6.3 homework answers algebra 1

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• Pearson Algebra 1 Common Core, 2015

Building Quadratic Functions from Geometric Patterns

• How many small squares will there be in each of these steps?
• Write an equation to represent the relationship between the step number, \(n\) , and the number of small squares, \(y\) , in each step.
• Explain how your equation relates to the pattern.

For access, consult one of our IM Certified Partners .

Which expression represents the relationship between the step number \(n\) and the total number of small squares in the pattern?

Each figure is composed of large squares and small squares. The side length of the large square is \(x\) . Write an expression for the area of the shaded part of each figure.

Here are a few pairs of positive numbers whose difference is 5.

Find the product of each pair of numbers. Then, plot some points to show the relationship between the first number and the product.

• Is the relationship between the first number and the product exponential? Explain how you know.

Here are some lengths and widths of a rectangle whose perimeter is 20 meters.

Complete the table. What do you notice about the areas?

• Without calculating, predict whether the area of the rectangle will be greater or less than 25 square meters if the length is 5.25 meters.

On the coordinate plane, plot the points for length and area from your table.

Do the values change in a linear way? Do they change in an exponential way?

Description: <p>A blank graph, origin O, with a grid. Horizontal axis, length, meters, scale 0 to 15, by 3’s. Vertical axis, area, square meters, scale 0 to 30, by 5’s.</p>

Here is a pattern of dots.

• Complete the table.
• How many dots will there be in Step 10?
• How many dots will there be in Step \(n\) ?

Mai has a jar of quarters and dimes. She takes at least 10 coins out of the jar and has less than \$2.00.

• Write a system of inequalities that represents the number of quarters, \(q\) , and the number of dimes, \(d\) , that Mai could have.

Is it possible that Mai has each of the following combinations of coins? If so, explain or show how you know. If not, state which constraint—the amount of money or the number of coins—it does not meet.

• 3 quarters and 12 dimes
• 4 quarters and 10 dimes
• 2 quarters and 5 dimes

A stadium can seat 63,026 people. For each game, the amount of money that the organization brings in through ticket sales is a function of the number of people, \(n\) , in attendance.

If each ticket costs \$30.00, find the domain and range of this function.

Big Ideas Math - Algebra 1, A Common Core Curriculum

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Big Ideas Math Algebra 1 Answers Chapter 6 Exponential Functions and Sequences

Exponential Functions and Sequences Chapter of Big Ideas Math Algebra 1 Answers provided helps students to learn the fundamentals associated with Exponential Functions and Sequences. Each of them is quite simple and is sequenced as per the BIM Textbook Ch 6 Exponential Functions and Sequences. In fact, our experts have covered questions belonging to Exercises, Cumulative Assessments, Chapter Tests, Review Test, Quiz, Practice Tests, etc.

Explore our library of Big Ideas Math Algebra 1 Answers Chapter 6 Exponential Functions and Sequences and become proficient in all the concepts underlying. Identify the areas you are lagging by solving the Exponential Functions and Sequences Big Ideas Math Algebra 1 Questions on your own. Improvize on the needed areas and attempt the exam with full confidence and score better grades.

Big Ideas Math Book Algebra 1 Answer Key Chapter 6 Exponential Functions and Sequences

You can improve your subject knowledge by taking the help of the BIM Algebra 1 Solution Key and stand out from the crowd. All you have to do is practice the respective concept by availing the quick links present below. Allot time to the areas you feel difficulty and solve accordingly. Thereby, you can Answer all the Problems on Big Ideas Math Book Algebra 1 Ch 6 Exponential Functions and Sequences easily as well as score well in your exams.

• Exponential Functions and Sequences Maintaining Mathematical Proficiency – Page 289 
• Exponential Functions and Sequences Mathematical Practices – Page 290
• Lesson 6.1 Properties of Exponents – Page(291-298) 
• Properties of Exponents 6.1 Exercises – Page(296-298)
• Lesson 6.2 Radicals and Rational Exponents – Page(299-304)
• Radicals and Rational Exponents 6.2 Exercises – Page(303-304)
• Lesson 6.3 Exponential Functions – Page(305-312)
• Exponential Functions 6.3 Exercises – Page(310-312)
• Lesson 6.4 Exponential Growth and Decay – Page(313-322)
• Exponential Growth and Decay 6.4 Exercises – Page(319-322)
• Exponential Functions and Sequences Study Skills: Analyzing Your Errors – Page 323
• Exponential Functions 6.1 – 6.4 Quiz – Page 324
• Lesson 6.5 Solving Exponential Functions – Page(325-330)
• Solving Exponential Functions 6.5 Exercises – Page(329-330)
• Lesson 6.6 Geometric Sequences – Page(331-338)
• Geometric Sequences 6.6 Exercises – Page(336-338)
• Lesson 6.7 Recursively Defined Sequences – Page(339-346)
• Recursively Defined Sequences 6.7 Exercises – Page(344-346)
• Exponential Functions and Sequences Performance Task: The New Car – Page 347
• Exponential Functions and Sequences Chapter Review – Page(348-350)
• Exponential Functions and Sequences Chapter Test – Page 351
• Exponential Functions and Sequences Cumulative Assessment – Page(352-353)

Exponential Functions and Sequences Maintaining Mathematical Proficiency

Evaluate the expression.

Question 1. 12(\(\frac{14}{2}\)) – 3 3 + 15 – 9² Answer: Given, 12(\(\frac{14}{2}\)) – 3 3 + 15 – 9² = 12(7) – 27 + 15 – 81 = 84 – 27 + 15 – 81 = -9

Question 2. 5 3 • 8 ÷ 2 2 + 20 • 3 – 4 Answer: Given 5 3 • 8 ÷ 2 2 + 20 • 3 – 4 = 125 . 8 ÷ 4 + 20 . 3 – 4 = 306

Question 3. -7 + 16 ÷ 2 4 + (10 – 4 2 ) Answer: Given, -7 + 16 ÷ 2 4 + (10 – 4 2 ) = -7 + (16/16) + (10 – 16) = -7 + 1 – 6 = -12

Find the square root(s).

Question 4. \(\sqrt{64}\) Answer: \(\sqrt{64}\) = \(\sqrt{8²}\) = 8

Question 5. –\(\sqrt{4}\) Answer: –\(\sqrt{4}\) = –\(\sqrt{2²}\) = -2

Question 6. –\(\sqrt{25}\) Answer: –\(\sqrt{25}\) = –\(\sqrt{5²}\) = -5

Question 7. ±\(\sqrt{21}\) Answer: ±\(\sqrt{21}\) = ±2.645

Question 8. 12, 14, 16, 18, . . . Answer: 14 – 12 = 2 So, the arithmetic sequence is 2.

Question 9. 6, 3, 0, -3, . . . Answer: Given, 6, 3, 0, -3, . . . 3 – 6 = -3 So, the arithmetic sequence is -3.

Question 10. 22, 15, 8, 1, . . . Answer: Given, 15 – 22 = -7 8 – 15 = -7 So, the arithmetic sequence is -7.

Question 11. ABSTRACT REASONING Recall that a perfect square is a number with integers as its square roots. Is the product of two perfect squares always a perfect square? Is the quotient of two perfect squares always a perfect square? Explain your reasoning. Answer: No, the quotient of two perfect squares is not always a perfect square.

Exponential Functions and Sequences Mathematical Practices

Monitoring Progress

Question 1. A rabbit population over 8 consecutive years is given by 50, 80, 128, 205, 328, 524, 839, 1342. Find the population in the tenth year. Answer: Given, A rabbit population over 8 consecutive years is given by 50, 80, 128, 205, 328, 524, 839, 1342. 80/50 = 1.6 128/80 = 1.6 205/128 ≈ 1.6 328/205 = 1.6 524/328 ≈ 1.6 839/524 ≈ 1.6 The population of a year is 60% greater that the population of the previous year. To find the population in tenth year, multiply the population of 8th year by 1.6 two times. population in ninth year = 1342 × 1.6 = 2147 population in tenth year = 2147 × 1.6 = 3435 So, the population in the tenth year is 3435.

Question 2. The sums of the numbers in the first eight rows of Pascal’s Triangle are 1, 2, 4, 8, 16, 32, 64, 128. Find the sum of the numbers in the tenth row. Answer: Given, The sums of the numbers in the first eight rows of Pascal’s Triangle are 1, 2, 4, 8, 16, 32, 64, 128. 2/1 = 2 4/2 = 2 The ninth term will be 128 × 2 = 256 The tenth term will be 256 × 2 = 512

Lesson 6.1 Properties of Exponents

Essential Question How can you write general rules involving properties of exponents?

Communicate Your Answer

Question 2. How can you write general rules involving properties of exponents? Answer: 1. In the product with equal bases the exponents are added. 2. A base with a double exponent, the exponents multiply. 3. A product raised to an exponent, each factor is raised to that exponent. 4. In the quotient with equal bases the exponents are subtracted. 5. A ratio raised to an exponent, each term is raised to that exponent. 6. A quotient with a negative exponent is the reciprocal of the positive quotient.

Answer: The small cube has side of 3 units V = s³ = 3³ = 27 The large cube is 3 times larger on each side. V = 27³ = 729 cu. units

Question 1. (-9)° Answer: Any number to the power 0 is always 1. -9° = 1

Question 2. 3 -3

Answer: 3 -3 = 1/3³ = 1/27

Simplify the expression. Write your answer using only positive exponents.

Question 5. 10 4 • 10 -6

Answer: Given, 10 4 • 10 -6 = 10 4 /10 6 = 1/10² = 1/100

Question 6. x 9 • x -9

Answer: Given, x 9 • x -9 = x 9 /x 9 = 1

Question 7. \(\frac{-5^{8}}{-5^{4}}\) Answer: Given, \(\frac{-5^{8}}{-5^{4}}\) =\(\frac{5^{8}}{5^{4}}\) = 5 4

Question 8. \(\frac{y^{6}}{y^{7}}\) Answer: Given, \(\frac{y^{6}}{y^{7}}\) = 1/y

Question 9. (6 -2 ) -1

Answer: Given, (6 -2 ) -1 = 6²

Question 10. (w 12 ) 5

Answer: Given, (w 12 ) 5 = w 60

Question 11. (10y) -3

Answer: (10y) -3 = 1/(10y)³ = 1/1000y³

Question 12. (\(-\frac{4}{n}\)) 5

Answer: (\(-\frac{4}{n}\)) 5 = -(4) 5 /(n) 5 = – 1024/n 5 = -1025n -5

Question 13. (\(\frac{1}{2 k^{2}}\)) 5

Answer: (\(\frac{1}{2 k^{2}}\)) 5 = 1/(2 5 . (k²) 5 = 1/32 . k 10

Question 14. (\(\frac{6 c}{7}\)) -2

Answer: (\(\frac{6 c}{7}\)) -2 = (7/6c)² = 7²/(6c)² = 49/36c²

Question 15. Write two expressions that represent the area of a base of the cylinder in Example 5. Answer: Area of circle with radius r is πr² r = h/2 πr² = π(h/2)² = πh²/4 By using the negative exponent property i.e, 1/a^n = a^-n we can write the above expression as 2 -2 π²h²

Question 16. It takes the Sun about 2.3 × 10 8 years to orbit the center of the Milky Way. It takes Pluto about 2.5 × 10 2 years to orbit the Sun. How many times does Pluto orbit the Sun while the Sun completes one orbit around the center of the Milky Way? Write your answer in scientific notation. Answer: Given, It takes the Sun about 2.3 × 10 8 years to orbit the center of the Milky Way. It takes Pluto about 2.5 × 10 2 years to orbit the Sun. 2.3 × 10 8 /2.5 × 10² = 0.92 × 10 6 /9.2 × 10 5

Properties of Exponents 6.1 Exercises

Vocabulary and Core Concept Check

In Exercises 5–12, evaluate the expression. (See Example 1.)

