Algebra II Module 3, Topic B, Lesson 12: Teacher Version
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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 12 M3 ALGEBRA II Lesson 12: Properties of Logarithms Student Outcomes ο§ Students justify properties of logarithms using the definition and properties already developed. Lesson Notes In this lesson, students work exclusively with logarithms base 10; generalization of these results to a generic base π occurs in the next lesson. The opening of this lesson, which echoes homework from Lesson 11, is meant to launch a consideration of some properties of the common logarithm function. The centerpiece of the lesson is the theoretical approach to demonstrating six basic logarithm properties, as opposed to the numerical approach used in previous lessons. In the Problem Set, students apply these properties to calculating logarithms, rewriting logarithmic expressions, and solving base 10 exponential equations (A-SSE.A.2, F-LE.A.4). Scaffolding: Classwork Opening Exercise (5 minutes) Students should work in groups of two or three on this Opening Exercise. This exercise serves to remind students of the “most important property” of logarithms and prepare them for justifying the properties later in the lesson. Verify that students are using the property to break up the logarithm and evaluating the logarithm at known values (e.g., 0.1, 10, 100). Opening Exercise Use the approximation π₯π¨π (π) ≈ π. ππππ to approximate the values of each of the following logarithmic expressions. a. π₯π¨π (ππ) π₯π¨π (ππ) = π₯π¨π (ππ β π) = π₯π¨π (ππ) + π₯π¨π (π) ≈ π + π. ππππ ≈ π. ππππ b. π₯π¨π (π. π) π₯π¨π (π. π) = π₯π¨π (π. π β π) = π₯π¨π (π. π) + π₯π¨π (π) ≈ −π + π. ππππ ≈ −π. ππππ Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 ο§ Prompt students who are struggling with any part of the Opening Exercise with a question, such as “How is the number in parentheses related to 2?” Follow that with the question, “So, how might you find its logarithm given that you know log(2)?” ο§ Ask students to factor each number into powers of 10 and factors of 2 before splitting the factors using log(π₯π¦) = log(π₯) + log(π¦). Students still struggling can be given additional products to break down before finding the approximations of their logarithms. 4=2⋅2 40 = 101 ⋅ 2 ⋅ 2 0.4 = 10−1 ⋅ 2 ⋅ 2 400 = 102 ⋅ 2 ⋅ 2 0.04 = 10−2 ⋅ 2 ⋅ 2 Challenge advanced students to find a general formula for log(2π ) by looking for a pattern in log(24 ), log(25 ), log(26 ), and so on. 160 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 12 M3 ALGEBRA II c. π₯π¨π (ππ ) π₯π¨π (ππ ) = π₯π¨π (π β π β π β π) = π₯π¨π (π β π) + π₯π¨π (π β π) = π₯π¨π (π) + π₯π¨π (π) + π₯π¨π (π) + π₯π¨π (π) ≈ π β (π. ππππ) ≈ π. ππππ Discussion (4 minutes) Discuss the properties of logarithms used in the Opening Exercise. ο§ In all three parts of the Opening Exercise, we used the property log(π₯π¦) = log(π₯) + log(π¦). ο§ What are some other properties we used? οΊ We also used