algebra 1 lesson 8-1 zero and negative exponents objectives: 1. to simplify expressions with zero...
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Algebra 1Algebra 1Lesson 8-1
Zero and Negative ExponentsZero and Negative Exponents
Objectives: 1. to simplify expressions with zero and negative exponents
2. to evaluate exponential expressions
Algebra 1Algebra 1
Simplify.
a. 3–2
Simplify.19=
Lesson 8-1
(–22.4)0b. Use the definition of zero as an exponent.
= 1
Zero and Negative ExponentsZero and Negative Exponents
=Use the definition of negative exponent.
1 32
Algebra 1Algebra 1
Simplify
a. b.
Simplify.3ab 2=
= 1 1x 3
Use the definition of negative exponent.
= x 3
Identity Property of Multiplication
Lesson 8-1
= 1 • x 3
Multiply by the reciprocal of , which is x 3.
1 x3
Rewrite using a division symbol.
= 1 x –3
Zero and Negative ExponentsZero and Negative Exponents
3ab –2 1 b2
Use the definition of negative exponent.
= 3a 1 x –3
Algebra 1Algebra 1
Evaluate 4x 2y –3 for x = 3 and y = –2.
Method 1: Write with positive exponents first.
Substitute 3 for x and –2 for y.4(3)2
(–2)3=
36–8 –4
12= = Simplify.
Lesson 8-1
Zero and Negative ExponentsZero and Negative Exponents
4x 2y –3 = Use the definition of negative exponent.4x 2
y 3
Algebra 1Algebra 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
12= = Simplify.
Lesson 8-1
Zero and Negative ExponentsZero and Negative Exponents
Algebra 1Algebra 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.
3000 • 2m = 3000 • 20 Substitute 0 for m.
= 3000 • 1 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.
Lesson 8-1
Zero and Negative ExponentsZero and Negative Exponents
a. Evaluate the expression for m = 0.
Algebra 1Algebra 1
(continued)
b. Evaluate the expression for m = –2.
3000 • 2m = 3000 • 2–2 Substitute –2 for m.
= 3000 • Simplify.
= 750
14
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.
Lesson 8-1
Zero and Negative ExponentsZero and Negative Exponents
Algebra 1Algebra 1Lesson 8-2
Scientific NotationScientific Notation
Objectives: 1. to write numbers in scientific and standard notation
2. to use scientific notation
Algebra 1Algebra 1Lesson 8-2
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 NotationScientific Notation
Algebra 1Algebra 1
Write each number in scientific notation.
a. 234,000,000
Drop the zeros after the 4.
2.34 108
b. 0.0000636.3 10–5
Drop the zeros before the 6.
Lesson 8-2
Move the decimal point 8 places to the left and use 8 as an exponent. 234,000,000 =
Move the decimal point 5 places to the right and use –5 as an exponent.0.000063 =
Scientific NotationScientific Notation
Algebra 1Algebra 1
Write each number in standard notation.
a. elephant’s mass: 8.8 104 kg
= 88,000
b. ant’s mass: 7.3 10–5 kg
= 0.000073
Lesson 8-2
8.8 104 = A positive exponent indicates a number greater than 10. Move the decimal point 4 places to the right.
8.8000.
7.3 10–5 = A negative exponent indicates a number between 0 and 1. Move the decimal point 5 places to the left.
0.00007.3
Scientific NotationScientific Notation
Algebra 1Algebra 1
Jupiter
Earth
Neptune
Mercury
4.84 108 mi
9.3 107 mi
4.5 109 mi
3.8 107 mi
PlanetDistance from
the Sun
List the planets in order from least to greatest distance from
the sun.
Order the powers of 10. Arrange the decimals with the same power of 10 in order.
Lesson 8-2
Scientific NotationScientific Notation
Algebra 1Algebra 1
(continued)
3.8 107 9.3 107 4.84 108 4.5 109
Mercury Earth Jupiter Neptune
From least to greatest distance from the sun, the order of the planets is
Mercury, Earth, Jupiter, and Neptune.
