1. Zero and Negative Exponents
Lesson 8-1
Objectives: 1. to simplify expressions with zero and negative exponents
2. to evaluate exponential expressions

2. Zero and Negative Exponents
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

3. Zero and Negative Exponents
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

4. Zero and Negative Exponents
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.

5. Zero and Negative Exponents
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.

6. Zero and Negative Exponents
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.

7. Zero and Negative Exponents
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. Objectives: 1. to write numbers in scientific and standard notation
Scientific Notation
Lesson 8-2
Objectives: 1. to write numbers in scientific and standard notation
2. to use scientific notation

9. Is each number written in scientific notation? If not, explain.
Lesson 8-2
Is each number written in scientific notation? If not, explain.
a  104
No; is less than 1.
b  10–2
yes
c  106
No; is greater than 10.

10. Write each number in scientific notation.
Lesson 8-2
Write each number in scientific notation.
a. 234,000,000
Move the decimal point 8 places to the left and use 8 as an exponent.
234,000,000 =
Drop the zeros after the 4.
2.34  108 b
Move the decimal point 5 places to the right and use –5 as an exponent. =
6.3  10–5
Drop the zeros before the 6.

11. Write each number in standard notation.
Scientific Notation
Lesson 8-2
Write each number in standard notation.
a. elephant’s mass: 8.8  104 kg
8.8  104 =
A positive exponent indicates a number greater than 10. Move the decimal point 4 places to the right.
= 88,000
b. ant’s mass: 7.3  10–5 kg  
7.3  10–5 =
A negative exponent indicates a number between 0 and 1. Move the decimal point 5 places to the left. =

12. Scientific Notation
Lesson 8-2
List the planets in order from least to greatest distance from the sun.
Jupiter
Earth
Neptune
Mercury
4.84  108 mi
9.3  107 mi
4.5  109 mi
3.8  107 mi
Planet
Distance from
the Sun
Order the powers of 10. Arrange the decimals with the same power of 10 in order.

13. Mercury Earth Jupiter Neptune
Scientific Notation
Lesson 8-2
(continued)
Jupiter
Earth
Neptune
Mercury
4.84  108 mi
9.3  107 mi
4.5  109 mi
3.8  107 mi
Planet
Distance from
the Sun
3.8     109
Mercury Earth Jupiter Neptune
From least to greatest distance from the sun, the order of the planets is Mercury, Earth, Jupiter, and Neptune.

14. Write each number in scientific notation.
Lesson 8-2
Order  105, 6.03  104, 6103, and 63.1  103 from least to greatest.
Write each number in scientific notation.
   103
6.3     104
Order the powers of 10. Arrange the decimals with the same power of 10 in order.
6.3     104
Write the original numbers in order.
   103

15. Simplify. Write each answer using scientific notation.
Lesson 8-2
Simplify. Write each answer using scientific notation.
Use the Associative Property of Multiplication.
(6 • 8)  10–4 = a.
6(8  10–4)
= 48  10–4
Simplify inside the parentheses.
= 4.8  10–3
Write the product in scientific notation.
Use the Associative Property of Multiplication.
(0.3 • 1.3)  103 = b.
0.3(1.3  103)
= 0.39  103
Simplify inside the parentheses.
= 3.9  102
Write the product in scientific notation.

16. Multiplication Properties of Exponents
Lesson 8-3
Objectives: 1. to multiply powers, with the same base
2. to work with scientific notation

17. Multiplication Properties of Exponents
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

18. Multiplication Properties of Exponents
Lesson 8-3
Simplify each expression. a.
p2 • p • p5
Add exponents of powers
with the same base. p =
= p 8
Simplify.
2q • 3p3 • 4q4 b.
Commutative and
Associative Properties of
Multiplication
(2 • 3 • 4)(p3)(q • q 4) =
Multiply the coefficients.
Write q as q1.
= 24(p3)(q1• q 4)
Add exponents of powers
with the same base.
= 24(p3)(q1 + q 4)
Simplify.
= 24p3q5

19. Multiplication Properties of Exponents
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.

20. Multiplication Properties of Exponents
Lesson 8-3
The speed of light is 3  108 m/s. If there are 1  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.

21. Multiplication Properties of Exponents
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.

22. More Multiplication Properties of Exponents
Lesson 8-4
Objectives: 1. to raise a power to a power
2. to raise a product to a power

23. More Multiplication Properties of Exponents
Lesson 8-4
Simplify (a3)4.
Multiply exponents when raising a power to a power.
(a3)4 = a3 • 4
Simplify.
= a12

24. More Multiplication Properties of Exponents
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.

25. More Multiplication Properties of Exponents
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.

26. More Multiplication Properties of Exponents
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.

27. More Multiplication Properties of Exponents
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.

28. Division Properties of Exponents
Lesson 8-5
Objectives: 1. to divide powers with the same base
2. to raise a quotient to a power

29. Division Properties of Exponents
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 =

30. Division Properties of Exponents
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.

31. Division Properties of Exponents
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.

32. Division Properties of Exponents
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 =

33. Division Properties of Exponents
Lesson 8-5
(continued) –2 . 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 =

34. Objectives: 1. to form geometric sequences
Lesson 8-6
Objectives: 1. to form geometric sequences
2. to use equations/rules when describing geometric sequences

35. (–5) (–5) (–5)  Find the common ratio of each sequence.
Geometric Sequences
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

36. Find the next three terms of the sequence 5, –10, 20, –40, . . .
Geometric Sequences
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.

37.  Determine whether each sequence is arithmetic or geometric.
Geometric Sequences
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.