log(10) = 1 and log(0.1) = −1. Example (6 minutes) Recall that, by definition, πΏ = log(π₯) means 10πΏ = π₯. Consider some possible values of π₯ and πΏ, noting that π₯ cannot be a negative number. What is πΏ … ο§ when π₯ = 1? οΊ ο§ πΏ=0 when π₯ = 0? οΊ The logarithm πΏ is not defined. There is no exponent of 10 that yields a value of 0. ο§ when π₯ = 109 ? ο§ when π₯ = 10π ? οΊ οΊ ο§ πΏ=9 πΏ=π 3 when π₯ = √10? οΊ πΏ= 1 3 Exercises 1–6 (15 minutes) Students should work in groups of two or three on each exercise. The first three should be straightforward in view of the definition of base 10 logarithms. Exercise 4 may look somewhat odd, but it, too, follows directly from the definition and presents an important property of logarithms. Exercises 5 and 6 are more difficult, which is why the hints are supplied. When all properties have been established, groups might be asked to show their explanations to the rest of the class as time permits. Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 161 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Exercises For Exercises 1–6, explain why each statement below is a property of base-ππ logarithms. 1. Scaffolding: Property 1: π₯π¨π (π) = π Establishing the logarithmic properties relies on the exponential laws. Make sure that students have access to the exponential laws either through a poster displayed in the classroom or through notes in their notebooks. π³ Because π³ = π₯π¨π (π) means ππ = π, then when π = π, π³ = π. 2. Property 2: π₯π¨π (ππ) = π. Because π³ = π₯π¨π (π) means πππ³ = π, then when π = ππ, π³ = π. 3. Property 3: For all real numbers π, π₯π¨π (πππ ) = π. Because π³ = π₯π¨π (π) means πππ³ = π, then when π = πππ , π³ = π. 4. Scaffolding: Property 4: For any π > π, πππ₯π¨π (π) = π. Students in groups that struggle with Exercises 3–6 should be encouraged to check the property with numerical values for π, π₯, π, and π. The check may suggest a general explanation. Because π³ = π₯π¨π (π) means πππ³ = π, then π = πππ₯π¨π (π). MP.3 5. Property 5: For any positive real numbers π and π, π₯π¨π (π β π) = π₯π¨π (π) + π₯π¨π (π). Hint: Use an exponent rule as well as property 4. By the rule ππ β ππ = ππ+π , πππ₯π¨π (π) β πππ₯π¨π (π) = πππ₯π¨π (π)+π₯π¨π (π). By property 4, πππ₯π¨π (π) β πππ₯π¨π (π) = π β π. Therefore, π β π = πππ₯π¨π (π)+π₯π¨π (π). Again, by property 4, π β π = πππ₯π¨π (πβπ). Then, πππ₯π¨π (πβπ) = πππ₯π¨π (π)+π₯π¨π (π); so, the exponents must be equal, and π₯π¨π (π β π) = π₯π¨π (π) + π₯π¨π (π). 6. Property 6: For any positive real number π and any real number π, π₯π¨π (ππ ) = π β π₯π¨π (π). Hint: Use an exponent rule as well as property 4. π By the rule (ππ )π = πππ, πππ π₯π¨π (π) = (πππ₯π¨π (π) ) . π By property 4, (πππ₯π¨π (π) ) = ππ . π Therefore, ππ = πππ π₯π¨π (π). Again, by property 4, ππ = πππ₯π¨π (π ). π Then, πππ₯π¨π (π ) = πππ π₯π¨π (π); so, the exponents must be equal, and π₯π¨π (ππ ) = π β π₯π¨π (π). Exercises 7–10 (8 minutes) The next set of exercises bridges the gap between the abstract