Question 8. (-2) -5

Question 12. \(\frac{(-8)^{-2}}{3^{-4}}\) Answer:

In Exercises 13–22, simplify the expression. Write your answer using only positive exponents.

Question 14. y° Answer: 1

In Exercises 23–32, simplify the expression. Write your answer using only positive exponents.

ERROR ANALYSIS In Exercises 35 and 36, describe and correct the error in simplifying the expression.

In Exercises 37–44, simplify the expression. Write your answer using only positive exponents.

Question 44. (\(\frac{1}{2 r^{6}}\)) -6

In Exercises 47–50, simplify the expression. Write your answer using only positive exponents.

In Exercises 51–54, evaluate the expression. Write your answer in scientific notation and standard form.

REWRITING EXPRESSIONS In Exercises 59–62, rewrite the expression as a power of a product.

Maintaining Mathematical Proficiency

Classify the real number in as many ways as possible.

Lesson 6.2 Radicals and Rational Exponents

Answer: a. Volume = 27 ft³ V = s³ s = \(\sqrt[3]{V}\) s = \(\sqrt[3]{27}\) s = 3 ft

b. Volume = 125 cm³ V = s³ s = \(\sqrt[3]{V}\) s = \(\sqrt[3]{125}\) s = 5 cm

c. Volume = 3375 in³ V = s³ s = \(\sqrt[3]{V}\) s = \(\sqrt[3]{3375}\) s = 15 in.

d. Volume = 3.375 m³ V = s³ s = \(\sqrt[3]{V}\) s = \(\sqrt[3]{3.375}\) s = 1.5 m

e. Volume = 1 yd³ V = s³ s = \(\sqrt[3]{V}\) s = \(\sqrt[3]{1}\) s = 1 yd

f. Volume = \(\frac{125}{8}\) mm³ V = s³ s = \(\sqrt[3]{V}\) s = \(\sqrt[3]{125/8}\) s = \(\frac{5}{2}\)

b. \(\sqrt[2]{0.5}\) \(\sqrt[2]{0.5}\) = 0.707

c. \(\sqrt[5]{2.5}\) \(\sqrt[5]{2.5}\) = 1.2

d. \(\sqrt[3]{65}\) \(\sqrt[3]{65}\) = 4.02

e. \(\sqrt[3]{55}\) \(\sqrt[3]{55}\) = 3.80

f. \(\sqrt[6]{20000}\) \(\sqrt[6]{20000}\) = 5.21

Question 3. How can you write and evaluate an nth root of a number? Answer: Let us say you wanted to find the cube root of 8. \(\sqrt[3]{8}\) = 8 1/3 \(\sqrt[n]{x}\) = x 1/n

Question 4. The body mass m (in kilograms) of a dinosaur that walked on two feet can be modeled by m = (0.00016)C 2.73 where C is the circumference (in millimeters) of the dinosaur’s femur. The mass of a Tyrannosaurus rex was 4000 kilograms. Use a calculator to approximate the circumference of its femur. Answer: Given, The body mass m (in kilograms) of a dinosaur that walked on two feet can be modeled by m = (0.00016)C 2.73 where C is the circumference (in millimeters) of the dinosaur’s femur. The mass of a Tyrannosaurus rex was 4000 kilograms. 4000 = (0.00016)C 2.73 C 2.73 = (4000/0.00016) C = (4000/0.00016) 1/2.73 C = (25,000,000) 1/2.73 C = 512.71 mm So, the approximate circumference of a dinosaur femur is 512.71 mm.

Find the indicated real nth root(s) of a.

Question 1. n = 3, a = -125 Answer: Given, n = 3, a = -125 \(\sqrt[n]{x}\) = x 1/n –\(\sqrt[3]{125}\)= -5

Question 2. n = 6, a = 64 Answer: Given n = 6, a = 64 \(\sqrt[n]{x}\) = x 1/n \(\sqrt[6]{64}\)= 2

Question 3. \(\sqrt[3]{-125}\) Answer: Given \(\sqrt[3]{-125}\) = –\(\sqrt[3]{125}\) = -5

Question 4. (-64) 2 / 3

Answer: Given (-64) 2 / 3 = –\(\sqrt[3]{64}\)² = (-4)² = 16

Question 5. 9 5 / 2

Answer: Given 9 5 / 2 = \(\sqrt[2]{9}\) 5 = (3) 5 = 3 × 3 × 3 × 3 × 3 = 243

Question 6. 256 3 / 4

Answer: Given 256 3 / 4 = \(\sqrt[4]{256}\) 5 = 4 5 = 4 × 4 × 4 × 4 × 4 = 1024

Question 7. WHAT IF? In Example 4, the volume of the beach ball is 17,000 cubic inches. Find the radius to the nearest inch. Use 3.14 for π. Answer: Given, The volume of the beach ball is 17,000 cubic inches. V = 4/3 × 3.14 × r³ 17,000 = 4/3 × 3.14 × r³ r³ = (17000 × 3)/4 × 3.14 r = 15.95 inch.

Question 8. The average cost of college tuition increases from $8500 to $13,500 over a period of 8 years. Find the annual inflation rate to the nearest tenth of a percent. Answer: Given, The average cost of college tuition increases from $8500 to $13,500 over a period of 8 years. 13500 – 8500 = $5000 5000/8500 × 100 = 58.8% So, the annual inflation rate to the nearest tenth of a percent is 58.8%

Radicals and Rational Exponents 6.2 Exercises

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, rewrite the expression in rational exponent form.

Question 4. \(\sqrt[5]{34}\) Answer: \(\sqrt[5]{34}\) = 34 1 / 5 .

In Exercises 5 and 6, rewrite the expression in radical form.

Question 6. 140 1 / 8 Answer: 140 1 / 8 = \(\sqrt[8]{140}\)

In Exercises 7–10, find the indicated real nth root(s) of a.

Question 8. n = 4, a = 81 Answer: The index n = 4 is even and a > 0 81 has two real square roots. 9² = 81 (-9)² = 81, the square roots of 81 are ±\(\sqrt[2]{81}\) = 9

Question 10. n = 9, a = -512 Answer: The index n = 9 is odd So 512 has one real cube root. 2 9 = 512, the ninth root of 512. \(\sqrt[9]{512}\) = 2

MATHEMATICAL CONNECTIONS In Exercises 11 and 12, find the dimensions of the cube. Check your answer.

In Exercises 13–18, evaluate the expression.

Question 14. \(\sqrt[3]{-216}\) Answer: 6 Explanation: \(\sqrt[3]{-216}\) = –\(\sqrt[3]{6.6.6}\) = -6

Question 16. –\(\sqrt[5]{1024}\) Answer: 4 Explanation: –\(\sqrt[5]{1024}\) = –\(\sqrt[5]{4.4.4.4.4}\) = -4

Question 18. (-64) 1 / 2 Answer: 8i Explanation: (-64) 1 / 2 = (\(\sqrt[2]{-64}\)) = 8i

In Exercises 19 and 20, rewrite the expression in rational exponent form.

Question 20. (\(\sqrt[5]{-21}\)) 6 Answer: (\(\sqrt[5]{-21}\)) 6 =-21 5/6

In Exercises 21 and 22, rewrite the expression in radical form.

Question 22. 9 5 / 2 Answer: 9 5 / 2 = (\(\sqrt[2]{9}\)) 5

In Exercises 23–28, evaluate the expression.

Question 24. 125 2 / 3 Answer: 125 2 / 3 = (\(\sqrt[3]{125}\)) 2

Question 26. (-243) 2 / 5 Answer: (-243) 2 / 5 = (\(\sqrt[5]{-243}\)) 2

Question 28. 343 4 / 3 Answer: 343 4 / 3 = (\(\sqrt[3]{343}\)) 4

In Exercises 31–34, evaluate the expression.

Question 32. (\(\frac{1}{64}\)) 1 / 6 Answer: (\(\frac{1}{64}\)) 1 / 6 = \(\sqrt[6]{1/64}\) = \(\frac{1}{2}\)

Question 34. (9)- 5 / 2 Answer: (9)- 5 / 2 = \(\sqrt[2]{9}\) -5

Question 36. PROBLEM SOLVING The volume of a cube-shaped box is 27 5 cubic millimeters. Find the length of one side of the box. Answer: Given, The volume of a cube-shaped box is 27 5 cubic millimeters. 27 5 = 27 × 27 × 27 × 27 × 27 = 1,43,48,907 V = s³ s = \(\sqrt[3]{V}\) = \(\sqrt[3]{14348907}\) = 243 So, the length of one side of the box is 243 millimeters.

Question 38. MODELING WITH MATHEMATICS The volume of a sphere is given by the equation V = \(\frac{1}{6 \sqrt{\pi}}\)S 3 / 2 , where S is the surface area of the sphere. Find the volume of a sphere, to the nearest cubic meter, that has a surface area of 60 square meters. Use 3.14 for π. Answer: V = 1/6(√3.14)S 3 / 2 V = 1/6(1.77)S 3 / 2 V = 1/10.62 (60) 3 / 2 V = 1/10.62 (7.75)³ V = 1/10.62 (465.5) V = 43.84 cu. meter

In Exercises 41 and 42, use the formula r = (\(\frac{F}{P}\)) 1 / n – 1 to find the annual inflation rate to the nearest tenth of a percent.

Question 42. The cost of a gallon of gas increases from $1.46 to $3.53 over a period of 10 years. Answer: Given, The cost of a gallon of gas increases from $1.46 to $3.53 over a period of 10 years. r = [(\(\frac{F}{P}\)) 1 / n – 1] × 100 r = [(\(\frac{3.53}{1.46}\)) 1 / 10  – 1] × 100 r = [2.417) 1 / 10  – 1] × 100 r = [0.0922] × 100 r = 9.22

Question 44. MAKING AN ARGUMENT Your friend says that for a real number a and a positive integer n, the value of \(\sqrt[n]{a}\) is always positive and the value of –\(\sqrt[n]{a}\) is always negative. Is your friend correct? Explain. Answer:

In Exercises 45–48, simplify the expression.

Question 46. (y • y 1 / 3 ) 3 / 2 Answer: = (y 4/3 ) 3 / 2 = y²

Question 48. (x 1 / 3 • y 1 / 2 ) 9 • \(\sqrt{y}\) Answer: (x 1 / 3 • y 1 / 2 ) 9 • \(\sqrt{y}\) = (x 1 / 3 ) 9 • (y 1 / 2 ) 9 • \(\sqrt{y}\) = x³ • \(\sqrt{y}\) 9 • \(\sqrt{y}\) = x³ • \(\sqrt{y}\) 10

Question 50. THOUGHT PROVOKING Find a formula (for instance, from geometry or physics) that contains a radical. Rewrite the formula using rational exponents. Answer: Volume = s³ s = \(\sqrt[3]{V}\) rewrite the formula using rational exponents s = V^1/3

ABSTRACT REASONING In Exercises 51–56, let x be a non negative real number. Determine whether the statement is always, sometimes, or never true. Justify your answer.

Question 52. x 1 / 3 = x -3 Answer: The statement x 1 / 3 = x -3 is always true.

Question 54. x = x 1 / 3 • x 3 Answer: x = x 1 / 3 • x 3 When bases are equal powers should be added. x = x 1 / 3 + 3 x = x 10/ 3

Question 55.

Evaluate the function when x = −3, 0, and 8. (Section 3.3)

Question 58. w(x) = -5x – 1 Answer: w(x) = -5x – 1 x = -3 w(-3) = -5(-3) – 1 = 15 – 1 = 14 x = 0 w(0) = -5(0) – 1 = 0 – 1 = -1 x = 8 w(8) = -5(8) – 1 = -40 – 1 = -41

Question 60. g(x) = 8x + 16 Answer: g(x) = 8x + 16 x = 0 g(0) = 8(0) + 16 = 10 x = -3 g(-3) = 8(-3) + 16 = -8 x = 8 g(8) = 8(8) + 16 = 80

Lesson 6.3 Exponential Functions

Question 4. What are some of the characteristics of the graph of an exponential function? Answer: 1. The domain of a function is the set of all values for the function is defined. 2. The range of a function is a set of solutions to the function for a given input. Characteristics of the exponential function is 1. The graph is increasing. 2. The graph is asymptotic to the x-axis as x approaches negative infinity. 3. The graph increases without bounds as x approaches positive infinity. 4. The graph is continuous. 5. The graph is smooth.

