Unit 1 Sequences, Series, Exponential and Logarithmic Functions SpringBoard Precalculus, Pages 1-116

Overview In this unit you will study recursive and explicit representations of arithmetic and geometric sequences. You will also study exponential, logarithmic, and power functions and explore the key features of their graphs. In addition, you will look at transformations, compositions, and inverses of functions.

Standards: MAFS.912.F-LE.1.1 Distinguish between situations that can be modeled with linear functions and with exponential functions. MAFS.912.F-LE.1.2 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or 2 input-output pairs (include reading these from a table). MAFS.912.F-LE.1.3 Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function. MAFS.912.F-LE.1.4 For exponential models, express as a logarithm the solution to abct =d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology. MAFS.912.F-BF.2.5 Understand the inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents. MAFS.912.F-BF.1.1 Write a function that describes a relationship between two quantities. a. Determine an explicit expression, a recursive process, or steps for calculation from a context. b. Combine standard function types using arithmetic operations. c. Compose functions. MAFS.912.F-BF.2.3 Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them. MAFS.912.F-BF.2.4 Find inverse functions. a. Solve an equation of the form f(x)=c for a simple function f that has an inverse and write an expression for the inverse. b. Verify by composition that one function is the inverse of another. c. Read values of an inverse function from a graph or a table, given that the function has an inverse. d. Produce an invertible function from a non-invertible function by restricting the domain. MAFS.912.A-APR.1.1 Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials. MAFS.912.A-APR.2.3 Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial. MAFS.912.F-IF.3.7 Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. a. Graph linear and quadratic functions and show intercepts, maxima, and minima. b. Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. c. Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior. d. Graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing end behavior. e. Graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude, and using phase shift.

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Embedded Assessment 1 SequencesThe Old Square Craft Pages 45-46 (center) List terms of a sequence (Lesson 1-1, 2-1) Identify types of sequences (Lesson 1-1, 2-1) Express a sequence recursively and explicitly (Lesson 3-1, 3-2) place on separate line Express the sum of a sequence in sigma notation (Lesson1-2) Calculate the sum of a sequence (Lesson 1-2, 2-2)

Embedded Assessment 2: Exponential and Logarithmic FunctionsPopulation Explosion Pages 75-76 Write an exponential equation for a real world application problem (Lesson 4-1) Graph an exponential equation (Lesson4-2) Find the y-intercept of an exponential equation and state what it represents (Lesson 4-2) Solve an exponential equation using logarithms and stating the properties used (Lesson 5-2) Write and solve a logarithmic equation (Lesson 5-3)

Find the sum of an infinite series (Lesson 2-3)

Embedded Assessment 3 Transformations, Compositions, and InversesFeeding Frenzy Pages 115-116 Graph a power function (Lesson 7-2) State the domain and range of a power function (Lesson 7-2) State whether a function increases or decreases (Lesson 7-2) Describe the transformation of a graph (Lesson 6-1) Find the inverse of a function (Lesson 8-2) Composition of functions (Lesson 8-1)

Vocabulary (Academic/Math): 

Converge, depreciation, diverge, conjecture, sigma notation, sequence of partial sums, mathematical induction, polar grid, common ratio, series, nth partial sum, infinite sequence, infinite series, iteration, recursive, explicit form, exponential function, interest rate, exponential growth factor, exponential decay factor, half-life, logarithm, common logarithm, strictly monotonic, parent function, even function, odd function, composition, inverse function

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Support for Lesson1-1 Learning Targets for lesson are found on page 3. Main Ideas for success in lesson 1-1: 

 Write an expression for a sequence.



Use subscript notation.  Vocabulary used in this lesson includes: sequence, subscript notation  Standards for Mathematical Practice to be demonstrated are Make use of structure, Model with mathematics, Construct viable arguments, Critique the reasoning of others, Make sense of problems

Practice Support for Lesson1-1  This example is a model to help solve Practice problem 15. EXAMPLE: {An} is an arithmetic sequence with a1 = 3 and d = 3.5. a. Write the first five terms of this sequence. a1 is the first term in the sequence and d is the common difference from a1 to a 2 . To find a 2 , add 3 + 3.5 = 6.5

a3  6.5  3.5  10 a4  10  3.5  13.5 a5  13.5  3.5  17 b. Write an expression for An in terms of n, the term number. The formula for the nth term of an arithmetic sequence is An  a1  (n  1)d so 3  (n  1)3.5 use the distributive property, 3  3.5n  3.5 combine like terms 3.5n  .5 c. Write an expression for An+1 using the expression for An.

An1 represents the next consecutive term and An1  An  d therefore, An1  3.5n  .5  3.5 combine like terms An1  3.5n  3.

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Practice Support for Lesson 1-1Continued:  This example is a model to help solve Practice problem 16. EXAMPLE Tim rents 2 DVDs each week at Bopper’s . Write an algebraic expression for Wn, the total number of points earned in the nth week. Using the same formula Wn  a1  (n  1)d where a1 is the total points earned for the first week from the chart on page 3. The first rental is 3 points and the second is 2 points. The total for the first week is 5 points and the common difference d = 4. The formula with the numbers substituted is Wn  5  (n  1)4, use the distributive property (n-1)4 = 4n – 1 Wn  5  4n  4, combine like terms Wn  4n  1.  This example is a model to help solve Practice problem 17. EXAMPLE: What does the notation {an} represent? {an} represents the entire sequence A1 , A2 , A3 ,....,An th

An represents the n term of the sequence

 This example is a model to help solve Practice problem 18. EXAMPLE: For the sequence 8, −3, −14, −25,…, determine the value of A10. Find the common difference (d) by subtracting 3  8  11 and the first term a1  8. To find the 10 term, th

substitute the values into the formula An  a1  (n  1)d and An  8  (10  1)( 11)  91.

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Additional Practicefor Lesson 1-1: 1. Marena makes 16 bracelets per week for charity. She has already made 112 bracelets for charity. a. List the first five terms of the sequence. b. Write an algebraic equation in terms of n for the total number of bracelets made.