properties of logarithms and computational problems like those in the Problem Set. Allow students to work alone, in pairs, or in small groups. Circulate to ensure that students are applying the properties correctly. Calculators are not needed for these exercises and should not be used. In Exercises 9 and 10, students need to know that the logarithm is well defined; that is, for positive real numbers π and π, if π = π, then log(π) = log(π). This is why we can “take the log of both sides” of an equation in order to bring down an exponent and solve the equation. In these last two exercises, students need to choose an appropriate base for the logarithm to use to solve the equation. Any logarithm works to solve the equations if applied properly, so students may find equivalent answers that appear to be different from those listed here. Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 162 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 12 M3 ALGEBRA II 7. Apply properties of logarithms to rewrite the following expressions as a single logarithm or number. a. π π π₯π¨π (ππ) + π₯π¨π (π) π₯π¨π (π) + π₯π¨π (π) = π₯π¨π (ππ) b. π π π₯π¨π (π) + π₯π¨π (ππ) π₯π¨π (π) + π₯π¨π (ππ ) = π₯π¨π (ππ) c. π π₯π¨π (π) + π₯π¨π (π. π) π₯π¨π (πππ) + π₯π¨π (π. π) = π₯π¨π (πππ) = π 8. Apply properties of logarithms to rewrite each expression as a sum of terms involving numbers, π₯π¨π (π), and π₯π¨π (π), where π and π are positive real numbers. a. π₯π¨π (πππ ππ ) π₯π¨π (π) + π π₯π¨π (π) + π π₯π¨π (π) b. π₯π¨π (√ππ ππ ) π π π₯π¨π (π) + π₯π¨π (π) π π 9. In mathematical terminology, logarithms are well defined because if πΏ = π, then π₯π¨π (πΏ) = π₯π¨π (π) for πΏ, π > π. This means that if you want to solve an equation involving exponents, you can apply a logarithm to both sides of the equation, just as you can take the square root of both sides when solving a quadratic equation. You do need to be careful not to take the logarithm of a negative number or zero. Use the property stated above to solve the following equations. a. πππππ = πππ π₯π¨π (πππππ ) = π₯π¨π (πππ) πππ = π π π= π b. πππ−π = π π+π ππ π₯π¨π (πππ−π ) = −π₯π¨π (πππ+π ) π − π = −(π + π) ππ = π π=π c. πππππ = ππππ−π π₯π¨π (πππππ ) = π₯π¨π (ππππ−π ) ππ π₯π¨π (πππ) = (ππ − π) ππ = ππ − π π = −π Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 163 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II 10. Solve the following equations. a. πππ = ππ π₯π¨π (πππ ) = π₯π¨π (ππ ) π = π π₯π¨π (π) b. πππ π +π = ππ π +π π₯π¨π (πππ ) = π₯π¨π (ππ) π π + π = π₯π¨π (ππ) π = ±√π₯π¨π (ππ) − π c. ππ = ππ π₯π¨π (ππ ) = π₯π¨π (ππ ) π π₯π¨π (π) = π π₯π¨π (π) π π₯π¨π (π) π= π₯π¨π (π) Closing (2 minutes) Point out that for each property 1–6, it has been established that the property holds, so these properties can be used in future work with logarithms. The Lesson Summary might be posted in the classroom for at least the rest of the module. Property 7 was established in Lesson 11 through numerical observation; students are now asked to verify both properties 7 and 8 using properties 1–6. Lesson Summary We have established the following properties for base-ππ logarithms, where π and π are positive real numbers and π is any real number: 1. π₯π¨π (π) = π 2. π₯π¨π (ππ) = π 3. π₯π¨π (πππ ) = π 4. πππ₯π¨π (π) = π 5. π₯π¨π (π β π) = π₯π¨π (π) + π₯π¨π (π) 6. π₯π¨π (ππ ) = π β π₯π¨π (π) Additional properties not yet established are the following: 7. 