Lesson 8-2
Scientific NotationScientific Notation
Jupiter
Earth
Neptune
Mercury
4.84 108 mi
9.3 107 mi
4.5 109 mi
3.8 107 mi
PlanetDistance from
the Sun
Algebra 1Algebra 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 6.03 104 6103 63.1 103
6.3 102 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
Lesson 8-2
Scientific NotationScientific Notation
Algebra 1Algebra 1
Simplify. Write each answer using scientific notation.
Lesson 8-2
6(8 10–4)a.
= 48 10–4 Simplify inside the parentheses.
= 4.8 10–3 Write the product in scientific notation.
0.3(1.3 103)b.
= 0.39 103 Simplify inside the parentheses.
= 3.9 102 Write the product in scientific notation.
Scientific NotationScientific Notation
Use the Associative Property of
Multiplication.(6 • 8) 10–4=
Use the Associative Property of
Multiplication.(0.3 • 1.3) 103=
Algebra 1Algebra 1Lesson 8-3
Multiplication Properties of ExponentsMultiplication Properties of Exponents
Objectives: 1. to multiply powers, with the same base
2. to work with scientific notation
Algebra 1Algebra 1
Rewrite each expression using each base only once.
Lesson 8-3
73 • 72
= 75 Simplify the sum of the exponents.
44 • 41 • 4–2
= 43 Simplify the sum of the exponents.
= 60 Simplify the sum of the exponents.
Use the definition of zero as an exponent.= 1
a.
b.
68 • 6–8c.
Multiplication Properties of ExponentsMultiplication Properties of Exponents
Add exponents of powers with thesame base.
73 + 2=
Think of 4 + 1 – 2 as 4 + 1 + (–2) to add the exponents.
44 + 1 – 2=
Add exponents of powers with thesame base.68 + (–8)=
Algebra 1Algebra 1Lesson 8-3
Simplify each expression.
p2 • p • p5a.
= p 8 Simplify.
Multiply the coefficients. Write q as q1.
= 24(p3)(q1• q 4)
Simplify. = 24p3q5
2q • 3p3 • 4q4 b.
Multiplication Properties of ExponentsMultiplication Properties of Exponents
Commutative and Associative Properties of Multiplication
(2 • 3 • 4)(p3)(q • q 4) =
Add exponents of powers with the same base.
p 2 + 1 + 5=
Add exponents of powers with the same base.
= 24(p3)(q1 + q 4)
Algebra 1Algebra 1
Simplify (3 10–3)(7 10–5). Write the answer in
scientific notation.
= 21 10–8 Simplify.
= 2.1 101 • 10–8 Write 21 in scientific notation.
= 2.1 101 + (– 8)Add exponents ofpowers with the samebase.
= 2.1 10–7 Simplify.
Lesson 8-3
(3 10–3)(7 10–5) =Commutative and Associative Properties of Multiplication
(3 • 7)(10–3 • 10–5)
Multiplication Properties of ExponentsMultiplication Properties of Exponents
Algebra 1Algebra 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 seconds
kilometersmeters
secondshour
• • Use dimensional analysis.
= (3 108) • (1 10–3) • (3.6 103) ms
kmm
sh
Substitute.
= (3 • 1 • 3.6) (108 • 10–3 • 103) Commutative and Associative Properties of Multiplication
= 10.8 (108 + (– 3) + 3) Simplify.
Lesson 8-3
Multiplication Properties of ExponentsMultiplication Properties of Exponents
Algebra 1Algebra 1
= 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.
(continued)
Lesson 8-3
Multiplication Properties of ExponentsMultiplication Properties of Exponents
Algebra 1Algebra 1Lesson 8-4
More Multiplication Properties of ExponentsMore Multiplication Properties of Exponents
Objectives: 1. to raise a power to a power
2. to raise a product to a power
Algebra 1Algebra 1
Simplify (a3)4.
Multiply exponents when raising a power to a power.
(a3)4 = a3 • 4
Simplify.= a12
Lesson 8-4
More Multiplication Properties of ExponentsMore Multiplication Properties of Exponents
Algebra 1Algebra 1
Simplify b2(b3)–2.
b2(b3)–2 = b2 • b3 • (–2) Multiply exponents in (b3)–2.
= b2 + (–6) Add exponents when multiplying powers of the same base.