Answers to Additional Practice1-1: 1. a. 128, 144, 160, 176, 192 b. Bn = 112 + 16n or Bn = 128 + 16(n – 1)

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Support for Lesson1-2 Learning Targets for lesson are found on page 8. Main Ideas for success in lesson 1-2:

 Use sigma notation to represent a series.  Write the algebraic form of an arithmetic sequence. Calculate the nth term or nth partial sum of an arithmetic series. Vocabulary used in this lesson includes sigma notation, arithmetic sequence, arithmetic series, and sequence of partial sums  Standards for Mathematical Practice to be demonstrated are Reason abstractly, Construct viable arguments, Model with Mathematics.

Practice Support for Lesson1-2  This example is a model to help solve Practice problems 11. EXAMPLE: Write the general term for the arithmetic sequence with a1 = 6 and d=4. Use the formula for finding the nth term of a arithmetic sequence An  a1  (n  1)d , substitute the values of a1 and d, so An  6  (n  1)4. Distribute (n – 1)4 = 4n – 4, An = 6 + 4n – 4, combine like terms An  4n  2  This example is a model to help solve Practice problems 12. EXAMPLE: Write the associated arithmetic series with a1 = 6 and d=4 using the first four terms, and express that sum using sigma notation. A series is the sum of the terms of a sequence. The first four terms of the sequence is 6, 6 + 4 =10, 10 + 4 = 14, 14 + 4 = 18. The sum of the first four terms can be written as 6 + 10 + 14 + 18 = 48 4

Sigma notation is

 (4k  2), k=1 means that the series starts with the first term and the 4 on top of sigma is the 4th k 1

term in this series, 4k+2 is the formula for the nth term of this sequence.

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Practice Support for Lesson 1-2 Continued:  This example is a model to help solve Practice problems 13. EXAMPLE: Write the sequence of the first three partial sums of the arithmetic sequence with a1 = 6 and d=4. Find the first three partial sums by substituting 1,2 and 3 into the formula and adding consecutive answers. The first partial sum is 4(1) + 2 = 6, the second partial sum is 6 + 4(2) + 2 = 16, the third partial sum is 16 + 4(3) + 2 = 30. The sequence would be 6, 16, 30.

 This example is a model to help solve Practice problem 14 EXAMPLE: Matt’s job starts at $18,000 per year and he is guaranteed a raise of $1500 every year for the next 3 years. What will the total amount of money that he made in all three years? Use the formula An  a1  (n  1)d to find the first three terms where the common difference (d) is 1500 and the first term (A1) is given in the problem as 18,000. A2  18,000  (2  1)1500  19,500,

A3  18,000  (3  1)1500  21,000. The sum is 18,000 + 19,500 + 21,000 = $58,000  This example is a model to help solve Practice problems 15. EXAMPLE: 5

3

5

k 1

k 1

3

Is  (2k  1)   (2k  1)   (2k  1), ? Verify your answer. 5

Substitute the values of k into the equation  (2k  1), and add results. k 1

substitute k = 1, 2(1)  1  3, substitute k = 2, 2(2) + 1 = 5 substitute k = 3, 2(3) + 1 = 7 substitute k = 4, 2(4) + 1 = 9 substitute k = 5, 2(5) + 1 = 11 the sum of 3 + 5 + 7 + 9 + 11 = 35 3

 (2k  1)  3  5  7  15, use the values of k = 1, 2, 3 k 1

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 (2k  1)  7  9  11  26, use the values of k = 3, 4, 5 k 3 5

3

5

k 1

k 1

3

 (2k  1)   (2k  1)   (2k  1), 15 + 26 = 41 and this is not equal to 35

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Practice Support for Lesson 1-2 Continued:   This example is a model to help solve Practice problems 17. EXAMPLE Rewrite the following series using sigma notation: (−22) + (−3) + (16) + 35 First find the general equation using An  a1  (n  1)d where the first term a1 = -22, the common difference d = 19 by subtracting a term by the previous term (-3) – (-22) = -3 + 22 = 19, substitute a1 and d into the formula An  22  (n  1)19, distribute 19(n – 1) = 19n – 19 An = -22 + 19n – 19, subtract -22 -– 19 = -41 An = 19n -41 4

show k=1 to 4 terms.

 (19k  41), k 1

  This example is a model to help solve Practice problems 18. EXAMPLE: How many terms of the arithmetic sequence −6, −3, 0, … must be added to arrive at a sum of 306? First find the common difference (d) by subtracting the first term from the second term d  3  (6)  3 Substitute the values of a1 and d into the formula for the nth term An  a1  (n  1)d and simplify

An  6  (n  1)3  6  3n  3  3n  9 The formula for finding the sum of an arithmetic series with the number of n terms is S n  Substitute a1 and the expression for An and the sum into the Sn formula, 306 

n (a1  a n ), 2

n (6  3n  9), 2

Now solve for n multiply both sides of the equation by 2 and combine like terms 612  n(3n  15), 2

2

Distribute n(3n – 15) = 3n – 15n and the equation is 612 = 3n – 15n, Subtract 612 from both sides, 3n 2  15n  612  612  612  0, Divide by 3 on both sides,

, simplify n 2  5n  204  0

Factor by finding two numbers that multiply to equal -204 and add to equal -5, The numbers are -17

12 = -204 qnd -17 + 12 = -5, the factors are (n  17)( n  12)  0,

Set n  17  0 and n  12  0, the values of n=17 and n=-12, but there cannot be a negative number of terms and -12 is extraneous Therefore the number of terms is 17.     

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Additional Practice 1-2:

1. 1

2.

3.

Answers to Additional Practice1-2: 1. $153,043.03

4.

5.

2. She will not have 245 pencils because the values for the sequence are the total number of pencils each week. 3. D 4. 3015 5. n

6

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Support for Lesson 1-3 Learning Targets for lesson are found on page 13. Main Ideas for success in 1-3:

 Understand the method of mathematical induction.  Use mathematical induction to prove statements.  Vocabulary used in this lesson includes mathematical induction.  Standards for Mathematical Practice to be demonstrated are: Attend to precision, Construct viable arguments, Make sense of problems, Make use of structure.