8. π π π π₯π¨π ( ) = π₯π¨π (π) − π₯π¨π (π) π π₯π¨π ( ) = −π₯π¨π (π) Also, logarithms are well defined, meaning that for πΏ, π > π, if πΏ = π, then π₯π¨π (πΏ) = π₯π¨π (π). Exit Ticket (5 minutes) Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 164 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Name Date Lesson 12: Properties of Logarithms Exit Ticket In this lesson, we have established six logarithmic properties for positive real numbers π₯ and π¦ and real numbers π. 1. log(1) = 0 2. log(10) = 1 3. log(10π ) = π 4. 10log(π₯) = π₯ 5. log(π₯ β π¦) = log(π₯) + log(π¦) 6. log(π₯ π ) = π β log(π₯) 1 π₯ 1. Use properties 1–6 of logarithms to establish property 7: log ( ) = −log(π₯) for all π₯ > 0. 2. Use properties 1–6 of logarithms to establish property 8: log ( ) = log(π₯) − log(π¦) for π₯ > 0 and π¦ > 0. π₯ π¦ Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 165 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Exit Ticket Sample Solutions In this lesson, we have established six logarithmic properties for positive real numbers π and π and real numbers π. 1. 1. π₯π¨π (π) = π 2. π₯π¨π (ππ) = π 3. π₯π¨π (πππ ) = π 4. πππ₯π¨π (π) = π 5. π₯π¨π (π β π) = π₯π¨π (π) + π₯π¨π (π) 6. π₯π¨π (ππ ) = π β π₯π¨π (π) π π Use properties 1–6 of logarithms to establish property 7: π₯π¨π ( ) = −π₯π¨π (π) for all π > π. By property 6, π₯π¨π (ππ ) = π β π₯π¨π (π). π π Let π = −π; then for π > π, π₯π¨π (π−π ) = (−π) β π₯π¨π (π), which is equivalent to π₯π¨π ( ) = −π₯π¨π (π). π π Thus, for any π > π, π₯π¨π ( ) = −π₯π¨π (π). 2. π π Use properties 1–6 of logarithms to establish property 8: π₯π¨π ( ) = π₯π¨π (π) − π₯π¨π (π) for π > π and π > π. By property 5, π₯π¨π (π β π) = π₯π¨π (π) + π₯π¨π (π). By Problem 1 above, for π > π, π₯π¨π (π−π ) = (−π) β π₯π¨π (π). Therefore, π π π₯π¨π ( ) = π₯π¨π (π) + π₯π¨π ( ) π π = π₯π¨π (π) + (−π)π₯π¨π (π) = π₯π¨π (π) − π₯π¨π (π). π π Thus, for any π > π and π > π, π₯π¨π ( ) = π₯π¨π (π) − π₯π¨π (π). Problem Set Sample Solutions Problems 1–7 give students an opportunity to practice using the properties they have established in this lesson. In the remaining problems, students apply base-10 logarithms to solve simple exponential equations. 1. Use the approximate logarithm values below to estimate the value of each of the following logarithms. Indicate which properties you used. a. π₯π¨π (π) = π. ππππ π₯π¨π (π) = π. ππππ π₯π¨π (π) = π. ππππ π₯π¨π (π) = π. ππππ π₯π¨π (π) Using property 5, π₯π¨π (π) = π₯π¨π (π) + π₯π¨π (π) ≈ π. ππππ. Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 166 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II b. π₯π¨π (ππ) Using property 5, π₯π¨π (ππ) = π₯π¨π (π) + π₯π¨π (π) ≈ π. ππππ. c. π₯π¨π (ππ) Using properties 5 and 6, π₯π¨π (ππ) = π₯π¨π (π) + π₯π¨π (ππ ) = π₯π¨π (π) + π π₯π¨π (π) ≈ π. ππππ. d. π₯π¨π (πππ ) Using property 3, π₯π¨π (πππ ) = π. e. π π π₯π¨π ( ) Using property 7, π π π₯π¨π ( ) = −π₯π¨π (π) ≈ −π. ππππ. f. π π π₯π¨π ( ) Using property 8, π π₯π¨π ( ) = π₯π¨π (π) − π₯π¨π (π) ≈ −π. πππ. π g. π π₯π¨π (√π) Using property 6, π π π₯π¨π (√π) = π₯π¨π (ππ ) = 2. π π₯π¨π (π) ≈ π. ππππ. π Let π₯π¨π (πΏ) = π, π₯π¨π (π) = π, and π₯π¨π (π) = π. Express each of the following in terms of π, π, and π. a. πΏ π b. π₯π¨π ( ) π+π π−π c. π₯π¨π (πΏπ ) d. ππ e. π₯π¨π (ππ) π π₯π¨π (√π) π π π f. π π₯π¨π ( √ ) π π₯π¨π (πΏππ ππ ) π + ππ + ππ π−π π Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 167 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 12 M3 ALGEBRA II 3. Use the properties of logarithms to rewrite each expression in an equivalent form containing a single logarithm. a. π₯π¨π ( ππ π ) + π₯π¨π ( ) π π ππ π₯π¨π ( ) π b. π π π π π₯π¨π ( ) − π₯π¨π ( ) π π₯π¨π ( ) π c. π π π π₯π¨π (ππ) + π₯π¨π (π) + π₯π¨π ( π ) π₯π¨π (π) 4. Use the properties of logarithms to rewrite each expression in an equivalent form containing a single logarithm. a. π π π₯π¨π (√π) + π₯π¨π ( ) + π π₯π¨π (π) π π π₯π¨π (ππ ) b. π π₯π¨π ( π√π) + π₯π¨π (√ππ ) π₯π¨π (π) c. π π π₯π¨π (π) + π π₯π¨π (π) − π₯π¨π (π) π₯π¨π ( d. π π πππ √π (π₯π¨π (π) − π π₯π¨π (π) + π₯π¨π (π)) π π₯π¨π ( √ e. ) ππ ) ππ π(π₯π¨π (π) − π₯π¨π (ππ)) + π(π₯π¨π (π) − π π₯π¨π (π)) π π π π ππ π₯π¨π (( ) ) + π₯π¨π (( π ) ) = π₯π¨π ( π π ) ππ π ππ π 5. In each of the following expressions, π, π, and π represent positive real numbers. Use properties of logarithms to rewrite each expression in an equivalent form containing only π₯π¨π (π), π₯π¨π (π), π₯π¨π (π), and numbers. a. πππ ππ ) √π π₯π¨π ( π₯π¨π (π) + π π₯π¨π (π) + π π₯π¨π (π) − Lesson 12: π π₯π¨π (π) π Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 168 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II π b. π₯π¨π ( ππ √πππ ) ππ π π π π₯π¨π (ππ) − π₯π¨π (π) + π₯π¨π (π) − π₯π¨π (π) π π c. πππππ ) ππ π₯π¨π ( π + π π₯π¨π (π) − π π₯π¨π (π) d. ππ ππ ) πππ π₯π¨π (√ π (πβπ₯π¨π (π) + π π₯π¨π (π) − π − π₯π¨π (π)) π e. π₯π¨π ( π ) ππππ π −π − π π₯π¨π (π) − π₯π¨π (π) 6. π π Express π₯π¨π ( − π π π ) + (π₯π¨π ( ) − π₯π¨π ( )) as a single logarithm for positive numbers π. π+π π π+π π π π π π π π π₯π¨π ( − ) + (π₯π¨π ( ) − π₯π¨π ( )) = π₯π¨π ( ) + π₯π¨π ( ) − π₯π¨π ( ) π π+π π π+π π(π + π) π π+π = −π₯π¨π (π(π + π)) − π₯π¨π (π) + π₯π¨π (π + π) = −π₯π¨π (π) − π₯π¨π (π + π) − π₯π¨π (π) + π₯π¨π (π + π) = −π π₯π¨π (π) 7. Show that π₯π¨π (π + √ππ − π) + π₯π¨π (π − √ππ − π) = π for π ≥ π. π₯π¨π (π + √ππ − π) + π₯π¨π (π − √ππ − π) = π₯π¨π ((π + √ππ − π) (π − √ππ − π)) π = π₯π¨π (ππ − (√ππ − π) ) = π₯π¨π (ππ − ππ + π) = π₯π¨π (π) =π 8. If ππ = πππ.ππ for some positive real numbers π and π, find the value of π₯π¨π (π) + π₯π¨π (π). ππ = πππ.