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

39. first term: A(1) = –3(2)1 – 1 = –3(2)0 = –3(1) = –3
Geometric Sequences
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

40. The first term is 2 meters, which is 200 cm.
Geometric Sequences
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.

41. A rule for the sequence is A(n) = 200 • 0.75n – 1.
Geometric Sequences
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.

42. Exponential Functions
Lesson 8-7
Objectives: 1. to evaluate exponential functions
2. to graph exponential functions

43. Exponential Functions
Lesson 8-7
Evaluate each exponential function.
a. y = 3x for x = 2, 3, 4
x y = 3x y
2 32 = 9 9
3 33 = 27 27
Edit text. See page 130, Example 1, part b. PHM3579_AES_Ch 08_MQ
4 34 = 81 81
b. p(q) = 3 • 4q for the domain {–2, 3}
q p(q) = 3 • 4q p(q)
–2 3 • 4–2 = 3 • = 1 16 3
3 3 • 43 = 3 • 64 =

44. Exponential Functions
Lesson 8-7
Suppose two mice live in a barn. If the number of mice quadruples every 3 months, how many mice will be in the barn after 2 years?
ƒ(x) = 2 • 4x
ƒ(x) = 2 • 48
In two years, there are 8 three-month time periods.
ƒ(x) = 2 • 65,536
Simplify powers.
ƒ(x) = 131,072
Simplify.

45. Exponential Functions
Lesson 8-7
Graph y = 2 • 3x.
2 2 • 32 = 2 • 9 = 18 (2, 18)
x y = 2 • 3x (x, y)
–2 2 • 3–2 = = (–2, ) 2
3 2 9
–1 2 • 3–1 = = (–1, ) 2 31 3
0 2 • 30 = 2 • 1 = 2 (0, 2)
1 2 • 31 = 2 • 3 = 6 (1, 6)

46. Exponential Functions
Lesson 8-7
The function ƒ(x) = 1.25x models the increase in size of an image being copied over and over at 125% on a photocopier. Graph the function.
x ƒ(x) = 1.25x (x, ƒ(x))
= (1, 1.3)
= (2, 1.6)
= (5, 3.1)
= (4, 2.4)
= (3, 2.0)

47. Exponential Growth and Decay
Lesson 8-8
Objectives: 1. to model exponential growth
2. to model exponential decay

48. Exponential Growth and Decay
Lesson 8-8
In 1998, a certain town had a population of about 13,000 people. Since 1998, the population has increased about 1.4% a year.
a. Write an equation to model the population increase.
Relate:  y = a • bx
Use an exponential function.
Define:
Let x = the number of years since 1998.
Let y = the population of the town at various times.
Let a = the initial population in 1998, 13,000 people.
Let b = the growth factor, which is
100% + 1.4% = 101.4% =
Write: y = 13,000 • 1.014x

49. Exponential Growth and Decay
Lesson 8-8
(continued)
b. Use your equation to find the approximate population in 2006.
y = 13,000 • 1.014x
y = 13,000 •
2006 is 8 years after 1998, so substitute 8 for x.
Use a calculator. Round to the nearest
whole number.
14,529
The approximate population of the town in 2006 is 14,529 people.

50. Exponential Growth and Decay
Lesson 8-8
Suppose you deposit $1000 in a college fund that pays 7.2% interest compounded annually. Find the account balance after 5 years.
Relate:  y = a • bx
Use an exponential function.
Define:
Let x = the number of interest periods.
Let y = the balance.
Let a = the initial deposit, $1000
Let b = 100% + 7.2% = 107.2% =
Write: y = 1000 • 1.072x
= 1000 •
Once a year for 5 years is 5 interest periods. Substitute 5 for x.
Use a calculator. Round to the nearest cent.
The balance after 5 years will be $

51. Exponential Growth and Decay
Lesson 8-8
Suppose the account in the above problem paid interest compounded quarterly instead of annually. Find the account balance after 5 years.
Relate:  y = a • bx
Use an exponential function.
Define:
Let x = the number of interest periods.
Let y = the balance.
Let a = the initial deposit, $1000
Let b = 100% + There are 4 interest periods in
1 year, so divide the interest into 4 parts.
= = 1.018
7.2% 4

52. Exponential Growth and Decay
Lesson 8-8
(continued)
Write: y = 1000 • 1.018x
= 1000 •
Four interest periods a year for 5 years is 20 interest periods. Substitute 20 for x.
Use a calculator.
Round to the nearest cent.
The balance after 5 years will be $

53. Exponential Growth and Decay
Lesson 8-8
Technetium-99 has a half-life of 6 hours. Suppose a lab has 80 mg of technetium-99. How much technetium-99 is left after 24 hours?
In 24 hours there are four 6-hour half lives.
After one half-life, there are 40 mg.
After two half-lives, there are 20 mg.
After three half-lives, there are 10 mg.
After four half-lives, there are 5 mg.

54. Exponential Growth and Decay
Lesson 8-8
Suppose the population of a certain endangered species has decreased 2.4% each year. Suppose there were 60 of these animals in a given area in 1999.
a. Write an equation to model the number of animals in this species that remain alive in that area.
Relate:  y = a • bx
Use an exponential function.
Define:
Let x = the number of years since 1999
Let y = the number of animals that remain
Let a = 60, the initial population in 1999
Let b = the decay factor, which is 100% % = 97.6% = 0.976
Write: y = 60 • 0.976x

55. Exponential Growth and Decay
Lesson 8-8
(continued)
b. Use your equation to find the approximate number of animals remaining in 2005.
y = 60 • 0.976x
y = 60 •
2005 is 6 years after 1999, so substitute 6 for x. 52
Use a calculator. Round to the nearest whole number.
The approximate number of animals of this endangered species remaining in the area in 2005 is 52.