Practice Support for Lesson 1-3  This example is a model to help solve Practice problems 7-8. EXAMPLE:  Verify the following formula using mathematical induction

n

 3k  2  k 1

k (3k  1) . 2

The first step to verify is to show that the property is true for the initial value which is k=1 Substitute the value of 1 into both sides of the equation to see if they are equal. n

 3(1)  2  3  2  1, and k 1

1(3(1)  1) 3  1 2    1 , they are both equal. 2 2 2

The next step is to show if the property is true for n = k, then it must be true for the next value of n where n = k + 1

k (3k  1) is true. Add k + 1 the (k + 1) term to both sides of the equation to see if they are equal. 2 k (3k  1) , Adding the next term to the left side of the equation 3(k  1)  2  2 Assume that 3k  2 

Simplify by using the distributive property 3(k  1)  2 = 3k  3  2, Subtract 3 – 2 = 1, so it equals 3k + 1. The left side of the equation is (3k  1) 

k (3k  1) , 2

To add the rational expressions, they must have common denominators. The common denominator is 2 and if the 2(3k  1) k (3k  1)  , denominator is multiplied by 2 then the numerator must be multiplied by 2. 2 2

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Practice Support for Lesson 1-3 Continued:  2

Distribute 2(3k + 1) = 6k + 2 and k(3k – 1) = 3k – k and the expression is,

Combine like terms in the numerator, 6k – k = 5k and the expression is

Factor the numerator

6k  2 3k 2  k  , 2 2

3k 2  5k  2 , 2

(3k  2)( k  1) , 2

Substitute k + 1 into the right side of the equation

Distribute 3(k +1) =3k +3 and the expression is

Subtract 3 – 1 =2 and the expression is

k (3k  1) (k  1)[3(k  1)  1]  , 2 2

(k  1)[( 3k  3)  1] , 2

(k  1)(3k  2) , 2

The left side of the equation is equal to the right side of the equation and the property has been verified.   This example is a model to help solve Practice problem 9. EXAMPLE What value of n is used in the initial step of a mathematical induction proof? The value of n in the initial step of a mathematical induction proof is the initial value which can be any value.

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Additional Practice 1-3:

1.

2.

3. Answers to Additional Practice1-3:

4.

5.

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Support for Lesson 2-1 Learning Targets for lesson are found on page 19. Main Ideas for success inplace lesson # here:  Identify a geometric sequence.  Determine the common ratio of a geometric sequence.  Vocabulary used in this lesson includes: geometric sequence, common ratio, polar grid, and pole  Standards for Mathematical Practice to be demonstrated are Construct viable arguments and Attend to precision.

Practice Support for Lesson 2-1  This example is a model to help solve Practice problem 10. EXAMPLE: When looking at consecutive terms of a geometric sequence, the common ratio can be found by dividing any term by the immediate previous term. For this example we will create a proportion (setting two different expressions for the ratio equal to each other) and then solve for the value of x.

Find x such that x – 3, x, and 3x – 6 are three consecutive terms in a geometric sequence. create a proportion that represents the common ratio for this sequence cross multiply multiply by distribution set equal to 0 and combine like terms factor to solve for x and

set each factor equal to 0 and solve for x

 This example is a model to help solve Practice problem 11. EXAMPLE: The lesson develops a partial formula in problem number 6 that relates any term of a geometric sequence to the first term and the common ratio. That formula is where is any term of the geometric sequence, is the first term and r is the common ratio.

Determine the first term of a geometric sequence with r = 1.3 and

.

plug all the given information into the formula evaluate divide both sides by 3.71293 to solve for the first term

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Practice Support for Lesson 2-1 Continued:  This example is a model to help solve Practice problem 12. EXAMPLE Using the same formula for finding the

term of a geometric sequence can be used in multiple ways.

Calculate n for a geometric sequence with

, r = 3 and

.

substitute all the given information into the formula multiply both sides of the equation by 81 to isolate the power from Algebra II, create common bases ( 6=n–1

)

set the exponents equal since the bases are the same solve for n

 This example is a model to help solve Practice problem 13. EXAMPLE: In this problem, we will use the same formula for finding the term of a geometric sequence, however, the initial cost of the car is not because the first term will represent the price of the car after one year. Thus, we will use the term to represent the initial price of the car.

A new car costs $25,000. The depreciation rate is 25% per year. What is the value of the car after 4 years? Round to the nearest dollar. the initial price of the car Since the car depreciates 25% each year, 75% of the price of the car remains. multiply 25,000 by 0.75 (75%) We want the value of the car after 4 years or

.

the common ratio is 0.75 since 75% of the value of the car remains each year evaluate multiply $7,910

round to the nearest dollar

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Additional Practice:

Answers to additional practice:

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Support for Lesson 2-2 Learning Targets for lesson are found on page 23. Main Ideas for success in place lesson # here:  Write the algebraic form of a geometric sequence.  Calculate the sum of a finite geometric series.  Vocabulary used in this lesson includes: partial sum, series, and geometric series  Standards for Mathematical Practice to be demonstrated are Attend to precision, Use appropriate tools strategically, Make use of structure, and Attend to precision.

Practice Support for Lesson 2-2  This example is a model to help solve Practice problems 13. EXAMPLE: The student should be looking for a pattern for the sum of the first 2 terms, first 3 terms and first 4 terms. The next example will utilize the formula that was discovered in item number 8page #.

For the geometric sequence

,

,

,

, predict

and

.

Notice that to get from to , you need to multiply the numerator by 3 and add 1 to get the numerator of and multiply the denominator by 3 to get the denominator of . The next two sums are,

and

.

 This example is a model to help solve Practice problem 14. EXAMPLE: Using the given information for the

term from the above example,

, the sequence can be generated as the

following:

nd

st

Thus, by taking the 2 term and dividing by the 1 term, it is clear that the common ratio is as given by the

term

from the example.

Write a general term,

, for the geometric sequence above.