Simplify. = b–4
= b2 • b–6 Simplify.
1 b4= Write using only positive exponents.
Lesson 8-4
More Multiplication Properties of ExponentsMore Multiplication Properties of Exponents
Algebra 1Algebra 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.
Lesson 8-4
More Multiplication Properties of ExponentsMore Multiplication Properties of Exponents
Algebra 1Algebra 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.
Lesson 8-4
More Multiplication Properties of ExponentsMore Multiplication Properties of Exponents
Algebra 1Algebra 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)2Raise 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 1018Simplify.Write in scientific notation.
Lesson 8-4
More Multiplication Properties of ExponentsMore Multiplication Properties of Exponents
Algebra 1Algebra 1Lesson 8-5
Division Properties of ExponentsDivision Properties of Exponents
Objectives: 1. to divide powers with the same base
2. to raise a quotient to a power
Algebra 1Algebra 1
Simplify each expression.
Lesson 8-5
x4
x9a.
Simplify the exponents.= x–5
Rewrite using positive exponents. 1 x5=
p3 j –4
p–3 j 6
= p6 j –10 Simplify.
Rewrite using positive exponents.p6
j10=
b.
=Subtract exponents when dividing powers with the same base.
x4 – 9
=Subtract exponents when dividing powers with the same base.
p3 – (–3)j –4 – 6
Division Properties of ExponentsDivision Properties of Exponents
Algebra 1Algebra 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?
64 million beats 530 thousand min
6.4 107 beats
5.3 105 min= Write in scientific notation.
6.45.3 107–5=
Subtract exponents when dividing powers with the same base.
6.45.3 102= 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.
Lesson 8-5
Division Properties of ExponentsDivision Properties of Exponents
Algebra 1Algebra 1
Simplify . 3 y 3
4
34
y 12= Multiply the exponent in the denominator.
81y 12= Simplify.
Lesson 8-5
Division Properties of ExponentsDivision Properties of Exponents
3 y 3
4 34
(y 3)4=
Raise the numerator and the denominator to the fourth power.
Algebra 1Algebra 1
a. Simplify .23
–3
33
23=Raise the numerator and the denominator to the third power.
Simplify.278 3
38or=
Lesson 8-5
Division Properties of ExponentsDivision Properties of Exponents
23
–3 32
3= Rewrite using the reciprocal of .
23
Algebra 1Algebra 1
(continued)
b. Simplify
Raise the numerator and denominator to the second power.
(–c)2
(4b)2=
Simplify.c2
16b2=
Lesson 8-5
Division Properties of ExponentsDivision Properties of Exponents
4bc
–
4bc
––2 c
4b–
2= Rewrite using the reciprocal of .4b
c–
Write the fraction with a negative numerator.
c 4b
– 2=
–2
.
Algebra 1Algebra 1Lesson 8-6
Geometric SequencesGeometric Sequences
Objectives: 1. to form geometric sequences
2. to use equations/rules when describing geometric sequences
Algebra 1Algebra 1
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. 3,32
34
38, , , ...
332
34
38
12 1
2 12
The common ratio is .12
Lesson 8-6
Geometric SequencesGeometric Sequences
Algebra 1Algebra 1
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.
Lesson 8-6
Geometric SequencesGeometric Sequences
Algebra 1Algebra 1
Determine whether each sequence is arithmeticor geometric.
a. 162, 54, 18, 6, . . .
62 54 18 6
13 1
3 13
The sequence has a common ratio.
Lesson 8-6
The sequence is geometric.
Geometric SequencesGeometric Sequences
Algebra 1Algebra 1
(continued)
b. 98, 101, 104, 107, . . .
The sequence has a common difference.
98 101 104 107
+ 3 + 3 + 3
The sequence is arithmetic.
Lesson 8-6
Geometric SequencesGeometric Sequences
Algebra 1Algebra 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
Lesson 8-6
Geometric SequencesGeometric Sequences
Algebra 1Algebra 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.
Lesson 8-6
Geometric SequencesGeometric Sequences
Algebra 1Algebra 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.