ππ π. ππ = π₯π¨π (ππ) π₯π¨π (ππ) = π. ππ π₯π¨π (π) + π₯π¨π (π) = π. ππ Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 169 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II 9. Solve the following exponential equations by taking the logarithm base ππ of both sides. Leave your answers stated in terms of logarithmic expressions. a. π πππ = πππ π π₯π¨π (πππ ) = π₯π¨π (πππ) ππ = π₯π¨π (πππ) π = ±√π₯π¨π (πππ) b. π πππ = πππ π π₯π¨π (πππ ) = π₯π¨π (πππ) π = π₯π¨π (πππ β π) π π = π + π₯π¨π (π) π π = ππ + π π₯π¨π (π) c. ππππ = πππ π₯π¨π (ππππ ) = π₯π¨π (πππ) ππ ⋅ π₯π¨π (ππ) = π₯π¨π (πππ β π) ππ ⋅ π = π + π₯π¨π (π) π π = (π + π₯π¨π (π)) π d. πππ = πππ π₯π¨π (πππ ) = π₯π¨π (πππ) ππ ⋅ π₯π¨π (π) = π₯π¨π (πππ) + π₯π¨π (π) π + π₯π¨π (π) π₯π¨π (π) π + π₯π¨π (π) π= π π₯π¨π (π) ππ = e. ππ± = π−ππ+π π₯π¨π (ππ ) = π₯π¨π (π−ππ+π ) π π₯π¨π (π) = (−ππ + π)π₯π¨π (π) π π₯π¨π (π) + ππ π₯π¨π (π) = π π₯π¨π (π) π(π₯π¨π (π) + π π₯π¨π (π)) = π π₯π¨π (π) π= π π₯π¨π (π) π₯π¨π (ππ) π₯π¨π (ππ) = = π₯π¨π (π) + π π₯π¨π (π) π₯π¨π (π) + π₯π¨π (πππ) π₯π¨π (ππππ) (Any of the three equivalent forms given above are acceptable answers.) 10. Solve the following exponential equations. a. πππ = π π = π₯π¨π (π) Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 170 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II b. πππ = ππ π = π₯π¨π (ππ) c. πππ = πππ π³ = π₯π¨π (πππ) d. Use the properties of logarithms to justify why π, π, and π form an arithmetic sequence whose constant difference is π. Since π = π₯π¨π (ππ), π = π₯π¨π (ππ β π) = π + π₯π¨π (π) = π + π. Similarly, π = π + π₯π¨π (π) = π + π. Thus, the sequence π, π, π is the sequence π₯π¨π (π), π + π₯π¨π (π), π + π₯π¨π (π), and these numbers form an arithmetic sequence whose first term is π₯π¨π (π) with constant difference π. 11. Without using a calculator, explain why the solution to each equation must be a real number between π and π. a. πππ = ππ ππ is greater than πππ and less than πππ , so the solution is between π and π. b. πππ = ππ ππ is greater than πππ and less than πππ , so the solution is between π and π. c. ππππ = ππππ ππππ = πππππ, so π, πππ is between ππππ and πππ , so the solution is between π and π. d. ( π π ) = π. ππ ππ π πππ e. is between π π π π π π ( ) = and π , so the solution is between π and π. πππ π π π π ( ) = , and f. π ππ π π is between π π π and , so the solution is between π and π. π πππ = ππππ πππ = ππππ. Since π, πππ is less than π, πππ and greater than ππ, the solution is between π and π. 12. Express the exact solution to each equation as a base-ππ logarithm. Use a calculator to approximate the solution to the nearest ππππππ. a. πππ = ππ π₯π¨π (πππ ) = π₯π¨π (ππ) π π₯π¨π (ππ) = π₯π¨π (ππ) π= π₯π¨π (ππ) π₯π¨π (ππ) π ≈ π. πππ Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 171 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II b. πππ = ππ π= π₯π¨π (ππ) π₯π¨π (ππ) π ≈ π. πππ c. ππππ = ππππ π= π₯π¨π (ππππ) π₯π¨π (πππ) π ≈ π. πππ d. ( π π ) = π. ππ ππ π π=− π₯π¨π ( π ) ππ π ≈ π. πππ e. π π π ( ) = π π π π₯π¨π ( ) π π= π π₯π¨π ( ) π