Use the formula generated in item 8,

. plug everything in to the general formula for the sum of the first n terms

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Practice Support for Lesson 2-2 Continued:  evaluate the denominator when dividing by a fraction, multiply by the reciprocal multiply and to get distribute the (Note:

and cannot be multiplied to

due to order

of operations)mention the product of one half and one is equivalent to 1 raised to any power is 1 multiply fractions by multiplying the numerator and denominator multiply the numerator and denominator of the first fraction by

to get a

common denominator combine the terms in to one fraction (Note: due to order of operations)

cannot be multiplied into

 This example is a model to help solve Practice problem 15. EXAMPLE: This is a direct use of the formula for the sum of the first n terms of a geometric sequence. Find the sum of the first seven terms of the geometric sequence whose first term is 1.4 and ratio is 2. , evaluate

and to 128 reference PS lesson# to evaluate exponents

evaluate 1 – 128 to -127 and 1 – 2 to -1 use = instead of “to” evaluate 1.4 (-127) to -177.8 the sum of the first seven terms of the given geometric series

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Practice Support for Lesson 2-2 Continued:  This example is a model to help solve Practice problem 16. EXAMPLE The sigma notation was utilized in activity 1pg # or reference the PS lesson#. The expression in this summation represents a geometric sequence when generated. However, when k = 0 is evaluated, this generates the first term of the sequence or . Still use the formula for the sum of n terms of a geometric sequence.**include when evaluating an exponent of 0 the answer =1….which is why =4

Evaluate

and round to the nearest hundredth.

evaluate when k = 0 r = 1.3

evaluate several terms of the sequence to see (4, 5.2, 6.76, …)

n=8

there will be 8 terms from k = 0 to k = 7 substitute the given information into the formula evaluate

and 1 – 1.3 refer to lesson_ on how to evaluate exponents

evaluate 1 – 8.1573 evaluate 4(-7.1573) evaluate round to the nearest hundreth

 This example is a model to help solve Practice problem 17. EXAMPLE: Recall from lesson 2-1 the formula for finding the

term of a geometric sequence as

.

Express the sum for the example to Practice problem 15 using sigma notation. using the formula for the

term starting when k = 1 OR

manipulate the expression by separating the power using the product rule for exponents is equivalent to using the rules of negative exponents multiply 1.4 and

show/mention the step of multiplying the decimals or

fractions 6

Additional Practice:

Answers to additional practice:

7

Support for Lesson 2-3 Learning Targets for lesson are found on page 27. Main Ideas for success inlesson # here:  Determine if a sequence converges or diverges.  Find the sum of an infinite geometric series.  Vocabulary used in this lesson includes: converge, diverge, infinite sequence and infinite series.  Standards for Mathematical Practice to be demonstrated are Reason quantitatively, Express regularity in repeated reasoning, Critique the reasoning of others, Make use of structure, and Model with mathematics.

Practice Support for Lesson 2-3  This example is a model to help solve Practice problem 15. EXAMPLE: It is easy to quickly create an expression for the term of a geometric sequence using the formula discovered in lesson 2-2 ( ) using the first term and the common ratio.

Write the sum 10 + 2 + 0.4 + 0.08 + … using summation notation.

the common ratio can be found by

or

or

show the step

this will create the series by first plugging in k = 0 to get the first term, k = 1 to get the second term and so on

 This example is a model to help solve Practice problem 16. EXAMPLE: This problem is similar to Try These A (b) on p. 30.

Express the repeating decimal 0.22222… as a fraction. This decimal can be written as an infinite geometric series: 0.2 + 0.02 + 0.002 + …put this 2nd The common ratio for this series is r = 0.1 and this series can also be written as follows: you may want to put this step first, then put the answers after it… 0

1

2

0.2(0.1) + 0.2(0.1) + 0.2(0.1) + … The summation notation for this series is as follows:

We use the formula that was derived in 10(c) for finding the infinite sum:

write the pg #

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Practice Support for Lesson 2-3 Continued:  This example is a model to help solve Practice problem 17. EXAMPLE The formula derived in item 10 pg# in this lesson will be used for this example:

Solve for r:

and the common ratio is represented by r. The first term, value of 1 for n. The infinite sum, S = 4.

, is calculated by simply substituting in a

substitute the values in to the formula for the infinite sum 2r = 4(1 – r)

multiply both sides of the equation by 1 – r

2r = 4 – 4r

distribute the 4 to both terms mention what is being multiplied

6r = 4

add 4r to both sides of the equation divide both sides of the equation by 6 simplify the fraction

 This example is a model to help solve Practice problem 18. EXAMPLE: In item number 6 pg# in the lesson, we are asked to write a sequence of the partial sums. In sequence II from item 3 pg#, the sum of the first term is 100, the first two terms is 150, the first three terms is 175, etc… This sequence (which is the sum of the original converging sequence) appears to converge to 200 which means that the infinite sum converges to 200 (the series converges to 200).If you choose to include this blurb you should include the problem/example in written form that you are explaining… An infinite series converges to 35 with a common ratio of 0.4. What is Using the formula for the infinite sum again

?

will give us the following setup:

Substitute the given values into the formula for the infinite sum. evaluate 1 – 0.4 multiply both sides of the equation by 0.6

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Additional Practice:

Answers to additional practice:

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Support for Lesson 3-1 Learning Targets for lesson are found on page 33. Main Ideas for success in lesson 3-1:  Represent arithmetic and geometric sequences recursively  Determine the explicit form of a recursive sequence.  Vocabulary used in this lesson includes: arithmetic sequence, geometric sequence, explicit form, iteration and recursive sequence.  Standards for Mathematical Practice to be demonstrated are: Use appropriate tools strategically, Reason quantitatively, Model with mathematics, Construct viable arguments.

Practice Support for Lesson 3-1 This example is a model to help solve Practice problem 16. EXAMPLE: The number of bacteria in a Petri dish grows by 5 percent every hour. After the growth, about 50 bacteria die. a. Suppose initially there are 100 bacteria. How many bacteria are alive after 1 hour? Use the formula un = un-1(1 + r) + d where un = total, un-1 = previous total or original amount = 100, r = percent as a decimal = 5/100 = .05, and d is the amount added or subtracted each time = -50. un = 100(1 + .05) – 50, add 1 + 0.05 = 1.05, = 100(1.05) – 50, multiply 100(1.05) = 105 = 105 – 50, subtract 105 – 50 = 55 b. How many bacteria are alive after 2 hours? Use the same formula except u2 = 55, un = 55(1 + .05) – 50 = 55(1.05) – 50 = 7.75 c. Write expressions for u0. and un for this situation. U0 = 100 the original amount, Use the formula for un from the problem “a” above and un = un-1(1 + r) + d = un-1(1 + .05) – 50 = un-1(1.05) – 50. d. What is happening to the bacteria population? The population is decreasing each hour.

