algebra 1 lesson 8-1 zero and negative exponents objectives: 1. to simplify expressions with zero...

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


=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


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


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


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


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



Scientific NotationScientific Notation

Objectives: 1. to write numbers in scientific and standard notation

2. to use scientific notation


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.


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 =


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


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


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


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


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.


Use the Associative Property of

Multiplication.(6 • 8) 10–4=

Use the Associative Property of

Multiplication.(0.3 • 1.3) 103=


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



Use the definition of zero as an exponent.= 1

a.

b.

68 • 6–8c.


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)=


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.


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)


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


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



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


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


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


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


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



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


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


Algebra 1Algebra 1

Simplify . 3 y 3

4

34

y 12= Multiply the exponent in the denominator.

81y 12= Simplify.

Lesson 8-5


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


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


4bc

–

4bc

––2 c

4b–

2= Rewrite using the reciprocal of .4b

c–

Write the fraction with a negative numerator.

c 4b

– 2=

–2

.


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


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


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.


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


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


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


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



Exponential FunctionsExponential Functions

Objectives: 1. to evaluate exponential functions

2. to graph exponential functions


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


–2 3 • 4–2 = 3 • = 1

163

16 316


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.


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


–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 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


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


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.


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



Suppose the account in the above problem paid interest

compounded quarterly instead of annually. Find the account balance

after 5 years.


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



(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.



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.


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.


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


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


(continued)

algebra 1 lesson 8-1 zero and negative exponents objectives: 1. to simplify expressions with zero...

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