50, 25, ...
8-6

36. Common difference: 2 17 – 19 = –2 Common difference: –2
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Solutions
1. 1, 3, 5, 7, , 17, 15, 13, ...
7 – 5 = 2, 5 – 3 = 2, 3 – 1 = – 15 = –2, 15 – 17 = –2,
Common difference: – 19 = –2
Common difference: –2
3. 1.3, 0.1, –1.1, –2.3, , 21.5, 25, 28.5, ...
–2.3 – (–1.1) = –1.2, – – 25 = 3.5, 25 – 21.5 = 3.5,
– 0.1 = –1.2, 0.1 – 1.3 = – – 18 = 3.5
Common difference: –1.2 Common difference: 3.5
8-6

37. Solutions (continued) 5. 2, 4, 8, 16, ... 6. 4, 12, 36, ...
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Solutions (continued)
5. 2, 4, 8, 16, , 12, 36, ...
2(2) = 4, 4(2) = 8, 8(2) = 16, 4(3) = 12, 12(3) = 36,
16(2) = 32, 32(2) = (3) = 108, 108(3) = 324
Next two numbers: 32, 64 Next two numbers: 108, 324
7. 0.2, 0.4, 0.8, 1.6, , 100, 50, 25, ...
(0.2)2 = 0.4, 0.4(2) = 0.8, 0.8(2) = 200  2 = 100, 100  2 = 50,
1.6, 1.6(2) = 3.2, 3.2(2) =  2 = 25, 25  2 = 12.5,
Next two numbers: 3.2,  2 = 6.25
8-6

38. (–5) (–5) (–5)  Geometric Sequences
ALGEBRA 1 LESSON 8-6
Find the common ratio of each sequence.
a. 3, –15, 75, –375, . . .
3 –15 75 –375
(–5) (–5) (–5)
The common ratio is –5.
b. 3, 3 2 4 8 ,
, ... 3 2 4 8  1
The common ratio is . 1 2
8-6

39. Find the next three terms of the sequence 5, –10, 20, –40, . . .
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Find the next three terms of the sequence 5, –10, 20, –40, . . .
5 –10 20 –40
(–2) (–2) (–2)
The common ratio is –2.
The next three terms are –40(–2) = 80, 80(–2) = –160, and –160(–2) = 320.
8-6

40.  Geometric Sequences Determine whether each sequence is arithmetic
ALGEBRA 1 LESSON 8-6
Determine whether each sequence is arithmetic
or geometric.
a. 162, 54, 18, 6, . . . 1 3 
The sequence has a common ratio.
The sequence is geometric.
8-6

41. The sequence has a common difference.
Geometric Sequences
ALGEBRA 1 LESSON 8-6
(continued)
b. 98, 101, 104, 107, . . .
+ 3
The sequence has a common difference.
The sequence is arithmetic.
8-6

42. first term: A(1) = –3(2)1 – 1 = –3(2)0 = –3(1) = –3
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

43. The first term is 2 meters, which is 200 cm.
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

44. A rule for the sequence is A(n) = 200 • 0.75n – 1.
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 • – 1
Substitute 4 for n to find the height of the third bounce.
= 200 • 0.753
Simplify exponents.
= 200 •
Evaluate powers. =
Simplify.
The height of the third bounce is cm.
8-6

45. 2. Find the next three terms of the sequence 243, 81, 27, 9, . . .
Geometric Sequences
ALGEBRA 1 LESSON 8-6
1. Find the common ratio of the geometric sequence –3, 6, –12, 24, . . .
2. Find the next three terms of the sequence 243, 81, 27, 9, . . .
3. Determine whether each sequence is arithmetic or geometric.
a. 37, 34, 31, 28, . . .
b. 8, –4, 2, –1, . . .
4. Find the first, fifth, and ninth terms of the sequence that has the rule A(n) = 4(5)n–1.
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%.) –2
3, 1, 1 3
arithmetic
geometric
4, 2500, 1,562,500
A(n) = 6(1.2)n-1; about 14.9 in.
8-6