Question 5. Sketch the graph of each exponential function. Does each graph have the characteristics you described in Question 4? Explain your reasoning. a. y = 2 x b. y = 2(3) x c. y = 3(1.5) x d. y = (\(\frac{1}{2}\)) x e. y = 3(\(\frac{1}{2}\)) x f. y = 2(\(\frac{3}{4}\)) x

Answer: a. y = 2 x

Does the table represent a linear or an exponential function? Explain.

Evaluate the function when x = −2, 0, and \(\frac{1}{2}\).

Question 3. y = 2(9) x

Answer: Given, y = 2(9) x x = -2 y = 2(9) x y = 2(9) -2 = 2\(\sqrt[2]{9}\) = 2(3) = 6 x = 0 y = 2(9) x y = 2(9) 0 = 2(1) = 2 x = \(\frac{1}{2}\) y = 2(9) x y = 2(9) \(\frac{1}{2}\) = 2\(\sqrt[2]{9}\) = 2(3) = 6

Question 4. y = 1.5(2) x

Answer: y = 1.5(2) x x = -2 y = 1.5(2) x = y = 1.5(2) -2  = 1.5\(\sqrt[2]{2}\) x = 0 y = 1.5(2) 0  = 1.5(1) = 1.5 x = \(\frac{1}{2}\) y = 1.5(2) \(\frac{1}{2}\) =1.5\(\sqrt[2]{2}\)

Graph the function. Compare the graph to the graph of the parent function. Describe the domain and range of f.

Question 5. f(x) = -2(4) x

Question 6. f(x) = 2(\(\frac{1}{4}\)) x

Graph the function. Describe the domain and range.

Question 7. y = -2(3) x + 2 – 1

Question 9. WHAT IF? In Example 6, the dependent variable of g is multiplied by 3 for every 1 unit the independent variable x increases. Graph g when g(0) = 4. Compare g and the function f from Example 3 over the interval x = 0 to x = 2. Answer:

Question 10. A bacterial population y after x days can be represented by an exponential function whose graph passes through (0, 100) and (1, 200). (a) Write a function that represents the population. (b) Find the population after 6 days. (c) Does this bacterial population grow faster than the bacterial population in Example 7? Explain. Answer: a. Plot the points (0, 100) and (1, 200)

y = 200^x b. Substitute x = 5 in y = 200^x y = 200^5 y = 3.2 × 10^11 There will be 3.2 × 10^11 bacteria after 6 days c. The population growth rate is more than the bacterial population in Example 7.

Exponential Functions 6.3 Exercises

Question 2. REASONING Why is a the y-intercept of the graph of the function y = ab x ? Answer: The y-intercept occurs on a graph when x = 0. So on a graph of y = ab^x when x = 0. y = ab^0 = 1. Thus the y-intercept is at the point (0, 1).

In Exercises 5–10, determine whether the equation represents an exponential function. Explain.

Question 6. y = -6x Answer: The equation y = -6x does not represent an exponential function because it does not fit the pattern y = ab^x.

Question 8. y = -3x Answer: The equation y = -3x does not represent an exponential function because it does not fit the pattern y = ab^x.

Question 10. y = \(\frac{1}{2}\)(1) x Answer: The equation y = \(\frac{1}{2}\)(1) x represents an exponential function, because it fits the equation y = ab x .

In Exercises 11–14, determine whether the table represents a linear or an exponential function. Explain. 

In Exercises 15–20, evaluate the function for the given value of x.

Question 16. f(x) = 3(2) x ; x = -1 Answer: f(x) = 3(2) x ; x = -1 f(x) = 3(2)^-1 = 3 × 1/2 = 3/2 = 0.5

Question 18. f(x) = 0.5 x ; x = -3 Answer: f(x) = 0.5 x x = -3 f(x) = 0.5 -3 = 8

Question 20. y = \(\frac{1}{4}\)(4) x ; x = \(\frac{3}{2}\) Answer: y = \(\frac{1}{4}\)(4) x ; x = \(\frac{3}{2}\) y = \(\frac{1}{4}\)(4) \(\frac{3}{2}\) ; y = \(\frac{1}{4}\) (2)³ y = \(\frac{1}{4}\) × 8 y = 2

USING STRUCTURE In Exercises 21–24, match the function with its graph.

Question 22. y = -2(0.5) x Answer: B The parent function of f(x) = -2(0.5) x is g(x) is (0.5) x . The graph of the parent function g increases as x increases 0 < b < 1.

In Exercises 25–30, graph the function. Compare the graph to the graph of the parent function. Describe the domain and range of f.

In Exercises 31–36, graph the function. Describe the domain and range.

In Exercises 37–40, compare the graphs. Find the value of h, k, or a.

In Exercises 43 and 44, graph the function with the given description. Compare the function to f (x) = 0.5(4) x over the interval x = 0 to x = 2.

Question 44. An exponential function h models a relationship in which the dependent variable is multiplied by \(\frac{1}{2}\) for every 1 unit the independent variable x increases. The value of the function at 0 is 32. Answer:

In Exercises 47–50, write an exponential function represented by the table or graph.

Question 52. PROBLEM SOLVING A sales report shows that 3300 gas grills were purchased from a chain of hardware stores last year. The store expects grill sales to increase 6% each year. About how many grills does the store expect to sell in Year 6? Use an equation to justify your answer. Answer: Given, A sales report shows that 3300 gas grills were purchased from a chain of hardware stores last year. The store expects grill sales to increase 6% each year. y = A(b)^x A: initial quantity of gas grills sold b: increase in sales every year t: the amount of time in years y = (3300)(1.06)^6 y = 4681.11 y = 4681 grills in 6 years

Question 56. OPEN-ENDED Write a function whose graph is a horizontal translation of the graph of h(x) = 4 x . Answer:

b. Describe the change in the stock price from Week 1 to Week 3. Answer: From the graph, we can observe that the price of the stock in week 1 is $40 and the price of the stock in week 3 is $10. Thus the change in stock price from week 1 to week 3 is 40 – 10 = 30.

Question 60. THOUGHT PROVOKING Write a function of the form y = ab x that represents a real-life population. Explain the meaning of each of the constants a and b in the real-life context. Answer:

Question 62. PROBLEM SOLVING A function g models a relationship in which the dependent variable is multiplied by 4 for every 2 units the independent variable increases. The value of the function at 0 is 5. Write an equation that represents the function. Answer:

Write the percent as a decimal.

Question 64. 4% Answer: 4% can be written in fraction form as 4/100. 4/100 = 0.04

Question 66. 128% Answer: 128% can be written as 128/100 128% = 128/100 = 1.28

Lesson 6.4 Exponential Growth and Decay

Essential Question What are some of the characteristics of exponential growth and exponential decay functions?

y2/y1 = 1.58 y3/y2 = 1.81 y4/y3 = 1.50 y5/y4 = 1.34 y6/y5 = 1.43 average growth rate = (1.58 + 1.81 + 1.5 + 1.34 + 1.43)/5 = 1.528 100000 = 1188(1.528)^x x = 10.5 So, after 10.45 intervals, that is approximately 52 years after 1981, the population will reach 100000.

EXPLORATION 2 Describing a Decay Pattern Work with a partner. A forensic pathologist was called to estimate the time of death of a person. At midnight, the body temperature was 80.5°F and the room temperature was a constant 60°F. One hour later, the body temperature was 78.5°F. a. By what percent did the difference between the body temperature and the room temperature drop during the hour? Answer: Given, A forensic pathologist was called to estimate the time of death of a person. At midnight, the body temperature was 80.5°F and the room temperature was a constant 60°F. One hour later, the body temperature was 78.5°F. 78.5/80.5 = 0.975 1 – 0.975 = 0.025 Thus the percentage of temperature drop will be 2.5%

b. Assume that the original body temperature was 98.6°F. Use the percent decrease found in part (a) to make a table showing the decreases in body temperature. Use the table to estimate the time of death. Answer: The percentage of temperature drop will be 2.5% The original body temperature was 98.6°F. 78.5/0.975 = 80.5 80.5/0.975 = 82.6 82.6/0.975 = 84.7 84.7/0.975 = 86.9 86.9/0.975 = 91.4 91.4/0.975 = 93.7 93.7/0.975 = 96.1 96.1/0.975 = 98.6 Thus we can see that the original temperature 98.6°F was at 4 p.m. Therefore the time of death is 4 p.m.

Question 3. What are some of the characteristics of exponential growth and exponential decay functions? Answer: These are the characteristics of the exponential growth and exponential decay functions. Properties of Exponential Growth Functions The function is an increasing function; y will increase as x will increase. Range: If a>0, the range is {positive real numbers} The graph is continually above the x-axis. Horizontal Asymptote: when b>1, the horizontal asymptote is the negative x-axis, as x will become a huge negative. Properties of Exponential Decay Functions The feature y=f(x) function represents decay if k < 0. The function is a reducing function; y decreases as x will increase. Range: If a>0, the range is {positive real numbers} The graph is always above the x-axis.

Question 4. Use the Internet or some other reference to find an example of each type of function. Your examples should be different than those given in Explorations 1 and 2. a. exponential growth b. exponential decay Answer: a. Exponential growth is a pattern of data that shows greater increases with passing time, creating the curve of an exponential function. For example, suppose a population of mice rises exponentially by a factor of two every year starting with 2 in the first year, then 4 in the second year, 8 in the third year, 16 in the fourth year, and so on. The population is growing by a factor of 2 each year in this case.

b. Examples of exponential decay are radioactive decay and population decrease.

Question 1. A website has 500,000 members in 2010. The number y of members increases by 15% each year. (a) Write an exponential growth function that represents the website membership t years after 2010. (b) How many members will there be in 2016? Round your answer to the nearest ten thousand. Answer: Given, A website has 500,000 members in 2010. The number y of members increases by 15% each year. a = 500,000(1.15)^t In 2016 a = 500,000(1.15)^6 a = 1,156,530 It is rounded to 1,160,000.

Determine whether the table represents an exponential growth function, an exponential decay function, or neither. Explain.

Determine whether the function represents exponential growth or exponential decay. Identify the percent rate of change.

Question 4. y = 2(0.92) t

Answer: The function represents exponential decay.

Question 5. f(t) = (1.2) t

Answer: The function represents exponential growth.

Rewrite the function to determine whether it represents exponential growth or exponential decay.

Question 6. f(t) = 3(1.02) 10t

Question 7. y = (0.95) t + 2

Question 8. You deposit $500 in a savings account that earns 9% annual interest compounded monthly. Write and graph a function that represents the balance y (in dollars) after t years. Answer: Given, You deposit $500 in a savings account that earns 9% annual interest compounded monthly. Balance = Amount(1 + rate) t = 9%/12 = 0.75% t is in years so has to be converted as well to months = t × 12 months y = 500(1 + 0.75%) 12t

Question 9. WHAT IF? The car loses 9% of its value every year. (a) Write a function that represents the value y (in dollars) of the car after t years. (b) Find the approximate monthly percent decrease in value. (c) Graph the function from part (a). Use the graph to estimate the value of the car after 12 years. Round your answer to the nearest thousand. Answer: y = (1 – 0.09) t y =a(b) x y = 21500(1 – 0.09) 12 y = 21,500(0.91) 12

Exponential Growth and Decay 6.4 Exercises

Question 2. VOCABULARY What is the decay factor in the exponential decay function y = a(1 – r) t ? Answer: Given, y = a(1 – r) t y – future value a – present value (1 – r) – decay factor t – number of decay periods

Question 4. WRITING When does the function y = ab x represent exponential growth? exponential decay? Answer: y = ab x If b is a number between 0 and 1, then the function represents exponential decay.

