Download Algebra Ch 8, exponents

Zero and Negative Exponents
Lesson 8-1
Algebra 1
Objectives: 1. to simplify expressions with zero and negative exponents
2. to evaluate exponential expressions
Zero and Negative Exponents
Lesson 8-1
Algebra 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.
Zero and Negative Exponents
Lesson 8-1
Algebra 1
Simplify
a. 3ab –2 = 3a
1
b2
Use the definition of negative
exponent.
=
3a
b2
Simplify.
b.
1
–3
=
1

x
–3
x
Rewrite using a division symbol.
1
=1 3
x
Use the definition of negative
exponent.
= 1 • x3
1
Multiply by the reciprocal of x3 ,
which is x 3.
= x3
Identity Property of Multiplication
Zero and Negative Exponents
Lesson 8-1
Algebra 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.
Zero and Negative Exponents
Lesson 8-1
Algebra 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.
Zero and Negative Exponents
Lesson 8-1
Algebra 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.
Zero and Negative Exponents
Lesson 8-1
Algebra 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.
Scientific Notation
Lesson 8-2
Algebra 1
Objectives: 1. to write numbers in scientific and standard notation
2. to use scientific notation
Scientific Notation
Lesson 8-2
Algebra 1
Is each number written in scientific notation? If not, explain.
a. 0.46  104
No;
0.46 is less than 1.
b. 3.25  10–2
yes
c. 13.2  106
No;
13.2 is greater than 10.
Scientific Notation
Lesson 8-2
Algebra 1
Write each number in scientific notation.
a. 234,000,000
234,000,000 = 2.34  108
Move the decimal point 8 places to the
left and use 8 as an exponent.
Drop the zeros after the 4.
b. 0.000063
0.000063 =
6.3  10–5
Move the decimal point 5 places to the
right and use –5 as an exponent.
Drop the zeros before the 6.
Scientific Notation
Lesson 8-2
Algebra 1
Write each number in standard notation.
a. elephant’s mass: 8.8  104 kg
8.8  104 = 8.8000.
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 = 0.00007.3
= 0.000073
A negative exponent indicates a
number between 0 and 1. Move the
decimal point 5 places to the left.
Scientific Notation
Lesson 8-2
Algebra 1
List the planets in order from least to greatest distance from
the sun.
Planet
Distance from
the Sun
Jupiter
4.84  108 mi
Earth
9.3  107 mi
Neptune
4.5  109 mi
Mercury
3.8  107 mi
Order the powers of 10. Arrange the decimals with the same power of 10
in order.
Scientific Notation
Lesson 8-2
Algebra 1
(continued)
3.8  107
Mercury
Planet
Distance from
the Sun
Jupiter
4.84  108 mi
Earth
9.3  107 mi
Neptune
4.5  109 mi
Mercury
3.8  107 mi
9.3  107
Earth
4.84  108
Jupiter
4.5  109
Neptune
From least to greatest distance from the sun, the order of the planets is
Mercury, Earth, Jupiter, and Neptune.
Scientific Notation
Lesson 8-2
Algebra 1
Order 0.0063  105, 6.03  104, 6103, and 63.1  103 from
least to greatest.
Write each number in scientific notation.
0.0063  105
6.3  102
6.03  104
63.1  103
6103
6.03  104
6.103  103
6.31  104
Order the powers of 10. Arrange the decimals with the same power of 10
in order.
6.3  102
6.103  103
6.03  104
6.31  104
Write the original numbers in order.
0.0063  105
6103
6.03  104
63.1  103
Scientific Notation
Lesson 8-2
Algebra 1
Simplify. Write each answer using scientific notation.
a. 6(8  10–4) = (6 • 8)  10–4
b.
Use the Associative Property of
Multiplication.
= 48  10–4
Simplify inside the parentheses.
= 4.8  10–3
Write the product in scientific notation.
0.3(1.3  103) = (0.3 • 1.3)  103
Use the Associative Property of
Multiplication.
= 0.39  103
Simplify inside the parentheses.
= 3.9  102
Write the product in scientific notation.
Multiplication Properties of Exponents
Lesson 8-3
Algebra 1
Objectives: 1. to multiply powers, with the same base
2. to work with scientific notation
Multiplication Properties of Exponents
Lesson 8-3
Algebra 1
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)
Multiplication Properties of Exponents
Lesson 8-3
Algebra 1
Simplify each expression.
a.
p2 • p • p5
= p2+1+5
Add exponents of powers
with the same base.
= p8
Simplify.
b. 2q • 3p3 • 4q4 = (2 • 3 • 4)(p3)(q • q 4)
= 24(p3)(q1• q 4)
= 24(p3)(q1 + q 4)
= 24p3q5
Commutative and
Associative Properties of
Multiplication
Multiply the coefficients.
Write q as q1.
Add exponents of powers
with the same base.
Simplify.
Multiplication Properties of Exponents
Lesson 8-3
Algebra 1
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)
= 21  10–8
= 2.1  101 • 10–8
= 2.1 
101 + (– 8)
= 2.1  10–7
Commutative and
Associative Properties of
Multiplication
Simplify.
Write 21 in scientific
notation.
Add exponents of
powers with the same
base.
Simplify.
Multiplication Properties of Exponents
Lesson 8-3
Algebra 1
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
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.
Multiplication Properties of Exponents
Lesson 8-3
Algebra 1
(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.
More Multiplication Properties of Exponents
Lesson 8-4
Algebra 1
Objectives: 1. to raise a power to a power
2. to raise a product to a power
More Multiplication Properties of Exponents
Lesson 8-4
Algebra 1
Simplify (a3)4.