Lesson 8-6
Geometric SequencesGeometric Sequences
Algebra 1Algebra 1Lesson 8-7
Exponential FunctionsExponential Functions
Objectives: 1. to evaluate exponential functions
2. to graph exponential functions
Algebra 1Algebra 1Lesson 8-7
Evaluate each exponential function.
a. y = 3x for x = 2, 3, 4
x y = 3x y2 32 = 9 93 33 = 27 274 34 = 81 81
b. p(q) = 3 • 4q for the domain {–2, 3}
q p(q) = 3 • 4q p(q)
3 3 • 43 = 3 • 64 = 192 192
Exponential FunctionsExponential Functions
–2 3 • 4–2 = 3 • = 1
163
16 316
Algebra 1Algebra 1Lesson 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.
Exponential FunctionsExponential Functions
Algebra 1Algebra 1
Graph y = 2 • 3x.
x y = 2 • 3x (x, y)
–1 2 • 3–1 = = (–1, ) 2 31
23
23
0 2 • 30 = 2 • 1 = 2 (0, 2)
1 2 • 31 = 2 • 3 = 6 (1, 6)
2 2 • 32 = 2 • 9 = 18 (2, 18)
Lesson 8-7
Exponential FunctionsExponential Functions
–2 2 • 3–2 = = (–2, ) 2 3 2
29
29
Algebra 1Algebra 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 1.3 (1, 1.3)
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)
Lesson 8-7
Exponential FunctionsExponential Functions
Algebra 1Algebra 1Lesson 8-8
Exponential Growth and DecayExponential Growth and Decay
Objectives: 1. to model exponential growth
2. to model exponential decay
Algebra 1Algebra 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 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 is100% + 1.4% = 101.4% = 1.014.
Write: y = 13,000 • 1.014x
Lesson 8-8
Exponential Growth and DecayExponential Growth and Decay
Algebra 1Algebra 1
(continued)
b. Use your equation to find the approximate population in 2006.
y = 13,000 • 1.014x
y = 13,000 • 1.0148 2006 is 8 years after 1998, so substitute 8 for x.
The approximate population of the town in 2006 is 14,529 people.
Use a calculator. Round to the nearestwhole number.
14,529
Lesson 8-8
Exponential Growth and DecayExponential Growth and Decay
Algebra 1Algebra 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 Use an exponential function.
Define: Let x = the number of interest periods.Let y = the balance.Let a = the initial deposit, $1000Let b = 100% + 7.2% = 107.2% = 1.072.
Write: y = 1000 • 1.072x
= 1000 • 1.0725 Once a year for 5 years is 5 interest periods. Substitute 5 for x.
The balance after 5 years will be $1415.71.
Use a calculator. Round to the nearest cent.
1415.71
Lesson 8-8
Exponential Growth and DecayExponential Growth and Decay
Algebra 1Algebra 1Lesson 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, $1000Let b = 100% + There are 4 interest periods in
1 year, so divide the interest into 4 parts.
= 1 + 0.018 = 1.018
7.2%4
Exponential Growth and DecayExponential Growth and Decay
Algebra 1Algebra 1Lesson 8-8
(continued)
Write: y = 1000 • 1.018x
= 1000 • 1.01820
Four interest periods a year for 5 years is 20 interest periods. Substitute 20 for x.
1428.75Use a calculator.Round to the nearest cent.
The balance after 5 years will be $1428.75.
Exponential Growth and DecayExponential Growth and Decay
Algebra 1Algebra 1Lesson 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.
Exponential Growth and DecayExponential Growth and Decay
Algebra 1Algebra 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 Use an exponential function.
Define:Let x = the number of years since 1999Let y = the number of animals that remainLet a = 60, the initial population in 1999Let b = the decay factor, which is 100% - 2.4 % = 97.6% = 0.976
Write: y = 60 • 0.976x
Lesson 8-8
Exponential Growth and DecayExponential Growth and Decay
Algebra 1Algebra 1
b. Use your equation to find the approximate number of animals remaining in 2005.
y = 60 • 0.976x
y = 60 • 0.9766 2005 is 6 years after 1999, so substitute 6 for x.
52Use 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.
Lesson 8-8
Exponential Growth and DecayExponential Growth and Decay
(continued)