This example is a model to help solve Practice problem 17. EXAMPLE: For the sequence 1, 3, 7, 13, … write expressions for u0 and un. u0 = 1 is the initial term in the sequence, u0 = 1, u1 = 3, u2 = 7, u3 = 13 Find the differences u1 – u0 = 3 – 1 = 2, u2 – u1 = 7 – 3 = 4, u3 – u2 = 13 – 7 = 6, they increase by 2 each time or 2n more each time, therefore un = un-1 + 2n

1

Practice Support for Lesson 3-1 Continued: This example is a model to help solve Practice problem 18. EXAMPLE An arithmetic sequence has a first term of 48 and a constant difference of 6. Write expressions for u0 and un to represent this sequence. The first term of the sequence is u0 = 48 and un = the previous term (un-1) plus the common difference (6), un = un-1 + 6 This example is a model to help solve Practice problem 19. EXAMPLE: Use a spreadsheet or calculator to determine how long it will take for the bacteria from Item 16 to die off completely. Using the pattern formed in problem #16 and a calculator, the bacteria will die off by 3 hours because un = un-1(1 + r) + d, u3 = 7.75(1 + .05) – 50 = 7.75(1.05) – 50 = 8.1375 – 50 = -41.8625 and the number of bacteria can not be negative.

2

Additional Practice3-1:

Answers to additional practice 3-1:

3

Support for Lesson 3-2 Learning Targets for lesson are found on page 39. Main Ideas for success in lesson 3-2:

Represent arithmetic and geometric sequences recursively. Determine the explicit form of a recursive sequence.  Vocabulary used in this lesson includes recursive sequence and explicit form.  Standards for Mathematical Practice to be demonstrated is Construct viable arguments.

Practice Support for Lesson 3-2  This example is a model to help solve Practice problem 16. EXAMPLE: Kim opens a savings account with $30 and deposits $10 each month. Her bank pays 2.5 percent interest, compounded monthly. Write a recursive expression for an. Use the formulas u0 = Initial amount, un = un-1(1+r) + d, un-1 is the previous term of un, r = the interest rate as a decimal (r = 2.5/100 = 0.025), d = amount added each month (d = 10) u0 = 30, un = un-1(1+0.025) + 10, add 1 + 0.025 = 1.025, un = un-1(1.025) + 10 This example is a model to help solve Practice problem 17. EXAMPLE: If a9 = 137.01, find a10. Use the formula from the previous problem and substitute 137.01 for u n-1. un = un-1(1.025) + 10 = 137.10(1.025) + 10 = 140.53 + 10 = 150.53 This example is a model to help solve Practice problem 17. EXAMPLE: Write an explicit expression for an. n

The explicit formula is un = a + b·r and use the values for u0 = 30 and u1 = 30(1.025) + 10 = 30.75 + 10 = 40.75 0

u0 = a + b·r , substitute u0 = 30 and r = 1.025 into the equation, 0

the new equation is 30 = a + b·1.025 , 0

1.025 =1 and

, so 30 = a + b

0

u1 = a + b·r , substitute u1=40.75 and r = 1.025 into the equation 1

40.75 = a + b·1.025 , simplify

4

Practice Support for Lesson 3-2 Continued: 1

1.025 = 1.025, so 40.75 = a + b·1.025 Now solve the system of equations by using the elimination method. Line up the equations and subtract them, 40.75 = a + 1.025b -30 = -a + -b 



10.75 = 0.025b

 Solve the equation for b by dividing both sides by 0.025,

, and b = 430

Next, substitute the value of b = 430 into the equation a + b =30, a + 430 = 30 Solve for a by subtracting 430 from both sides of the equation a + 430 – 430 = 30 – 430 , and a = -400 n

Lastly, substitute the values of a and b into the original formula un = a + b·r , un = -400 + 430·1.025

n

This example is a model to help solve Practice problem 19. EXAMPLE Use the explicit form for an to find a12. n

Use the formula developed in the previous problem with n =12, u n = -400 + 430·1.025 , 12

u12 = -400 + 430·1.025 , use a calculator and u12 = 178.30

This example is a model to help solve Practice problem 20. EXAMPLE Explain when it would be preferable to use a recursive formula rather than an explicit formula. The best time to use a recursive formula is when the previous term is known because the formula contains u n-1.

5

Additional Practice 3-2: 1.

2.

3.

4.

5.

Answers to additional practice 3-2: 36. an = 7.3(1.0114)

n

37. The explicit expression has the variable as an exponent 38. an = -3.5 + 11.5 39. a55 = -0.2105

n

6

Support for Lesson 4-1 Learning Targets for lesson are found on page 47. Main Ideas for success in lesson 4-1:

Write, graph, analyze, and model with exponential functions. Solve exponential equations.  Vocabulary used in this lesson includes: exponential function, principal, interest rate, and growth factor.  Standards for Mathematical Practice to be demonstrated are: Model with mathematics, Use appropriate tools strategically, Attend to precision, Reason quantitatively, Make use of structure, and Express regularity in repeated reasoning.

Practice Support for Lesson 4-1  This example is a model to help solve Practice problem 23. EXAMPLE: Amanda opened a savings account at an annual interest rate of 3%. At the end of 4 years, the account balance is $619.03. If Amanda did not add any other amounts to this account, how much was her initial deposit? nt

The formula for compound interest is A = P(1 + ) , where A = total amount (A = 619.03), P = principal (original amount unknown), r = interest rate as a decimal (r = 3/100=.03), n = number of compoundings per year (n =1), t = time in years (t = 4) Substitute the values into the formula 619.03 = P(1 + solve for P by simplifying (1 +

)

1·4

,

4

) = 1.03 and 1.03 = 1.1255,

the equation becomes 619.03 = 1.1255P, divide both sides of the equation by 1.1255, so P = 619.03/1.1255 = $550.00.

 This example is a model to help solve Practice problem 24. EXAMPLE: Find the annual interest rate when the amount of $1050 grows to $1097.25 in the first year and $1146.63 in the second year. Divide the second year by the first year to find the growth factor which is 1 + r and r is the interest rate, , then set 1.045 = 1 + r and solve for r by subtracting 1 from both sides and 1.045 – 1 = 0.045 Since this is the rate as a decimal, multiply 0.045 by 100 to make it a percent, 100(0.045) = 4.5%

 

7

Practice Support for Lesson 4-1 Continued:  This example is a model to help solve Practice problem 25. EXAMPLE: Write and solve an equation to determine the balance after 12 years in an account that had an initial investment of $8,000 at 6% interest, compounded annually. Use the compound interest formula A = P(1 + ) A = 8000(1 +

)

nt

and substitute P = 8000, r =

, n = 1, and t = 12

1·12

, use the calculator and A = $16,097.57

 This example is a model to help solve Practice problem 26. EXAMPLE: Any principal amount invested at 4% annual interest takes 18 years to double. How many years does it take for the principal amount to quadruple?