(a3)4 = a3 • 4
= a12
Multiply exponents when raising a
power to a power.
Simplify.
More Multiplication Properties of Exponents
Lesson 8-4
Algebra 1
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.
More Multiplication Properties of Exponents
Lesson 8-4
Algebra 1
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.
More Multiplication Properties of Exponents
Lesson 8-4
Algebra 1
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.
More Multiplication Properties of Exponents
Lesson 8-4
Algebra 1
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.
Division Properties of Exponents
Lesson 8-5
Algebra 1
Objectives: 1. to divide powers with the same base
2. to raise a quotient to a power
Division Properties of Exponents
Lesson 8-5
Algebra 1
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.
Division Properties of Exponents
Lesson 8-5
Algebra 1
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.
Division Properties of Exponents
Lesson 8-5
Algebra 1
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.
Division Properties of Exponents
Lesson 8-5
Algebra 1
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.
Division Properties of Exponents
Lesson 8-5
Algebra 1
(continued)
–2
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.
Geometric Sequences
Lesson 8-6
Algebra 1
Objectives: 1. to form geometric sequences
2. to use equations/rules when describing geometric sequences
Geometric Sequences
Lesson 8-6
Algebra 1
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
The common ratio is 2 .
1
2
Geometric Sequences
Lesson 8-6
Algebra 1
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.
Geometric Sequences
Lesson 8-6
Algebra 1
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.
Geometric Sequences
Lesson 8-6
Algebra 1
(continued)
b. 98, 101, 104, 107, . . .
98
101
+3
104
+3
107
+3
The sequence has a common difference.
The sequence is arithmetic.
Geometric Sequences
Lesson 8-6
Algebra 1
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
Geometric Sequences
Lesson 8-6
Algebra 1
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.
Geometric Sequences
Lesson 8-6
Algebra 1
(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.
Exponential Functions
Lesson 8-7
Algebra 1
Objectives: 1. to evaluate exponential functions
2. to graph exponential functions
Exponential Functions
Lesson 8-7
Algebra 1
Evaluate each exponential function.
a. y = 3x for x = 2, 3, 4
x
2
3
4
y = 3x
32 = 9
33 = 27
34 = 81
y
9
27
81
b. p(q) = 3 • 4q for the domain {–2, 3}
q
p(q) = 3 • 4q
1
16
p(q)
3
16
3
16
–2
3 • 4–2 = 3 •
3
3 • 43 = 3 • 64 = 192 192
=
Exponential Functions
Lesson 8-7
Algebra 1
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.
Exponential Functions
Lesson 8-7
Algebra 1
Graph y = 2 • 3x.
x
y = 2 • 3x
–2
2 • 3–2 = 3 2 =
–1
2 • 3–1 =
(x, y)
2
2
31
=
2
9
(–2, 29 )
2
3
(–1, 3 )
2
0
2 • 30 = 2 • 1 = 2
(0, 2)
1
2 • 31 = 2 • 3 = 6
(1, 6)
2
2 • 32 = 2 • 9 = 18 (2, 18)
Exponential Functions
Lesson 8-7
Algebra 1
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
1.251 = 1.25
2
1.252 = 1.5625
1.6
(2, 1.6)
3
1.253 = 1.9531
2.0
(3, 2.0)
4
1.254 = 2.4414
2.4
(4, 2.4)
5
1.255 = 3.0518
3.1
(5, 3.1)
1.3
(1, 1.3)
Exponential Growth and Decay
Lesson 8-8
Algebra 1
Objectives: 1. to model exponential growth
2. to model exponential decay
Exponential Growth and Decay
Lesson 8-8
Algebra 1
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
Define:
Write:
Use an exponential function.
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% = 1.014.
y = 13,000 • 1.014x
Exponential Growth and Decay
Lesson 8-8
Algebra 1
(continued)
b.
Use your equation to find the approximate population in 2006.
y = 13,000 • 1.014x
y = 13,000 • 1.0148
14,529
2006 is 8 years after 1998, so
substitute 8 for x.
Use a calculator. Round to the nearest
whole number.
The approximate population of the town in 2006 is 14,529 people.
Exponential Growth and Decay
Lesson 8-8
Algebra 1
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
Define:
Write:
Use an exponential function.
Let x = the number of interest periods.
Let y = the balance.
Let a = the initial deposit, $1000
Let b = 100% + 7.2% = 107.2% = 1.072.
y = 1000 • 1.072x
= 1000 • 1.0725
1415.71
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 $1415.71.
Exponential Growth and Decay
Lesson 8-8
Algebra 1
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
Define:
Use an exponential function.
Let x = the number of interest periods.
Let y = the balance.
Let a = the initial deposit, $1000
Let b = 100% + 7.2%
There are 4 interest periods in
4
1 year, so divide the interest into
4 parts.
= 1 + 0.018 = 1.018
Exponential Growth and Decay
Lesson 8-8
Algebra 1
(continued)
Write:
y = 1000 • 1.018x
= 1000 • 1.01820
1428.75
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 $1428.75.
Exponential Growth and Decay
Lesson 8-8
Algebra 1
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.
Exponential Growth and Decay
Lesson 8-8
Algebra 1
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
Define:
Write:
Use an exponential function.
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% - 2.4 % = 97.6% = 0.976
y = 60 • 0.976x
Exponential Growth and Decay
Lesson 8-8
Algebra 1
(continued)
b. Use your equation to find the approximate number of animals
remaining in 2005.
y = 60 • 0.976x
y = 60 • 0.9766
52
2005 is 6 years after 1999, so
substitute 6 for x.
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.