In Exercises 5–12, identify the initial amount a and the rate of growth r (as a percent) of the exponential function. Evaluate the function when t = 5. Round your answer to the nearest tenth.

Question 6. y = 10(1 + 0.4) t Answer: 1 + r = 1 + 0.4 r = 0.4 The initial amount is a = 10 and the rate of growth is r = 0.4 or 40%. y = 10(1 + 0.4) 5 y = 10(1.4) 5 y = 10(5.37) y = 53.7 So, the value of y is about 53.7 when t = 5

Question 8. y = 12(1.05) t Answer: 1 + r = 1.05 r = 1.05 – 1 r = 0.05 a = 12 and r = 0.05 or 5% t = 5 y = 12(1.05) 5 y = 12(1.276) y = 15.312 So, the value of y is about 15.312 when t = 5.

Question 10. h(t) = 175(1.028) t Answer: 1 + r = 1.028 r = 1.028 – 1 r = 0.028 a = 175 and r = 2.8% h(t) = 175(1.028) 5 h(t) = 175(1.148) h(t) = 200.9

Question 12. p(t) = 1.8 t Answer: 1 + r= 1.8 r = 1.8 – 1 r = 0.8 a = 1 and r = 0.8 t = 5 p(t) = 1.8 5 p(t) = 18.89

In Exercises 13–16, write a function that represents the situation.

Question 14. Your starting annual salary of $35,000 increases by 4% each year. Answer: The initial amount = $35,000 The rate of growth = 4% or 0.04 y = a(1 + r) t y = 35,000(1 + 0.04) t y = 35000(1.04) t

Question 16. An item costs $4.50, and its price increases by 3.5% each year. Answer: Given, An item costs $4.50, and its price increases by 3.5% each year. a = $4.50 r = 3.5% or 0.035 y = a(1 + r) t y = 4.50(1 + 0.035) t y = 4.50(1.035) t

In Exercises 19–26, identify the initial amount a and the rate of decay r (as a percent) of the exponential function. Evaluate the function when t = 3. Round your answer to the nearest tenth.

Question 20. y = 8(1 – 0.15) t Answer: a = 8 r = 0.15 t = 3 y = 8(1 – 0.15) 3 y = 8(0.85)³ y = 8 × 0.614 y = 4.912

Question 22. f(t) = 475(0.5) t Answer: a = 475 1 – r= 0.5 r = 1 – 0.5 r = 0.5 t = 3 f(t) = 475(0.5) t f(t) = 475(-0.5)³ f(t) = 475(0.125) f(t) = 59.375

Question 24. h(t) = 1250(0.865) t Answer: a = 1250 1 – r = 0.865 r = 1- 0.865 r = 0.135 t = 3 h(t) = 1250(0.135)³ h(t) = 1250(0.0024) h(t) = 3

Question 26. y = 0.5 (\(\frac{3}{4}\)) t Answer: a = 0.5 t = 3 1 – r = \(\frac{3}{4}\) r = 1 – \(\frac{3}{4}\) r = \(\frac{1}{4}\) y = 0.5 (\(\frac{1}{4}\))³ y = 0.5(0.015) y = 0.0075

In Exercises 27–30, write a function that represents the situation.

Question 28. A $900 sound system decreases in value by 9% each year. Answer: Given, A $900 sound system decreases in value by 9% each year. a = 900 1 – r = 9% 1 – r = 0.09 1 – 0.09 = r r = 0.91 y = a(1 – r) t y = 900(1 – 0.91) t

Question 30. A company profit of $20,000 decreases by 13.4% each year. Answer: Given, A company profit of $20,000 decreases by 13.4% each year. a = 20,000 1 – r = 13.4 1 – r = 0.134 1 – 0.134= r r = 0.866 y = a(1 – r) t y = 20,000(1 – 0.866) t

In Exercises 33–38, determine whether the table represents an exponential growth function, an exponential decay function, or neither. Explain.

b. How many people will have visited the website after it is online 47 days? Answer: The number of visitors is increasing by 1.1 the preceding number of visitor t(43) = 1.1 × t(42) t(44) = 1.1 × t(43) t(45) = 1.1 × t(44) t(46) = 1.1 × t(45) = 1.1 × 14641 = 16105.1 t(47) = 1.1 × t(46) t(46) = 16105.1 t(47) = 1.1 × 16105.1 = 17715.61 ≈ 17716

In Exercises 41–48, determine whether each function represents exponential growth or exponential decay. Identify the percent rate of change.

Question 42. y = 15(1.1) t Answer: y = 15(1.1) t y = a(1 – r) t 1 – r = 1.1 r = 1.1 – 1 r = 0.1 So, the rate of decay is 10%

Question 44. y = 5(1.08) t Answer: y = 5(1.08) t y = a(1 – r) t 1 – r = 1.08 r = 1.08 – 1 r = 0.08 So, the rate of decay is 8%

Question 46. s(t) = 0.65(0.48) t Answer: s(t) = 0.65(0.48) t y = a(1 – r) t 1 – r = 0.48 r = 1 – 0.48 r = 0.52 So, the rate of decay is 52%

Question 48. m(t) = (\(\frac{4}{5}\)) t Answer: m(t) = (\(\frac{4}{5}\)) t y = a(1 – r) t 1 – r =\(\frac{4}{5}\) r = 1 – \(\frac{4}{5}\) r = \(\frac{5}{5}\) – \(\frac{4}{5}\) So, the rate of decay is \(\frac{1}{5}\) or 0.2 or 2%

In Exercises 49–56, rewrite the function to determine whether it represents exponential growth or exponential decay.

Question 50. y = (1.4) t + 8 Answer: y = 1.4 t × (1.4) 8 y = 14.76(1.4) t The function is of the form y = a(1 + r) t 1 + r > 1 So, it represents exponential growth.

Question 52. y = 5(0.82) t/5 Answer: y = 5(0.82) t/5 The function is of the form y = a(1 – r) t 1 – r < 1 So, it represents exponential decay.

Question 54. f(t) = 0.4(1.16) t – 1 Answer: f(t) = 0.4 × (1.16) t /(1.16) f(t) = 0.4/1.16 × (1.16) t f(t) = 0.34((1.16) t The function is of the form y = a(1 + r) t 1 + r > 1 So, it represents exponential growth.

Question 56. r(t) = (0.88) 4r Answer: r(t) = (0.88) 4r The function is of the form y = a(1 – r) t 1 – r < 1 So, it represents exponential decay.

In Exercises 57–60, write a function that represents the balance after t years. 

Question 58. $1400 deposit that earns 10% annual interest compounded semiannually Answer: Given, $1400 deposit that earns 10% annual interest compounded semiannually n = 2 r = 10% = 0.1 P = 1400 y = P(1 + r/n) nt 1400(1 + (0.1)/2) 2t

Question 60. $3500 deposit that earns 9.2% annual interest compounded quarterly Answer: Given, $3500 deposit that earns 9.2% annual interest compounded quarterly. P = 3500 t = 4 P = 1400 y = P(1 + r/n) nt y = 3500(1.023) 4t

Question 66. COMBINING FUNCTIONS You deposit $9000 in a savings account that earns 3.6% annual interest compounded monthly. You also save $40 per month in a safe at home. Write a function C(t) = b(t) + h(t), where b(t) represents the balance of your savings account and h(t) represents the amount in your safe after t years. What does C(t) represent? Answer: Given, You deposit $9000 in a savings account that earns 3.6% annual interest compounded monthly. You also save $40 per month in a safe at home. Write a function C(t) = b(t) + h(t), where b(t) represents the balance of your savings account and h(t) represents the amount in your safe after t years. P = 9000 r = 3.6% n = 12 b(t) = P(1 + r/n) nt b(t) = 9000(1 + 0.036/12) 12t b(t) = 9000(1.003) 12t h(t) = 40(12t) h(t) = 480t C(t) = b(t) + h(t) C(t) = 9000(1.003) 12t + 480t

The value of computers decreases by 18 percent each year. r = 18/100 r = 0.18 1 – r = 1 – 0.18 b = 0.82 a = 500 y = ab x y = 500(0.82)¹ y = 500 × 0.82 y = 410 y = 410(0.82)² y = 275.6 Points are (0, 500) (1, 410) (2, 275) Graph B has these points.

A deposit earns 11% annual interest compounded yearly r = 11/100 r = 0.11 b = 1 + r b = 1.11 a = 50 y = ab x y = 50(1.11)¹ y = 55.5 y = 55.5(1.11)² y = 68.38 y = 68.38(1.11)³ y = 93.61 The graph is a.

r = 5.5/100 r = 0.055 b = 1 – r b = 0.845 a = 500 y = ab x y = 500(0.845)¹ y = 422.5 y = 422.5(0.845)² y = 422.5(0.714) y = 301.67 The graph c.

Question 70. THOUGHT PROVOKING Describe two account options into which you can deposit $1000 and earn compound interest. Write a function that represents the balance of each account after t years. Which account would you rather use? Explain your reasoning. Answer:

Solve the equation. Check your solution. (Section 1.3)

Question 74. 5 – t = 7t + 21 Answer: Given, 5 – t = 7t + 21 -t – 7t = 21 – 5 -8t = 16 t = -2

Find the slope and the y-intercept of the graph of the linear equation. (Section 3.5)

Question 76. y = -6x + 7 Answer: The equation of the line is y = mx + c m = -6 c = 7

Question 78. 3y = 6x – 12 Answer: The equation of the line is y = mx + c 3y = 6x – 12 y = 2x – 6 m = 2 c = -6

Exponential Functions and Sequences Study Skills: Analyzing Your Errors

6.1–6.4 What Did You Learn?

Core Vocabulary

Mathematical Practices

Question 1. How did you apply what you know to simplify the complicated situation in Exercise 56 on page 297?

Question 2. How can you use previously established results to construct an argument in Exercise 44 on page 304?

Question 3. How is the form of the function you wrote in Exercise 66 on page 322 related to the forms of other types of functions you have learned about in this course?

Study Skills

Analyzing Your Errors

Misreading Directions

• What Happens: You incorrectly read or do not understand directions.
• How to Avoid This Error: Read the instructions for exercises at least twice and make sure you understand what they mean. Make this a habit and use it when taking tests.