π ≈ π. πππ f. πππ = ππππ π= π₯π¨π (ππππ) π₯π¨π (ππ) π ≈ π. πππ 13. Show that for any real number π, the solution to the equation πππ = π β πππ is π₯π¨π (π) + π. Substituting π = π₯π¨π (π) + π into πππ and using properties of exponents and logarithms gives πππ = πππ₯π¨π (π)+ π = πππ₯π¨π (π)πππ = π ⋅ πππ . Thus, π = π₯π¨π (π) + π is a solution to the equation πππ = π ⋅ πππ. 14. Solve each equation. If there is no solution, explain why. a. π β ππ = ππ ππ = π π₯π¨π (ππ ) = π₯π¨π (π) π π₯π¨π (π) = π₯π¨π (π) π= π₯π¨π (π) π₯π¨π (π) Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 172 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II b. πππ−π = ππ π₯π¨π (πππ−π ) = π₯π¨π (ππ) π = π + π₯π¨π (ππ) c. πππ + πππ+π = ππ πππ (π + ππ) = ππ πππ = π π=π d. π − ππ = ππ −ππ = π ππ = −π There is no solution because ππ is always positive for all real π. 15. Solve the following equation for π: π¨ = π·(π + π)π . π¨ = π·(π + π)π π₯π¨π (π¨) = π₯π¨π [(π·(π + π)π ] π₯π¨π (π¨) = π₯π¨π (π·) + π₯π¨π [(π + π)π ] π₯π¨π (π¨) − π₯π¨π (π·) = π π₯π¨π (π + π) π₯π¨π (π¨) − π₯π¨π (π·) π₯π¨π (π + π) π¨ π₯π¨π ( ) π· π= π₯π¨π (π + π) π= The remaining questions establish a property for the logarithm of a sum. Although this is an application of the logarithm of a product, the formula does have some applications in information theory and can help with the calculations necessary to use tables of logarithms, which are explored further in Lesson 15. 16. In this exercise, we will establish a formula for the logarithm of a sum. Let π³ = π₯π¨π (π + π), where π, π > π. a. π π Show π₯π¨π (π) + π₯π¨π (π + ) = π³. State as a property of logarithms after showing this is a true statement. π π π₯π¨π (π) + π₯π¨π (π + ) = π₯π¨π (π (π + )) π π ππ = π₯π¨π (π + ) π = π₯π¨π (π + π) =π³ π π Therefore, for π,π > π, π₯π¨π (π + π) = π₯π¨π (π) + π₯π¨π (π + ). Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 173 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II b. Use part (a) and the fact that π₯π¨π (πππ) = π to rewrite π₯π¨π (πππ) as a sum. π₯π¨π (πππ) = π₯π¨π (πππ + πππ) πππ ) πππ = π₯π¨π (πππ) + π₯π¨π (π. ππ) = π₯π¨π (πππ) + π₯π¨π (π + = π + π₯π¨π (π. ππ) c. Rewrite πππ in scientific notation, and use properties of logarithms to express π₯π¨π (πππ) as a sum of an integer and a logarithm of a number between π and ππ. πππ = π. ππ × πππ π₯π¨π (πππ) = π₯π¨π (π. ππ × πππ ) = π₯π¨π (π. ππ) + π₯π¨π (πππ ) = π + π₯π¨π (π. ππ) d. What do you notice about your answers to (b) and (c)? Separating πππ into πππ + πππ and using the formula for the logarithm of a sum is the same as writing πππ in scientific notation and using the formula for the logarithm of a product. e. Find two integers that are upper and lower estimates of π₯π¨π (πππ). Since π < π. ππ < ππ, we know that π < π₯π¨π (π. ππ) < π. This tells us that π < π + π₯π¨π (π. ππ) < π, so π < π₯π¨π (πππ) < π. Lesson 12: Properties of Logarithms This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 174 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