 

Set up the proportion

because it takes 18 years to double and it takes x years to quadruple,

To solve the proportion cross multiply and set the products equal 18(4) = 2x, multiply 18(4) = 72, nt

So, 72 = 2x and divide both sides by 2, Start with P = any value, try P = 10 and r = A = 10(1 +

)

1·36

check your answer by using the formula A = P(1 + ) , , n = 1 and t = 36, simply the expression using a calculator

= 41.039 which is at least four times as much as 10.

If 35 years was used, then the total would only be 39.461 which is not quite four times as much.

 This example is a model to help solve Practice problem 27. EXAMPLE: Graph on a calculator and state the following for the equation:

-

x

)

a. increasing or decreasing on the interval (-,) increasing because the y-values increase as the x-values increase b. x-intercept none because the graph will never touch the x-axis c. y-intercept (0,-2) because when x = 0 is substituted into the equation

0

) and any number to the zero power is 1

and -2 (1) = -2 which is the y-intercept d. horizontal asymptote y = 0 because as the x-values approach  the y values get closer and closer to zero, but never get greater to zero e. domain (-,) because all values of x can be used in the equation and the graph is continuous from (-,) f. range (-,0) because for all x-values, the y-values are less than or equal to zero ( the negative 2 reflects the graph over the x-axis)

8

Additional Practice 4-1:

Answers to additional practice 4-1:

9

Support for Lesson 4-2 Learning Targets for lesson are found on page 52. Main Ideas for success in lesson 4-2: Write, graph, analyze, and model with exponential functions.  Calculate compound interest.  Solve exponential equations.  Vocabulary used in this lesson includes: compound interest, exponential equations, quarterly, monthly, and semiannually.  Standards for Mathematical Practice to be demonstrated are Make sense of problems and Attend to precision.

Practice Support for Lesson 4-2  This example is a model to help solve Practice problem 13. EXAMPLE: Write and solve an equation to determine the balance after 8 years in an account that had an initial investment of $12,000 at 4% interest, compounded monthly. The formula for compound interest is

nt

) , A = total amount (unknown), P = principal or original

amount (P = 12,000), r = interest rate as a decimal (r = per year ( monthly n = 12), t = time in years (t = 8).

, n = number of compoundings

12*8

Substitute the values into the formula and simplify using calculator

)

= $16,516.74

  This example is a model to help solve Practice problem 13. EXAMPLE: How much additional interest could $1800 earn in 6 years, compounded semi-annually, if the annual interest rate were % as opposed to 2.5%? Use the same formula as the last problem

nt

) , but use the two different interest rates to compare

how much additional interest is earned. The first time use P = 1800, r = n = 2 (semi-annually), t = 6 2*6

)

= $2120.52, now use r =

= .025,

% = 2.75% =

2*6

)

= .0275,

= $2089.36

Lastly, subtract the totals 21,120.52 – 2089.36 = $31.16

1

Practice Support for Lesson 4-2 Continued:  This example is a model to help solve Practice problem 15. EXAMPLE How much money needs to be deposited into an account that earns 5% annual interest, compounded quarterly, to have a balance of $8000 after 10 years? nt

Use the formula

) and this time you know A = 8000, P = unknown, r =

= 0.05, n = 4 (quarterly)

t = 10, substitute the values into the equation and simplify the right side, 4*10

)

40

40

, 8000 = P(1.0125 ), now divide both sides of the equation by 1.0125 ,

2

Additional Practice 4-2: 1 -5

1.

2.

Answers to additional practice 4-2: 3.

1. a. D b. A2(t) = 35,572(1,1169)

4.

t

2. $36,327; $980,702 3. 4739 years 4. Neither will reach the goal of $2 million. Instead of Investing, I would spend the money on college since a bachelor’s degree holder will earn at least a third more than a high school diploma holder.

5.

5. about 37 years

3

Support for Lesson 4-3 Learning Targets for lesson are found on page. Main Ideas for success in lesson 4-3:  Write, graph, analyze, and model with exponential functions. Calculate compound interest.  Solve exponential equations.  Vocabulary used in this lesson includes continuous compounding, depreciation and half-life.  Standards for Mathematical Practice to be demonstrated are: Model with mathematics, Make use of structure and Critique the reasoning of others.

Practice Support for Lesson 4-3  This example is a model to help solve Practice problem 8. EXAMPLE: An account that was invested at 4% with continuous compounding for 6 years now contains $2600. What was the initial investment? The formula for continuous compounding is , A = 2600 the total amount, P = principal or original amount (P = unknown), which is an approximation and the e on a calculator should be used when performing calculations, r = interest rate as a decimal (

, t = 6 time in years

, multiply 0.04(6) = 0.24 and the equation is Use a calculator to divide both sides of the equation by e

.24

,

and

This example is a model to help solve Practice problem 9. EXAMPLE: A new car is purchased for $14,500. It depreciates continuously at a rate of 15%. Write an exponential function that represents the value of the car after t years of ownership. When will the car have a value of $0. Explain.

 

Usethe continuous compounding formula because it depreciates which decreases in value.

where P = 14,500, and r =

the rate is negative

, this exponential function approaches a horizontal asymptote at y = 0 as x approaches infinity and therefore the function will equal values close to zero, but never equal to zero. To show an example, substitute a large value of time into the equation. For example use t = 100 years. = $0.0044

4

Practice Support for Lesson 4-3 Continued:  This example is a model to help solve Practice problem 10. EXAMPLE The half-life of the Magnesium-27 is about 9.45 minutes. This means after every 9.45minutes, the amount present is half −9.45r as much as before. Solve the equation 0.5 = e to find the decay rate r. Round your answer to five decimal places. 