Exponential Functions 6.1 – 6.4 Quiz

Simplify the expression. Write your answer using only positive exponents. (Section 6.1)

Question 1. 3 2 • 3 4

Answer: 3 2 • 3 4  Bases are equal so powers should be added 3 4+2 = 3 6

Question 2. (k 4 ) -3

Answer: (k 4 ) -3 = k -12

Evaluate the expression .(Section 6.2)

Question 5. \(\sqrt[3]{27}\) Answer: Given, \(\sqrt[3]{3³}\) = 3

Question 6. \(\frac{1}{16}\) 1/4

Answer: Given, \(\frac{1}{16}\) 1/4 =\(\sqrt[4]{1/16}\) = 1/4

Question 7. 512 2/3

Answer: Given, 512 2/3 (\(\sqrt[3]{512}\))² = 8² = 64

Question 8. \(\sqrt{4}\) 5

Answer: Given, \(\sqrt{4}\) 5 = 2 5 2 × 2 × 2 × 2 × 2 = 32

Graph the function. Describe the domain and range. (Section 6.3)

Question 9. y = 5 x

Question 10. y = -2(\(\frac{1}{6}\)) x

Determine whether the table represents an exponential growth function, an exponential decay function, or neither. Explain. (Section 6.4)

Determine whether the function represents exponential growth or exponential decay. Identify the percent rate of change. (Section 6.4)

Question 14. y = 3(1.88) t

Answer: y = 3(1.88) t y = a(1 + r) t 1 + r =1 + 0.88 r = 0.88 So, the rate of growth is 88%

Question 15. f(t) = \(\frac{1}{3}\)(1.26) t

Answer: f(t) = \(\frac{1}{3}\)(1.26) t y = a(1 + r) t 1 + r =1 + 0.26 r = 0.26 So, the rate of growth is 26%

Question 16. f(t) = 80(\(\frac{3}{5}\)) t

Answer: f(t) = 80(\(\frac{3}{5}\)) t y = a(1 – r) t 1 – r =\(\frac{3}{5}\) r = 1 – \(\frac{3}{5}\) r = 0.4 So, the rate of decay is 40%

b. How many times smaller is a milligram than a hectogram? Write your answer using only positive exponents. Answer: Given, A milligram and a hectogram 1 hectogram = 10 5 milligram So, a milligram is 10 5 times smaller than a hectogram.

c. Which is greater, 10,000 milligrams or 1000 decigrams? Explain your reasoning. Answer: 1 decigram = 100 milligrams 1000 decigrams = 100 × 1000 milligrams = 1,00,000 milligrams 1,00,000 milligrams is greater than 10,000 milligrams Thus 1000 decigrams is greater than 10,000 milligrams

Question 19. The function f(t) = 5(4) t represents the number of frogs in a pond after t years. (Section 6.3 and Section 6.4) a. Does the function represent exponential growth or exponential decay? Explain. Answer: The function f(t) = 5(4) t represents the number of frogs in a pond after t years. f(t) = 5(1 + 3) t y = a(1 + r) t The function represents exponential growth.

c. What is the yearly percent change? the approximate monthly percent change? Answer: f(t) = 5(1 + 3) t 1 + r = 4 r = 3 The yearly percent change is 300. 1 year = 12 months t = t/12 f(t) = 5(4) t f(1/2) = 5(4) t/12 = 5(1.12246) t = 5(1 + 0.12246) t 1 + r = 1 + 0.12246 r = 0.12246 r = 0.12 The monthly percent change is 12.

d. How many frogs are in the pond after 4 years? Answer: f(t) = 5(4) t t = 4 f(t) = 5(4) 4 f(4) = 5(256) = 1280 f(4) = 1280. After 4 years, there will be 1280 frogs in a pond.

Lesson 6.5 Solving Exponential Functions

Question 4. How can you solve an exponential equation graphically? Answer: 1. Use Steps for Solving an Exponential Equation with Different Bases. 2. Rewrite the problem using the same base. 3. Use the properties of exponents to simplify the problem. 4. Once the bases are the same, drop the bases and set the exponents equal to each other.

Question 5. A population of 30 mice is expected to double each year. The number p of mice in the population each year is given by p = 30(2 n ). In how many years will there be 960 mice in the population? Answer: Given, A population of 30 mice is expected to double each year. The number p of mice in the population each year is given by p = 30(2 n ) p = 960 960 = 30(2 n ) 960/30 = 2 n 32 = 2 n 2 5 = 2 n 5 = n So, it will take 5 years for the population of mice to reach 960.

Solve the equation. Check your solution.

Question 1. 2 2x = 2 6

Answer: 2 2x = 2 6 When bases are equal powers should be equated 2x = 6 x = 6/2 x = 3

Question 2. 5 2x = 5 x + 1

Answer: 5 2x = 5 x + 1 When bases are equal powers should be equated. 2x = x + 1 2x – x = 1 x = 1

Question 3. 7 3x + 5 = 7 x + 1

Answer: 7 3x + 5 = 7 x + 1 When bases are equal powers should be equated. 3x + 5 = x + 1 3x – x = 1 – 5 2x = -4 x = -2

Question 4. 4 x = 256

Answer: Given, 256 can be written in the fourth root as 4 4 4 x = 256 4 x = 4 4 x = 4

Question 5. 9 2x = 3 x – 6

Answer: Given, 9 2x = 3 x – 6 3 2(2x) = 3 x – 6 4x = x – 6 4x – x = -6 3x = -6 x = -2

Question 6. 4 3x = 8 x + 1

Answer: Given, 4 3x = 8 x + 1 2 2(3x) = 2 3(x + 1) 6x = 3x + 3 6x – 3x = 3 3x = 3 x = 1

Question 7. (\(\frac{1}{3}\)) x – 1 = 27 Answer: (\(\frac{1}{3}\)) x – 1 =3³ (\(\frac{1}{3}\)) x – 1 =(\(\frac{1}{3}\)) -3 x – 1 = -3 x = – 3 + 1 x = -2

Use a graphing calculator to solve the equation.

Solving Exponential Functions 6.5 Exercises

In Exercises 3–12, solve the equation. Check your solution.

Question 4. 7 x – 4 = 7 8 Answer: 7 x – 4 = 7 8 When bases are equal powers should be equated. x – 4 = 8 x = 8 + 4 x = 12

Question 6. 2 4x = 2 x + 9 Answer: 2 4x = 2 x + 9 When bases are equal powers should be equated. 4x = x + 9 4x – x = 9 3x = 9 x = 9/3 x = 3

Question 8. 3x = 243 Answer: 3x = 243 x = 243/3 x = 81

Question 10. 216 x = 6 x + 10 Answer: When bases are equal powers should be equated. 216 x = 6 x + 10 6³ x = 6 x + 10 3x = x + 10 3x – x = 10 2x = 10 x = 5

Question 12. 27 x = 9 x – 2 Answer: 27 x = 9 x – 2 When bases are equal powers should be equated. 27 can be written as 3³ 9 can be written as 3² 3x = 2x – 4 3x – 2x = -4 x = -4

In Exercises 13–18, solve the equation. Check your solution.

Question 14. (\(\frac{1}{4}\)) x = 256 Answer: (\(\frac{1}{4}\)) x = 256 256 can be written as 4 4 (\(\frac{1}{4}\)) x = 4 4 (\(\frac{1}{4}\)) x = 4 -4 x = -4

Question 16. 3 4x – 9 = \(\frac{1}{243}\) Answer: 3 4x – 9 = \(\frac{1}{3^5}\) 3 4x – 9 = 3 -5 4x – 9 = -5 4x = -5 + 9 4x = 4 x = 1

Question 18. (\(\frac{1}{27}\)) 4-x = 9 2x-1 Answer: (\(\frac{1}{27}\)) 4-x = 9 2x-1 When bases are equal powers should be equated. 27 can be written as 3³ 9 can be written as 3² (\(\frac{1}{3^3}\)) 4-x = 3 2(2x-1) 3 (4-x)-3 = 3 2(2x-1) -12 + 3x = 4x – 2 -12 + 2 = 4x – 3x -10 = x x = -10

ERROR ANALYSIS In Exercises 19 and 20, describe and correct the error in solving the exponential equation.

Question 22. 4 2x – 5 = 6 Answer: Graph D is correct. x ≈ 3.14

In Exercises 25–36, use a graphing calculator to solve the equation.

In Exercises 37–40, solve the equation by using the Property of Equality for Exponential Equations.

Question 38. 12 • 2 x – 7 = 24 Answer: 12 • 2 x – 7 = 24 2 x – 7 = 24/12 2 x – 7 = 2 Bases are equal, so powers should be equated. x – 7 = 1 x = 1 + 7 x = 8

Question 40. 2(4 2x + 1 ) = 128 Answer: 2(4 2x + 1 ) = 128 (4 2x + 1 ) = 128/2 (4 2x + 1 ) = 64 64 = 4³ (4 2x + 1 ) = 4³ Bases are equal, so powers should be equated. 2x + 1 = 3 2x = 3 – 1 2x = 2 x = 1

In Exercises 43–46, solve the equation.

Question 44. 3 4x + 3 = 81 x Answer: 3 4x + 3 = 81 x 81 = 3 4 3 4x + 3 = 3 4 x 4x + 3 = 4x 3 = 0 It has no solution.

Question 46. 5 8(x – 1) = 625 2x – 2 Answer: 5 8(x – 1) = 625 2x – 2 625 = 5 4 8(x-1) = 2x – 2 8x – 8 = 2x – 2 8x – 2x = -2 + 8 6x = 6 x = 1

Question 48. PROBLEM SOLVING There are a total of 128 teams at the start of a citywide 3-on-3 basketball tournament. Half the teams are eliminated after each round. Write and solve an exponential equation to determine after which round there are 16 teams left. Answer: Given, There are a total of 128 teams at the start of a citywide 3-on-3 basketball tournament. Half the teams are eliminated after each round. T = 128 × (1/2)n T = 16 16 = 128 × (1/2)^n 1/8 = (1/2)^n n = 3

Question 52. THOUGHT PROVOKING Is it possible for an exponential equation to have two different solutions? If not, explain your reasoning. If so, give an example. Answer:

USING STRUCTURE In Exercises 53–58, solve the equation.

Question 54. \(\sqrt{5}\) = 5 x + 4 Answer: \(\sqrt{5}\) = 5 x + 4 5 1/2 = 5 x + 4 1/2 = x + 4 x = 4 – 1/2 x = 3 1/2

Determine whether the sequence is arithmetic. If so, find the common difference. (Section 4.6)

Question 60. -20, -26, -32, -38, . . . Answer: a1 = -20 a2 = -26 d = -26 – (-20) = -26 + 20 = -6 So, the common difference is -6.

Question 62. -5, -8, -12, -17, . . . Answer: a1 = -5 a2 = -8 -8 – (-5) = -8 + 5 = -3 -8 – 3 = -11 The sequence does not have a common difference. So, it is not arithmetic.

Lesson 6.6 Geometric Sequences

Essential Question

How can you use a geometric sequence to describe a pattern?

In a geometric sequence, the ratio between each pair of consecutive terms is the same. This ratio is called the common ratio.

b. The thickness of the piece of paper is 0.1 millimeters. So, the thickness of the paper = T(2)^m

As we know only the thickness of the paper is given that is 0.1 millimeter and we do not know the length and width of the paper. We can find the greatest number of times a paper can fold and its thickness since the size of the paper becomes smaller and the thickness of the paper increases when we fold it.

c. m = 15 T = 0.1(2)^15 = 3276.8 mm 3276.8 mm = 3267.8/10 cm = 326.78 cm Yes, the thickness of the paper is taller than us since when we fold the paper of thickness 0.1 mm in half 15 times, we get 326.78 cm of the thickness of the paper.

Question 3. How can you use a geometric sequence to describe a pattern? Answer: You have to find the common ratio of the given sequence to find the geometric sequence. Geometric sequence an = a1 . (r)^(n-1) r = common ratio a1 = first term

Question 4. Give an example of a geometric sequence from real life other than paper folding. Answer: Compound Interest is the example of a geometric sequence. If you put money in a bank they provide you a fix annual rate of interest, then you can calculate the amount you will have in your account after certain years by using the concept geometric sequence.

Decide whether the sequence is arithmetic, geometric, or neither. Explain your reasoning.

Question 1. 5, 1, -3, -7, . . . Answer: Given, 5, 1, -3, -7, . . . 5 – 1 = 4 1 – (-3) = 4 a1 = 5 Common difference = 4 So, it represents an arithmetic sequence.

Question 2. 1024, 128, 16, 2, . . . Answer: Given, 1024, 128, 16, 2, . . . 128/1024 = 1/8 16/128 = 1/8 2/16 = 1/2 So, the common ratio is 1/8. So, it is a geometric sequence.

Question 3. 2, 6, 10, 16, . . Answer: Given, 2, 6, 10, 16, . . a1 = 2 a2 = 6 6 – 2 = 4 10 – 6 = 4 16 – 10 = 6 There is no common difference. It is neither arithmetic nor geometric sequence.

Write the next three terms of the geometric sequence. Then graph the sequence.

