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

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

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

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.

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.

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 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 (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.

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

Is each number written in scientific notation? If not, explain. Lesson 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.

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. 0.000063 Move the decimal point 5 places to the right and use –5 as an exponent. 0.000063 = 6.3  10–5 Drop the zeros before the 6.

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. 8.8000. = 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. 0.00007.3 = 0.000073

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.

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  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.

Write each number in scientific notation. Lesson 8-2 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

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.

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

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 4 + 1 – 2 as 4 + 1 + (–2) to add the exponents. 44 + 1 – 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

Multiplication Properties of Exponents Lesson 8-3 Simplify each expression. a. p2 • p • p5 Add exponents of powers with the same base. p 2 + 1 + 5 = = 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

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  101 + (– 8) Add exponents of powers with the same base. = 2.1  10–7 Simplify.

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.

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.

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

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

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.

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.

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.

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 • 102 + 16 Add exponents of powers with the same base. = 9  1018 Simplify. Write in scientific notation.

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

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 =

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.

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.

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 =

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 =

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

(–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

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.

 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, . . . 62 54 18 6 1 3  The sequence has a common ratio. The sequence is geometric.

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

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

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.

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 • 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 Objectives: 1. to evaluate exponential functions 2. to graph exponential functions

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 = 192 192

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.

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)

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 1.251 = 1.25 1.3 (1, 1.3) 2 1.252 = 1.5625 1.6 (2, 1.6) 5 1.255 = 3.0518 3.1 (5, 3.1) 4 1.254 = 2.4414 2.4 (4, 2.4) 3 1.253 = 1.9531 2.0 (3, 2.0)

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

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% = 1.014. Write: y = 13,000 • 1.014x

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 • 1.0148 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.

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% = 1.072. Write: y = 1000 • 1.072x = 1000 • 1.0725 Once a year for 5 years is 5 interest periods. Substitute 5 for x. Use a calculator. Round to the nearest cent. 1415.71 The balance after 5 years will be $1415.71.

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 + 0.018 = 1.018 7.2% 4

Exponential Growth and Decay Lesson 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. 1428.75 Use a calculator. Round to the nearest cent. The balance after 5 years will be $1428.75.

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.

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% - 2.4 % = 97.6% = 0.976 Write: y = 60 • 0.976x

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 • 0.9766 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.