Solve an equation with e and an unknown as an exponent by taking the natural logarithm of both sides of the −9.45r equation ln 0.5 = ln e , by using a property of logarithms the exponent -9.45r can be written as the product of -9.45r and ln e, the equation becomes ln 0.5 = -9.45r(ln e), ln e = 1 so now the equation is ln 0.5 = -9.45r, lastly, divide both sides of the equation by -9.45,

5

Additional Practice 4-3:

Answers to additional practice 4-3:

6

Support for Lesson 5-1 Learning Targets for lesson are found on page 59. Main Ideas for success in lesson 5-1: Explore the inverse relationship between exponents and logarithms. Graph logarithmic functions and analyze key features of the graphs.  Vocabulary used in this lesson includes logarithm, common logarithm, natural logarithm, and base.  Standards for Mathematical Practice to be demonstrated are: Express regularity in repeated reasoning, Critique the reasoning of others, Construct viable arguments.

Practice Support for Lesson 5-1  This example is a model to help solve Practice problems 22-23. EXAMPLE: Evaluate the logarithm log2 A logarithmic function can be written as an exponential function because a logarithm function is the inverse of an exponential function. The logarithmic function loga x = y is defined where “a” is the base, “y” is the exponent and y “x” is the answer. So, the equation is the same as a = x. Using the numbers in the problem log2

y

is the same as 2 =

.

To solve an exponential equation the bases on both sides of the equation must be the same. The expression

can be rewritten as

4

because 2 =

, but the expression is

and not just 16.

One of the properties of exponents is a base to the negative power is the reciprocal of the base to the positive value of the exponent. -4

Therefore, 2 =

y

-4

and the original problem is now 2 = 2 .

Since the bases are the same, the exponents must be equal and the answer is y = -4.

 This example is a model to help solve Practice problem 24. EXAMPLE: a. Evaluate the function

for the following values x =

.

Use the rules in the previous problem and converting to exponential equations to evaluate: because 3

-2 =

-1

because 3 = , 0

because 3 = 1, 1

because 3 = 3, 2

because 3 = 9

7

Practice Support for Lesson 5-1 Continued: 

b. Graph the function. Plot the values of x and y on the graph and connect them forming a curve.





c. Describe the key features of the function including the domain, range, x-intercept, asymptote, and end behavior. The domain is the values of x where the graph exists or is continuous from the left to the right. The graph does not exist for values of x that are less than or equal to zero. Therefore, the domain is 



The range is the values of y where the graph exists or is continuous. The graph continues up and down forever and the range is all real numbers or The x-intercept is the x-value where the y-value is equal to zero. When the equation was evaluated in part a, . Therefore, the x-intercept is (1,0). An asymptote is the values of x or y where a graph approaches or gets really close to, but never equals that value at its extremes. Since this graph does not exist at x = 0, but it does for values greater than 0, this function has a vertical asymptote at x =0. The end behavior of the function is the values of y the graph is approaching as x-values approach As x approaches Therefore,

, the values of y do not exist (DNE). As x approaches .

the values of y approaches

.

  This example is a model to help solve Practice problem 25. EXAMPLE The magnitude M of an earthquake whose seismographic reading measures x millimeters is calculated by this formula

, where x0 = a seismographic reading measures 0.001 millimeter at a distance of 100 kilometers -3

from the epicenter (this is called a zero-level earthquake x0 = 10 ). Find the magnitude of the earthquake San Francisco had in 1906 with a seismographic reading of 50,119 millimeters 100 kilometers from the center.

 

Substitute x = 50,119 and x0 = 0.001 into the formula and use a calculator to simplify

8

Practice Support for Lesson 5-1 Continued:   This example is a model to help solve Practice problem 26. EXAMPLE: x

Explain how the y-intercept of y = 2 and the x-intercept of y = log2 x are related. The two equations are inverses of each other. x

x

This means the domain of y = 2 is the same as the range of y = log2 x and the range of y = 2 is the same as the domain of y = log2 x (the x’s and the y’s switch places). x

0

The y-intercept of y = 2 is when x = 0 or y = 2 = 1, ordered pair (0,1). The x-intercept of y = log2 x is when y = 0 or 0 = log2 x, which x = 1, ordered pair (1,0). x

Therefore, the x’s and y’s switch places, proving y = 2 and y = log2 x are inverses.

9

Additional Practice 5-1: 1.

2.

Answers to additional practice 5-1: 1. 2.

3.

4.

3. 5.

4. 5.

10

Support for Lesson 5-2 Learning Targets for lesson are found on page 65. Main Ideas for success in lesson 5-2:  Apply the Change of Base Formula  Use properties of logarithms to evaluate and transform expressions.  Vocabulary used in this lesson includes exponents, logarithms, and Change of Base Formula.  Standards for Mathematical Practice to be demonstrated are: Reason quantitatively, Make use of structure, and Make sense of problems.

Practice Support for Lesson5-2  This example is a model to help solve Practice problem 9-10. EXAMPLE: Use a calculator and the Change of Base Formula to find an approximation for log 3 24. The Change of Base Formula is

,

substitute x = 24 and b = 3 into the formula and simplify with a calculator

 This example is a model to help solve Practice problems 11-12. EXAMPLE: Use the properties of logarithms to expand the expression

.

The first step is change When multiplying the values inside a logarithm, it can be expanded by adding the logarithms to each value: = log x + log

,

Next, use the power rule to completely expand log log

becomes

, the exponent multiplies by the logarithm of the value:

,

The final answer is:

11

Practice Support for Lesson 5-2 Continued:  This example is a model to help solve Practice problems 13-14. EXAMPLE Write the expression

as a single logarithm.

This problem is the same as condensing a logarithm or just the opposite of the last problem. The first step is to take the coefficient in front of 2 ln y and raise the y in ln y to the second power because of the 2. 2

2 ln y = ln y , the expression is now 2

Next, when logarithms are added and have the same base as these do, multiply the values inside ln xy – ln z When logarithms are subtracted and have the same base, divide the values inside The single logarithm is

.