Question 4. 1, 3, 9, 27, . . . Answer: a1 = 1 a2 = 3 a3 = 9 a4 = 27 r = 3 a5 = 27 . 3 = 81 a6 = 81 . 3 = 243 a7 = 243 . 3 = 729 So, the next three terms are 81, 243, 729

Question 5. 2500, 500, 100, 20, . . . Answer: a1 = 2500 a2 = 500 a3 = 100 a4 = 20 r = 1/5 a5 = 2500(1/5)^4 = 2500/625 = 4 a6 = 2500(1/5)^5 = 0.8 a7 = 2500(1/5)^6 = 2500/15625 = 0.16 So, the next three terms are 4, 0.8, 0.16

Question 6. 80, -40, 20, -10, . . . Answer: a1 = 80 a2 = -40 a3 = 20 a4 = -10 r = -1/2 a5 = -10/-2 = 5 a6 = 5/-2 = -2.5 a7 = -2.5/-2 = 1.25 So, the next three terms are 5, -2.5, 1.25

Question 7. -2, 4, -8, 16, . . . Answer: -2, 4, -8, 16, . . . a1 = -2 a2 = 4 a3 = -8 a4 = 16 r = -2 a5 = 16 × -2 = -32 a6 = -32 × -2 = 64 a7 = 64 × -2 = -128 So, the next three terms are -32, 64, -128.

Write an equation for the nth term of the geometric sequence. Then find a 7 .

Question 8. 1, -5, 25, -125, . . . Answer: Given, 1, -5, 25, -125, . . . a1 = 1 a2 = -5 r = a2/a1 = -5/1 = -5 r = a3/a2 = 25/-5 = -5 So, the common ratio is -5. an = a1 . (r)^n-1 n = 7 a7 = 1 . (-5)^7-1 a7 = (-5)^6 a7 = 15625

Question 9. 13, 26, 52, 104, . . . Answer: Given, 13, 26, 52, 104, . . . a1 = 13 a2 = 26 r = a2/a1 r = 26/13 r = 2 an = a1 . (r)^n-1 n = 7 a7 = 13 . (2)^7-1 a7 = 13(2)^6 a7 = 13 × 64 a7 = 832

Question 10. 432, 72, 12, 2, . . . Answer: Given, 432, 72, 12, 2, . . . a1 = 432 a2 = 72 r = a2/a1 = 72/432 = 1/6 an = a1 . (r)^n-1 n = 7 a7 = 432 . (1/6)^7-1 a7 = 432(1/6)^6 a7 = 432(1/46656) a7 = 0.0092

Question 11. 4, 10, 25, 62.5, . . . Answer: Given, 4, 10, 25, 62.5, . . . a1 = 4 a2 = 10 r = a2/a1 = 10/4 = 5/2 an = a1 . (r)^n-1 n = 7 a7 = 4 . (5/2)^7-1 a7 = 4 . (5/2)^6 a7 = 4(15625/64) a7 = 15625/16 a7 = 976.56

Question 12. WHAT IF? After how many clicks on the zoom-out button is the side length of the map 2560 miles? Answer:

a1 = 4 a2 = 12 r = 12/4 = 3 r = 36/12 = 3 a6 = 4 . (3)^6-1 a6 = 4 (3)^5 a6 = 4 × 243 a6 = 972

Geometric Sequences 6.6 Exercises

Question 2. CRITICAL THINKING Why do the points of a geometric sequence lie on an exponential curve only when the common ratio is positive? Answer: The points of any geometric sequence with a positive common ratio lie on an exponential curve. Because the base of the exponential function must be positive. If the common ratio is negative then the points are alternately positive and negative.

In Exercises 3–8, find the common ratio of the geometric sequence.

Question 4. 36, 6, 1, \(\frac{1}{6}\), . . . Answer: Given, 36, 6, 1, \(\frac{1}{6}\), . . . a1 = 36 a2 = 6 r = a2/a1 = 6/36 = \(\frac{1}{6}\) So, the common ratio is \(\frac{1}{6}\)

Question 6. 0.1, 1, 10, 100, . . . Answer: Given, 0.1, 1, 10, 100, . . . a1 = 0.1 a2 = 1 a3 = 10 a4 = 100 r = a2/a1 = 1/0.1 = 10 r = a3/a2 = 10/1 = 10 So, the common ratio is 10.

Question 8. -162, 54, -18, 6, . . . Answer: Given, -162, 54, -18, 6, . . . a1 = -162 a2 = 54 r = a2/a1 = 54/-162 = -1/3 So, the common ratio is -1/3

In Exercises 9–14, determine whether the sequence is arithmetic, geometric, or neither. Explain your reasoning.

Question 10. -1, 4, -7, 10, . . . Answer: Given, -1, 4, -7, 10, . . . a1 = -1 a2 = 4 d = a2 – a1 d = 4 – (-1) = 5 d = -7 – 4 = -11 There is no common difference. The sequence is neither arithmetic nor geometric.

Question 12. \(\frac{3}{49}\), \(\frac{3}{7}\), 3, 21, . . . Answer: Given, \(\frac{3}{49}\), \(\frac{3}{7}\), 3, 21, . . . a1 = \(\frac{3}{49}\) a2 = \(\frac{3}{7}\) r = a2/a1 r = \(\frac{3}{7}\)/\(\frac{3}{49}\) = 7 r = a4/a3 r = 21/3 = 7 So, the common ratio is 7. The sequence is geometric.

Question 14. -25, -18, -11, -4, . . . Answer: Given, -25, -18, -11, -4, . . . a1 = -25 a2 = -18 d = -18 – (-25) d = -18 + 25 = 7 The common difference is 7. So, the sequence is arithmetic.

In Exercises 15–18, determine whether the graph represents an arithmetic sequence, a geometric sequence, or neither. Explain your reasoning.

In Exercises 19–24, write the next three terms of the geometric sequence. Then graph the sequence.

Question 20. -3, 12, -48, 192, . . . Answer: Given, -3, 12, -48, 192, . . . a1 = -3 a2 = 12 r = a2/a1 r = 12/-3 r = -4 r = -48/12 = -4 The common ratio is -4. a5 = 192 × – 4 = -768 a6 = -768 × -4 = 3072 a7 = 3072 × -4 = -12288 The next three terms are -768, 3072, -12288.

Question 22. -375, -75, -15, -3, . . . Answer: Given, -375, -75, -15, -3, . . . a1 = -375 a2 = -75 a3 = -15 a4 = -3 r = -75/-375 = 1/5 r = -15/-75 = 1/5 The common ratio is 1/5 a5 = -3 × (1/5) = -3/5 a6 = -3/5 × 1/5 = -3/25 a7 = -3/25 × 1/5 = -3/75 So, the next three terms are -3/5, -3/25, -3/75.

Question 24. \(\frac{16}{9}\), \(\frac{8}{3}\), 4, 6, . . . Answer: Given, \(\frac{16}{9}\), \(\frac{8}{3}\), 4, 6, . . . a1 = \(\frac{16}{9}\) a2 = \(\frac{8}{3}\) r = \(\frac{8}{3}\)/\(\frac{16}{9}\) = \(\frac{3}{2}\) a3 = 4 a4 = 6 a5 = 6 × \(\frac{3}{2}\) = 9 a6 = 9 × \(\frac{3}{2}\) = \(\frac{27}{2}\) a7 = \(\frac{27}{2}\) × \(\frac{3}{2}\) = \(\frac{81}{4}\) So, the next three terms are 9, \(\frac{27}{2}\), \(\frac{81}{4}\)

In Exercises 25–32, write an equation for the nth termof the geometric sequence. Then find a 6 .

Question 26. 0.6, -3, 15, -75, . . . Answer: Given, 0.6, -3, 15, -75, . . . a1 = 0.6 a2 = -3 r = -3/0.6 = -5 r = 15/-3 = -5 a5 = -75 × – 5 = 375 a6 = 375 × -5 = -1875 So the 6th term of the geometric sequence is -1875.

Question 28. 0.1, 0.9, 8.1, 72.9, . . . Answer: Given, 0.1, 0.9, 8.1, 72.9, . . . a1 = 0.1 a2 = 0.9 a3 = 8.1 r = 0.9/0.1 = 9 r = 8.1/0.9 = 9 a4 = 72.9 a5 = 72.9 × 9 = 656.1 a6 = 656.1 × 9 = 5904.9 So, the 6th term is 5904.9

MATHEMATICAL CONNECTIONS In Exercises 39 and 40, (a) write a function that represents the sequence of figures and (b) describe the 10th figure in the sequence.

Question 42. REASONING Write a sequence that represents the perimeter of the graphing calculator screen in Exercise 34 after you zoom out n times. Determine whether the sequence is arithmetic, geometric, or neither. Explain your reasoning. Answer: The sequence is 96, 384, 1536,… a1 = 96 a2 = 384 r = a2/a1 r = 384/96 = 4 The given sequence is a geometric sequence.

Question 48. OPEN-ENDED Write a sequence that has a pattern but is not arithmetic or geometric. Describe the pattern. Answer: 9, -18, 27, -36, . . . a1 = 9 a2 = -18 r = -18/9 = -9 r = 27/-18 = -3/2 The sequence is neither arithmetic nor geometric sequence because it does not have any common difference.

Question 50. DRAWING CONCLUSIONS A college student makes a deal with her parents to live at home instead of living on campus. She will pay her parents $0.01 for the first day of the month, $0.02 for the second day, $0.04 for the third day, and so on. a. Write an equation that represents the nth term of the geometric sequence. Answer: Given, She will pay her parents $0.01 for the first day of the month, $0.02 for the second day, $0.04 for the third day, and so on. 0.01, 0.02, 0.04,… a1 = 0.01 a2 = 0.02 r = 0.02/0.01 = 2 r = 2 an = a1 . r^(n-1) an = 0.01 (2)^(n-1)

b. What will she pay on the 25th day? Answer: an = 0.01 (2)^(n-1) n = 25 a25 = 0.01 (2)^(25-1) a25 = 0.01(2)^24 a25 = 167772.16 So, she will pay %167772.16 on 25th day.

c. Did the student make a good choice or should she have chosen to live on campus? Explain. Answer: an = 0.01 (2)^(n-1) n = 30 a30 = 0.01 (2)^(30-1) a30 = 0.01(2)^29 a30 = 5368709.12 a30 = 5368709.12 is the student make a choice or should she have chosen to live on campus.

Use residuals to determine whether the model is a good fit for the data in the table. Explain.

Lesson 6.7 Recursively Defined Sequences

How can you define a sequence recursively? A recursive rule gives the beginning term(s) of a sequence and a recursive equation that tells how a n is related to one or more preceding terms.

Question 3. How can you define a sequence recursively? Answer: In order to define a sequence recursively, we must state the first term and then state a rule for how each successive term can be described from the one before it.

Question 4. Use the Internet or some other reference to determine the mathematician who first described the sequences in Explorations 1 and 2. Answer: A recursive sequence, also known as a recurrence sequence, is a set of numbers that is formed by solving a recurrence equation. The terms of a recursive series can be symbolically represented using a variety of notations, such as, or f[], where is a symbol for the sequence.

Write a recursive rule for the sequence.

Question 5. 8, 3, -2, -7, -12, . . . Answer: 8, 3, -2, -7, -12, . . . a1 = 8 a2 = 3 d = 8 – 3 = 5 an = a1 + (n – 1)d an = 8 + (n – 1)5 an = 8 + 5n – 5 an = 3 + 5n

Question 6. 1.3, 2.6, 3.9, 5.2, 6.5, . . . Answer: Given 1.3, 2.6, 3.9, 5.2, 6.5, . . . a1 = 1.3 a2 = 2.6 d = 1.3 an = a1 + (n – 1)d an = 1.3 + (n – 1)1.3 an = 1.3 + 1.3n – 1.3 an = 1.3n

Question 7. 4, 20, 100, 500, 2500, . . . Answer: Given, 4, 20, 100, 500, 2500, . . . a1 = 4 r = 5 an = a1. r^(n-1) an = 4 . 5^(n-1)

Question 8. 128, -32, 8, -2, 0.5, . . . Answer: Given, 128, -32, 8, -2, 0.5, . . . a1 = 128 r = -1/4 an = a1. r^(n-1) an = 128 . (-1/4)(^(n-1)

Write an explicit rule for the recursive rule.

Question 10. a 1 = -45, a n = a n – 1 + 20 Answer: Given a1 = 45 d = 20 an = a1 + (n – 1)d an = 45 + (n – 1)20 an = 45 + 20n – 20 an = 20n + 25

Question 11. a 1 = 13, a n = -3a n – 1

Answer: a1 = 13 r = -3 common ration  = -3 a n = -3a n – 1 an = 13 . (-3)^(n-1)

Write a recursive rule for the explicit rule.

Question 12. a n = -n + 1 Answer: a n = -n + 1 a1 = 0 common difference = -1

Question 13. a n = -2.5(4) n – 1

Answer: a n = -2.5(4) n – 1 a1 = 2.5 r = 4

Write a recursive rule for the sequence. Then write the next three terms of the sequence.

Question 14. 5, 6, 11, 17, 28, . . . Answer: Given, 5, 6, 11, 17, 28, . . . a1 = 5 a2 = 6 a3 = a1 + a2 = 5 + 6 = 11 a4 = a2 + a3 = 6 + 11 = 17 a5 = a3 + a4 = 11 + 17 = 28 a6 = a4 + a5 = 17 + 28 = 45 a7 = a5 + a6 = 28 + 48 = 73 a8 = a6 + a7 = 45 + 73 = 118 a9 = a7 + a8 = 73 + 118 = 191 So, the next three terms are 73, 118, 191.

Question 15. -3, -4, -7, -11, -18, . . . Answer: Given, -3, -4, -7, -11, -18, . . . a1 = -3 a2 = -4 a3 = -7 a4 = -11 a5 = -18 a6 = a4 + a5 = -11 – 18 = -29 a7 = a5 + a6 = -18 – 29 = -47 a8 = a6 + a7 = -29 – 47 = -76 a9 = a7 + a8 = -47 – 76 = -123 So, the next three terms are -47, -76, -123.

Question 16. 1, 1, 0, -1, -1, 0, 1, 1, . . . Answer: Given, 1, 1, 0, -1, -1, 0, 1, 1, . . . a1 = 1 a2 = 1 a3 = 0 The sequence has no common ratio a4 = a3 – a2= -1 a5 = -1 a6 = 0 a7 = 1 a8 = 1 a9 = a8 – a7 = 1 – 1 = 0 a10 = a9 – a8 = 0 – 1 = -1 a11 = a10 – a9 = -1

Question 17. 4, 3, 1, 2, -1, 3, -4, . . . Answer: Given, 4, 3, 1, 2, -1, 3, -4, . . . The sequence has no common ratio a1 = 4 a2 = 3 a3 = 1 a4 = 2 a5 = -1 a6 = 3 a7 = -4 a8 = a6 – a7 = 3 – (-4) = 7 a9 = a7 – a8 = -4 – 7 = -11 a10 = a8 – a9 = 7 – (-11) = 7 + 11 = 18 So, the next three terms are 7, -11, 18.

Recursively Defined Sequences 6.7 Exercises

In Exercises 3–6, determine whether the recursive rule represents an arithmetic sequence or a geometric sequence.

Question 4. a 1 = 18, a n = a n – 1 + 1 Answer: The rule a n = a n – 1 + 1 is in the arithmetic sequence.

Question 6. a 1 = 3, a n = -6a n – 1 Answer: The rule a n = -6a n – 1 is an = r . a(n-1). This represents a geometric sequence.

In Exercises 7–12, write the first six terms of the sequence. Then graph the sequence.

Question 12. a 1 = -7, a n = -4a n – 1 Answer: a 1 = -7, a n = -4a n – 1 a2 = -4a 2 – 1 = -4(-7) = 28 a3 = -4a 3 – 1 = -4(28) = -112 a4 = -4a 4 – 1 = -4(-112) = 448 a5 = -4a 5 – 1 = -4(448) = -1792 a6 = -4a 6 – 1 = -4(-1792) = -7168

In Exercises 13–20, write a recursive rule for the sequence.

Question 16. 3, 11, 19, 27, 35, . . . Answer: The sequence is arithmetic, with first term a1 = 3 and common difference d = 8. an = an-1 + d an = a(n-1) + 8

Question 18. 5, -20, 80, -320, 1280, . . . Answer: The sequence is geometric, with the first term a1 = 5 and common ratio = 4 an = r . an-1 an = -4 . an-1

In Exercises 23–28, write an explicit rule for the recursive rule.

Question 24. a 1 = 8, a n = a n – 1 – 12 Answer: The recursive rule represents an arithmetic sequence. a1 = 8 d = -12 a n = a n – 1 – 12 a n = 8n – 12

Question 26. a 1 = -2, a n = 9a n – 1 Answer: The recursive rule represents a geometric sequence. an = a1.r^(n-1) an = -2(9)^(n-1)

Question 28. a 1 = 5, a n = -5a n – 1 Answer: The recursive rule represents a geometric sequence. a1 = 5 r = -5 an = a1.r^(n-1) an = 5(-5)^(n-1)

In Exercises 29–34, write a recursive rule for the explicit rule.

Question 30. a n = -4n + 2 Answer: The explicit rule represents an arithmetic sequence a1 = -4 common difference = 2

Question 32. a n = 6n – 20 Answer: The explicit rule represents an arithmetic sequence a1 = 6(1) – 20 = -14 common difference = 6 So, the recursive rule for the sequence is a1 = -14, an = an-1 + 6

Question 34. a n = -81(\(\frac{2}{3}\)) n – 1 Answer: The explicit rule represents a geometric sequence with first sequence a1 = -81 r = 2/3 So, a recursive rule for the sequence is a1 = -81, an = \(\frac{2}{3}\)a(n-1)

In Exercises 35–38, graph the first four terms of the sequence with the given description. Write a recursive rule and an explicit rule for the sequence.

Question 36. The first term of a sequence is 16. Each term of the sequence is half the preceding term. Answer: Given, The first term of a sequence is 16. Each term of the sequence is half the preceding term. a1 = 16 r = 1/2 an = a1. r^(n-1) an = 16. (1/2)^(n-1)

Question 38. The first term of a sequence is 19. Each term of the sequence is 13 less than the preceding term. Answer: Given, The first term of a sequence is 19. Each term of the sequence is 13 less than the preceding term. a1 = 19 d = -13 an = a(n-1) + d an = 19n – 13

In Exercises 39–44, write a recursive rule for the sequence. Then write the next two terms of the sequence.

Question 40. 10, 9, 1, 8, -7, 15, . . . Answer:

Question 42. 6, 1, 7, 8, 15, 23, . . . Answer:

In Exercises 47–51, the function f represents a sequence. Find the 2nd, 5th, and 10th terms of the sequence.

Question 48. f(1) = -1, f(n) = 6f(n – 1) Answer: The function represents a geometric sequence a1 = -1 r = 6 an = a1 . r^(n-1) an = -1(6)^(n-1) a2 = -1(6)^1 = -1 × 6 = -6 a3 = -1(6)² = -36 a4 = -1(6)³ = -216

Question 50. f(1) = 4, f(2) = 5, f(n) = f(n – 2) + f(n – 1) Answer: f(1) = 4, f(2) = 5, f(n) = f(n – 2) + f(n – 1) f(3) = f(3 – 2) + f(3 – 1) f(3) = 4 + 5 = 9 f(4) = f(4 – 2) + f(4 – 1) f(4) = 9 + 5 = 14 f(5) = f(5 – 2) + f(5 – 1) f(5) = 14 + 9 = 23

Simplify the expression.

Question 60. 5x + 12x Answer: Combine the like terms 5x + 12x = 17x

Question 62. 2d – 7 – 8d Answer: 2d – 7 – 8d Combine the like terms 2d – 8d – 7 -6d – 7

Write a linear function f with the given values .(Section 4.2)

Question 64. f(2) = 6, f(-1) = -3 Answer: m = (6+3)/(2+1) = 9/3 = 3 y – 6 = m(x – 2) y – 6 = 3(x – 2) y – 6 = 3x – 6 y = 3x

Question 66. f(-3) = 5, f(-1) = 5 Answer: m = (5-5)/(-1+3) = 0 y – 5 = m(x + 3) y – 5 = 0(x – 2) y – 5 = 0 y = 5

Exponential Functions and Sequences Performance Task: The New Car

Core Concepts Section 6.5 Property of Equality for Exponential Equations, p. 326 Solving Exponential Equations by Graphing, p. 328

Section 6.6 Geometric Sequence, p. 332 Equation for a Geometric Sequence, p. 334

Section 6.7 Recursive Equation for an Arithmetic Sequence, p. 340 Recursive Equation for a Geometric Sequence, p. 340

Question 1. How did you decide on an appropriate level of precision for your answer in Exercise 49 on page 330?

Question 2. Explain how writing a function in Exercise 39 part (a) on page 337 created a shortcut for answering part (b).

Question 3. How did you choose an appropriate tool in Exercise 52 part (b) on page 345?

Performance Task

The New Car

Exponential Functions and Sequences Chapter Review

Question 1. y 3 • y -5

Answer: y 3 • y -5 When bases are equal powers should be added. y 3-5 y 3 • y -5 = y -2

Question 2. \(\frac{x^{4}}{x^{7}}\) Answer: \(\frac{x^{4}}{x^{7}}\) \(\frac{x^{m}}{x^{n}}\) = x m-n \(\frac{x^{4}}{x^{7}}\) = x 4-7 = x -3

Question 3. (x 0 y 2 ) 3

Answer: (x 0 y 2 )³ = (x 0 )³(y²)³ = y 6

Question 5. \(\sqrt[3]{8}\) Answer: \(\sqrt[3]{8}\) 8 = 2³ \(\sqrt[3]{8}\) = 2

Question 6. \(\sqrt[5]{-243}\) Answer: \(\sqrt[5]{-243}\) = -3

Question 7. 625 3 / 4

Answer: \(\sqrt[4]{625^3}\) = 5³ = 125

Question 8. (-25) 1 / 2

Answer: (-25) 1 / 2 –\(\sqrt[2]{25}\) = -5

Question 9. f(x) = -4 (\(\sqrt[1]{4}\)) x

Question 10. f(x) = 3 x + 2

Rewrite the function to determine whether it represents exponential growth or exponential decay. Identify the percent rate of change.

Question 15. f(t) = 4(1.25) t + 3

Answer: Given, f(t) = 4(1.25) t + 3 t + 3 > 0 b > 1 r + 1 = 1.25 r = 1.25 – 1 r = 0.25 r = 25% It represents exponential growth.

Question 16. y = (1.06) 8t

Answer: y = (1.06) 8t b > 1 r + 1 = 1.06 r = 0.06 r = 6% It represents exponential growth.

Question 17. f(t) = 6(0.84) t – 4

Answer: f(t) = 6(0.84) t – 4 b < 1 1 – r = 0.84 r = 1 – 0.84 r = 0.16 r = 16% It represents exponential decay.

Question 18. You deposit $750 in a savings account that earns 5% annual interest compounded quarterly. (a) Write a function that represents the balance after t years. (b) What is the balance of the account after 4 years? Answer: Given, You deposit $750 in a savings account that earns 5% annual interest compounded quarterly. P = 750 R = 5%/4 = 1.25% m = number of compounding = 4 FV = P(1 + r) nm FV = 750(1.0125) 4t

Question 19. The value of a TV is $1500. Its value decreases by 14% each year. (a) Write a function that represents the value y (in dollars) of the TV after t years. (b) Find the approximate monthly percent decrease in value. (c) Graph the function from part (a). Use the graph to estimate the value of the TV after 3 years. Answer: Given, The value of a TV is $1500. Its value decreases by 14% each year. 100 – 14 = 86 14/12 = 1.16 1500 × 0.86^t = x x = 1500 × 0.86³ x = 1500 × 0.636 x = 954.08 So, the function that represents this situation is 1500 × 0.86^t = x, the monthly percent decrease is 0.16% and after 3 years the car will be valued at $954.08

Solve the equation.