 This example is a model to help solve Practice problem 15. EXAMPLE Present a numerical example that illustrates the Multiplication Property of Logarithms. The property states that adding logarithms with the same base can be rewritten as the product of the values inside the logarithms. log2 2 + log2 8 can be written as log2 (2 So, log2 2 + log2 8 = x

To simplify log2 2 = x, write as an exponential equation 2 = 2, the value of x = 1 y

To simplify log2 8 = y, write as an exponential equation 2 = 8, the value of y = 3 y

To simplify log2 16 = y, write as an exponential equation 2 = 16, the value of y = 4 Therefore, 1 + 3 = 4 and the property has been illustrated

 This example is a model to help solve Practice problem 16. EXAMPLE The number of people infected by a virus doubles every day. There is 8 people with the virus today. Use a logarithm and the Change of Base Formula to predict how many days it will take for 1 million people to be infected. t

Use the formula for exponential growth A(t) = A 0(R) , where A(t) = 1,000,00 the total number of people infected, A0 = 8 the initial number of people infected, R = 2 the growth rate, t = time in days for this example. Substitute the values into the formula 1,000,000 = 8(2)

t

Solve this equation by isolating the base and the exponent, do this by dividing both sides of the equation by 8, t

simplify 125,000 = 2 , rewrite the exponential equation in logarithmic form log 2 125,000 = t, The Change of Base Formula from the first example of this lesson states that the equation can be rewritten as use a calculator to simplify and the answer is 16.932 days.

12

Additional Practice 5-2:

Answers to additional practice 5-2:

13

Support for Lesson 5-3 Learning Targets for lesson are found on page 69. Main Ideas for success in lesson 5-3:  Solve logarithmic equations by taking the logarithm of both sides.  Use properties of exponents and logarithms to solve logarithmic equations  Vocabulary used in this lesson includes: exponents, logarithms, and extraneous solution.  Standards for Mathematical Practice to be demonstrated are: Make use of structure, Attend to precision, and Reason quantitatively.

Practice Support for Lesson 5-3  This example is a model to help solve Practice problem 6. EXAMPLE: Solve 6

x+2

= 80

Since the equation cannot be rewritten with the same base on both sides of the equation and the exponent contains a variable, x+2

The first step is to take the log of both sides of the equation log 6

= log 80,

Use the power property of logarithms to bring the exponent to the front of the logarithm (x + 2)log 6 = log 80, Divide both sides of the equation by log 6 to isolate the x, the equation is

,

,

Subtract 2 from both sides of the equation and simplify using a calculator

= 0.446

 This example is a model to help solve Practice problem 7. EXAMPLE: 2x

x

Solve e – 6e + 8 = 0 x

2

Factor to solve this equation, but first substitute y for e into the equation y – 6y + 8 = 0 Find factors of 8 that have a sum of -6, -2(-4) = 8 and -2 + -4 = -6, 2

y – 6y + 8 = (y – 2)(y – 4) =0, set both parentheses equal to zero, y – 2 = 0, y – 4 = 0, solve both equations for y x

y = 2 and y = 4, now substitute e back in for y x

x

e = 2 and e = 4, to solve for x, take the ln of both sides because the ln has a base of e, x

x

ln e = ln 2 and ln e = ln 4, the power property brings the exponent to the front of the logarithm, x ln e = ln 2 and x ln e = ln 4, use the propery that ln e = 1 and use a calculator to evaluate the logarithm, x = ln 2 = 0.693 and x = ln 4 = 1.386

14

Practice Support for Lesson 5-3 Continued:  This example is a model to help solve Practice problems 8-9. EXAMPLE Solve log3 (x + 6) = 4 4

Write the logarithmic equation in exponential form 3 = x + 6 4

Evaluate 3 = 81, the equation is 81 = x + 6 Subtract 6 from both sides 81 – 6 = x + 6 – 6 and the solution is x = 75

 This example is a model to help solve Practice problem 10. EXAMPLE: Solve ln (x + 4) + ln x = ln 12 Condense the left side of the equation by using the multiplication property, (if logarithms are added with the same base, then multiply the values inside the logarithm and write as a single logarithm) 2

ln [(x + 4)x] = ln (x + 4x) = ln 12, both sides of the equation have the same base and the values inside the 2 logarithms must be equal, x + 4x = 12 set the equation equal to zero by subtracting 12 from both sides of the equation, 2

x + 4x – 12 = 0, factor the equation by finding factors of -12 that have a sum of 4, the factors are (x + 6)(x – 2) = 0 because 6(-2) = -12 and 6 + -2 = 4, set each factor equal to zero x + 6 = 0 and x – 2 = 0, solve both equations by adding the opposite to both sides x = -6 and x = 2, a logarithm cannot have a negative value inside and -6 is an extraneous solution (it cannot be substituted into the equation to provide a true statement). The solution is x = 2

 This example is a model to help solve Practice problem 11. EXAMPLE: The mouse population of a secluded island doubles every 3.5 days. The island starts with 20 mice. Write and solve an equation to determine the time in weeks required for the population to reach 150,000. kt

The formula for exponential growth is A(t) = A0(R) , where A(t) = 150,000 the total population, A0 = 20 the initial number of mice, R = 2 the growth rate, k =

the number of times the population doubles per week,

and t = time in weeks. 2t

Substitute the values into the formula 150,000 = 20(2) , divide both sides of the equation by 20 2t

to isolate the base,

, the equation becomes 7500 = 2 , take the log of both sides of the 2t

equation, log 7500 = log 2 , the power property allows the 2t to come in front of log 2, log 7500 = 2t log 2, divide both sides of the equation by 2log 2 to solve for t, , use a calculator to simplify

week

15

Practice Support for Lesson 5-3 Continued:  This example is a model to help solve Practice problem 12. EXAMPLE Hannah invested $5500 in an account that earns continuously compounded interest. After five years, she had $6634.27 in the account. Write and solve an equation to find the annual interest rate to the nearest hundredth of a percent.

 

rt

The formula for continuously compounded interest is A = Pe , where A = $6634.27 the total amount of money, P = $5500 the original amount, e 2.7183 the constant used for continuously compounded interest (use the value of e on the calculator for a more precise answer), r = unknown annual interest rate as a decimal, t = 5 the time in years. Substitute the values into the equation: , the equation is 1.2062 =

divide by 5500 on both sides of the equation , take the ln of both sides of the equation,

5r

ln 1.2062 = ln e , use the power property and the 5r comes to the front of the logarithm, ln 1.2062 = 5r ln e, the value of ln e = 1 ln 1.2062 = 5r, divide both sides of the equation by 5 , and r =

, turn the decimal into a percent by multiplying by 100,

r = 100(0.0375) = 3.75%

16

Additional Practice 5-3: 1.

2.

Answers to additional practice 5-3: 1. 3-5 2.

3.

4.

3. 5.

4. 5.

17