Question 20. 5 x = 5 3x – 2

Answer: 5 x = 5 3x – 2 When bases are equal powers should be equated x = 3x – 2 x – 3x = -2 -2x = – 2 x = 1

Question 21. 3 x – 2 = 1 Answer: 3 0 = 1 3 x – 2 = 3 0 x – 2 = 0 x = 2

Question 22. -4 = 6 4x – 3

Answer: -4 = 6 4x – 3 -4 = 6 4x /6³ -4 × 6³ = 6 4x -4 × 216 = 6 4x -864 = 6 4x

Question 23. (\(\frac{1}{3}\)) 2x + 3 = 5 Answer: (\(\frac{1}{3}\)) 2x + 3 = 5 (\(\frac{1}{3}\)) 2x × (\(\frac{1}{3}\))³ = 5 (\(\frac{1}{3}\)) 2x × \(\frac{1}{8}\) = 5 (\(\frac{1}{3}\)) 2x = 40

Question 24. (\(\frac{1}{16}\)) 3x = 64 2(x + 8)

Answer: 4 -6x = 4 2(x + 8)3 when bases are equal powers should be equated -6x = 6x + 48 -6x – 6x = 48 -12x = 48 x = -4

Question 25. 27 2x + 2 = 81 x + 4

Answer: 27 2x + 2 = 81 x + 4 when bases are equal powers should be equated 3 (2x + 2)3 = 3 (x + 4)4 6x + 6 = 4x + 16 6x – 4x = 16 – 6 2x = 10 x = 5

Decide whether the sequence is arithmetic, geometric, or neither. Explain your reasoning. If the sequence is geometric, write the next three terms and graph the sequence.

Question 27. 9, -18, 27, -36, . . . Answer: 9, -18, 27, -36, . . . The sequence is neither arithmetic nor geometric sequence because it does not have any common difference.

Question 28. 375, -75, 15, -3, . . . Answer: Given, 375, -75, 15, -3, . . . a1 = 375 common ratio = 375/-75 = -5 an = a1 × r n-1 n = 5 a5 = 375 × r 5-1 a5 = 375 × r 4 a5 = 375 × (-5) 4 = 234375 an = a1 × r n-1 n = 6 a6 = 375 × r 6-1 a5 = 375 × r 5 a5 = 375 × (-5) 5 = -1171875

Write an equation for the nth term of the geometric sequence. Then find a 9 .

Question 29. 1, 4, 16, 64, . . . Answer: Given, 1, 4, 16, 64, . . . an = a1 × r n-1 an = 1(4) n-1 n = 9 an = 4 9-1 = 4 8 = 65536

Question 30. 5, -10, 20, -40, . . . Answer: Given, 5, -10, 20, -40, . . . an = a1 × r n-1 an = 5(-2) 9-1 n = 9 an = 5 × (-2) 9-1 = 5 × 256 = 1280

Question 31. 486, 162, 54, 18, . . . Answer: a1 = 486 r = 486/162 = 3 an = a1 × r n-1 an = 486(3) 9-1 n = 9 an = 486 × (3) 8 = 486 × 6561 = 3188646

Write the first six terms of the sequence. Then graph the sequence.

Question 33. a 1 = -4, a n = -3a n – 1

Answer: a 1 = -4, a n = -3a n – 1 a1 = 4 a2 = -3(4) = -12 a3 = -3(-12) = 36 a4 = -3(36) = -108 a5 = -3(-108) = 324 a6 = -3(324) = -972

Question 34. a 1 = 32, a n = \(\frac{1}{4}\)a n − 1

Answer: a1 = 32 a 1 = 32, a n = \(\frac{1}{4}\)a n − 1 a2 = \(\frac{1}{4}\)a 2 − 1 = \(\frac{1}{4}\)(32) = 8 a3 = \(\frac{1}{4}\)a 3 − 1 = \(\frac{1}{4}\)(8) = 2 a4 = \(\frac{1}{4}\)a 3 − 1 = \(\frac{1}{4}\)(2) = 1/2 a5 = \(\frac{1}{4}\)a 4 − 1 = \(\frac{1}{4}\)(1/2) = 1/8 a6 = \(\frac{1}{4}\)a 5 − 1 = \(\frac{1}{4}\)(1/8) = 1/32

Question 35. 3, 8, 13, 18, 23, . . . Answer: a1 = 3 d = 8 – 3 = 5 a n = a n – 1 + 5

Question 36. 3, 6, 12, 24, 48, . . . Answer: a1 = 3 common ratio = 6/3 = 2 r = 2 a n = 2a n – 1

Question 37. 7, 6, 13, 19, 32, . . Answer: Given, 7, 6, 13, 19, 32, . . a1 = 7 a2 = 6 a3 = 13 a3 = a1 + a2 6 + 7 = 13 a4 = a2 + a3 = 6 + 13 = 19 an = a(n-1) + a(n – 2)

Question 38. The first term of a sequence is 8. Each term of the sequence is 5 times the preceding term. Graph the first four terms of the sequence. Write a recursive rule and an explicit rule for the sequence. Answer: Given, The first term of a sequence is 8. Each term of the sequence is 5 times the preceding term. a1 = 8 a(n+1) = 5 . an a2 = 8 × 5 = 40 a3 = 40 × 5 = 200 a4 = 200 × 5 = 1000 a5 = 1000 × 5 = 5000 an = 8(5)^n-1 an = 8/5 (5)^n

Exponential Functions and Sequences Chapter Test

Question 1. –\(\sqrt[4]{16}\) Answer: –\(\sqrt[4]{16}\) The fourth root of 16 is 2 16 can be written as 2 4 –\(\sqrt[4]{16}\) = -2

Question 2. 729^\(1 / 6\) Answer: Given, 729^\(1 / 6\) 729 can be written as 3 6 Sixth root of 729 is 3

Question 3. (-32)^\(7 / 5\) Answer: Given, (-32)^\(7 / 5\) 32 can be written as 2 5 Fifth root of 32 is 2 (-2) 7 = – 2 × -2× -2 × -2 × -2 × -2 × -2 = -128

Question 4. z -2 • z 4

Answer: z -2 • z 4 When Bases are equal powers should be added. z -2 + 4 = z²

Write and graph a function that represents the situation.

Question 7. Your starting annual salary of $42,500 increases by 3% each year. Answer: Given, Your starting annual salary of $42,500 increases by 3% each year. y = 42,500(1 + 3%) y = 42,500(1 + 0.03)

Question 8. You deposit $500 in an account that earns 6.5% annual interest compounded yearly. Answer: Given, You deposit $500 in an account that earns 6.5% annual interest compounded yearly. y = 500(1 + 6.5%) y = 500(1 + 0.065)

Write an explicit rule and a recursive rule for the sequence.

Question 11. 2 x = \(\frac{1}{128}\) Answer: 128 = 2 7 2 x = \(\frac{1}{2 7 }\) 2 x = 2 -7 x = -7

Question 12. 256 x + 2 = 16 3x – 1

Answer: 256 x + 2 = 16 3x – 1 16 2x + 4 =16 3x – 1 2x + 4 = 3x – 1 2x – 3x = -1 – 4 -1x = -5 x = 5

Question 14. \(\frac{5^{a}}{5^{b}}\) Answer: \(\frac{5^{a}}{5^{b}}\) = 5 a-b

Question 15. 9 a • 9 -b

Answer: When bases are equal powers should be added. 9 a • 9 -b = 9 a-b

Question 16. The first two terms of a sequence are a 1 = 3 and a 2 = -12. Let a 3 be the third term when the sequence is arithmetic and let b 3 be the third term when the sequence is geometric. Find a 3 – b 3 . Answer: The first two terms of a sequence are a 1 = 3 and a 2 = -12. d = a2 – a1 d = -12 – 3 = -15 a3 = a1 + 2d = 3 + 2(-15) = 3 – 30 = -27 a3 = -27 When the sequence is geometric r = a2/a1 = -12/3 = -4 r = -0.5 b3 = third term b3 = ar² = 3(-0.5)² = 3(0.25) = 0.75 a3 – b3 = -27 – 0.75 = -27.75

Question 17. At sea level, Earth’s atmosphere exerts a pressure of 1 atmosphere. Atmospheric pressure P (in atmospheres) decreases with altitude. It can be modeled by P =(0.99988) a , where a is the altitude (in meters). a. Identify the initial amount, decay factor, and decay rate. b. Use a graphing calculator to graph the function. Use the graph to estimate the atmospheric pressure at an altitude of 5000 feet. Answer: Given, Initial amount = 1 Decay factor = 0.99988 P =(0.99988) a , where a is the altitude P = (0.99988) 0 = 1 The pressure at sea level = 0 This is like the decay constant. From the model, P =(0.99988) a the decay constant/ factor is 0.99988 Decay rate = 1 – 0.99988 = 0.00012 Percentage = 0.12%.

Exponential Functions and Sequences Cumulative Assessment

Question 6. A data set consists of the heights y (in feet) of a hot-air balloon t minutes after it begins its descent. An equation of the line of best fit is y = 870 – 14.8t. Which of the following is a correct interpretation of the line of best fit? A. The initial height of the hot-air balloon is 870 feet. The slope has no meaning in this context. B. The initial height of the hot-air balloon is 870 feet, and it descends 14.8 feet per minute. C. The initial height of the hot-air balloon is 870 feet, and it ascends 14.8 feet per minute. D. The hot-air balloon descends 14.8 feet per minute. The y-intercept has no meaning in this context. Answer: The initial height of the hot-air balloon is 870 feet, and it descends 14.8 feet per minute. Option B is the correct answer.

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Math 230: Abstract and Discrete Mathematics

Homework 2 consists of the following exercises. Due Monday September 9: 8.2, 8.4, 8.8, 8.10, 8.15. 9.2, 9.5, 9.7, 9.9, 9.11. 10.1, 10.4, 10.9, 10.12, 10.13. 11.1, 11.2, 11.4, 11.7. 12.1, 12.5, 12.9, 12.12, 12.21, 12.24, 12.30. Homework 2 Solutions . Solutions to 10.12 and 10.13 .

Homework 3 consists of the following exercises. Due Monday September 23: Read Sections 17 and 22. 17.3, 17.4, 17.5, 17.8, 17.11, 17.16, 17.21, 17.37. 22.4, 22.5, 22.6, 22.7. Solutions .

Homework 4 consists of the following exercises. Due Wednesday October 2: Read Sections 14, 15 and 16. 14.1, 14.3 (no need to prove here), 14.5, 14.6, 14.10, 14.17 (proofs required in 14.6 and 14.10). 15.1, 15.3 (no proof required here), 15.7, 15.8, 15.14, 15.15. 16.1, 16.10, 16.15. Solutions .

Homework 5 consists of the following exercises. Due Friday October 11: 22.16, 22.17 20.4, 20.5, 20.9, 20.10, 20.13. 21.3, 21.7, 21.9. Solutions .

Homework 6 consists of the following exercises. Due on Monday October 26: 24.1, 24.2, 24.5, 24.6, 24.8, 24.14, 24.16, 24.17, 24.20. Solutions . (Note: Solutions to 24.6 have some errors. 24.6a should be "all the odd integers". 24.6f should be "[0,1]"

Homework 7 consists of the following exercises. Due on Friday November 6: 25.2, 25.6, 25.7, 25.9, 25.13, 26.7, 26.9, 26.10. Also do 26.1 to practice. Solutions .

Homework 8 consists of the following exercises. Due on Friday November 20: Homework 8 Homework 8 Solutions .

Homework 9 consists of the following exercises. Due on Wednesday December 4: 54.1, 54.2, 54.3, 54.4, 54.8. 55.1, 55.2, 55.5, 55.7. 56.1, 56.5. Suggested extra work : After 54.4, think about the width of the poset defined in 54.3 for any n (Hint: Consider n even and n odd separately). 54.9 is a very good exercise to get more practice with posets. 56.7 is "alphabetical" ordering in disguise (this ordering has the more technical name of "lexicographic ordering"). Solutions .