Lesson 13: Changing the Base
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Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Lesson 13: Changing the Base Student Outcomes ο§ Students understand how to change logarithms from one base to another. ο§ Students calculate logarithms with any base using a calculator that computes only logarithms base 10 and base π. ο§ Students justify properties of logarithms with any base. Lesson Notes The first example in this lesson demonstrates how to use a base-10 logarithm to calculate a base-2 logarithm, leading to the change of base formula for logarithms. The change of base formula allows students to generalize the properties of base-10 logarithms developed in the previous few lessons to logarithms with general base π. This lesson introduces the natural logarithm ln(π₯) = log π (π₯). Calculators are used briefly in this lesson to compute both common and natural logarithms, and one of the goals of the lesson is to explain why the calculator only has a LOG and an LN key. Students solve exponential equations by applying the appropriate logarithm (F-LE.A.4). Materials Students need access either to graphing calculators or computer software capable of computing logarithms with base 10 and base π, such as the WolframβAlpha engine. Classwork Example 1 (5 minutes) Scaffolding: The purpose of this example is to show how to find log 2 (π₯) using log(π₯). ο§ We have been working primarily with base-10 logarithms, but in Lesson 7 we defined logarithms for any base π. For example, the number 2 might be the base. When logarithms have bases other than 10, it often helps to be able to rewrite the logarithm in terms of base 10 logarithms. Let πΏ = log 2 (π₯), and show that πΏ = οΊ log(π₯) . log(2) Let πΏ = log 2 (π₯). Then 2πΏ = π₯. Taking the logarithm of each side, we get log(2πΏ ) = log(π₯) πΏ β log(2) = log(π₯) πΏ= Therefore, log 2 (π₯) = Lesson 13: ο§ Students who struggle with the first step of this example might need to be reminded of the definition of logarithm from Lesson 7: πΏ = log π (π₯) means π πΏ = π₯. Therefore, πΏ = log 2 (π₯) means 2πΏ = π₯. ο§ Advanced learners may want to immediately start with the second part of the example, converting log π (π₯) into log(π₯) log(π) . log(π₯) . log(2) log(π₯) . log(2) Changing the Base 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 175 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Remember that log(2) is a number, so this shows that log 2 (π₯) is a rescaling of log(π₯). ο§ The example shows how we can convert log 2 (π₯) to an expression involving log(π₯). More generally, suppose we are given a logarithm with base π. What is log π (π₯) in terms of log(π₯)? οΊ Let πΏ = log π (π₯). Then π πΏ = π₯. Taking the logarithm of each side, we get log(π πΏ ) = log(π₯) πΏ β log(π) = log(π₯) πΏ= Therefore, log π (π₯) = ο§ log(π₯) . log(π) log(π₯) . log(π) This equation not only allows us to change from log π (π₯) to log(π₯) but to change the base in the other direction as well: log(π₯) = log π (π₯) β log(π). Exercise 1 (3 minutes) The first exercise deals with the general formula for changing the base of a logarithm. It follows the same pattern as Example 1. Take time for students to share their results from Exercise 1 in a class discussion before moving on to Exercise 2 so that all students understand how the base of a logarithm is changed. Ask students to work in pairs on this exercise. Exercises 1. Scaffolding: If students have difficulty with Exercise 1, they should review the argument in Example 1, noting that it deals with base 10, whereas this exercise generalizes that base to π. Assume that π, π, and π are all positive real numbers, so that π ≠ π and π ≠ π. What is π₯π¨π π (π) in terms of π₯π¨π π(π)? The resulting equation allows us to change the base of a logarithm from π to π. Let π³ = π₯π¨π π (π). Then ππ³ = π. Taking the logarithm base π of each side, we get π₯π¨π π (ππ³) = π₯π¨π π (π) π³ β π₯π¨π π (π) = π₯π¨π π (π) π₯π¨π π (π) π³= . π₯π¨π π (π) Therefore, π₯π¨π π (π) = π₯π¨π π (π) . π₯π¨π π (π) Discussion (2 minutes) Ask a student to present the solution to Exercise 1 to the class to ensure that all students understand how to change the base of a logarithm and how the formula comes from the definition of the logarithm as an exponential equation. Be sure that students record the formula in their notebooks. Change of Base Formula for Logarithms If π₯, π, and π are all positive real numbers with π ≠ 1 and π ≠ 1, then log π (π₯) log π (π₯) = . log π (π) Lesson 13: Changing the Base 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 176 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 13 M3 ALGEBRA II Exercise 2 (2 minutes) In the second exercise, students practice changing bases. They need a calculator with the ability to calculate logarithms base 10. Later in the lesson, students need to calculate natural logarithms as well. Students should work in pairs on this exercise, with one student using the calculator and the other keeping track of the computation. Students should share their results for Exercise 2 in a class discussion before moving on to Exercise 3. 2. Approximate each of the following logarithms to four decimal places. Use the LOG key on your calculator rather than logarithm tables, first changing the base of the logarithm to ππ if necessary. a. π₯π¨π (ππ ) π₯π¨π (ππ ) = π₯π¨π (π) ≈ π. ππππ Therefore, π₯π¨π (ππ ) ≈ π. ππππ. Scaffolding: OR π₯π¨π (ππ ) = π π₯π¨π (π) ≈ π β π. ππππ ≈ π. ππππ Therefore, π₯π¨π (ππ ) ≈ π. ππππ. b. π₯π¨π π (ππ ) π₯π¨π π (ππ ) = Students who are not familiar with the LOG key on the calculator can check how it works by evaluating the following expressions: log(1), π π₯π¨π (π) =π π₯π¨π (π) log(10), Therefore, π₯π¨π π (ππ ) = π. ππππ. c. log(103 ). π₯π¨π π (ππ ) π₯π¨π π (ππ ) = π₯π¨π π (π) = π₯π¨π (π) ≈ π. ππππ π₯π¨π (π) Therefore, π₯π¨π π (ππ ) ≈ π. ππππ. Exercise 3 (8 minutes) 3. In Lesson 12, we justified a number of properties of base-ππ logarithms. Working in pairs, justify the following properties of base-π logarithms: a. π₯π¨π π (π) = π Because π³ = π₯π¨π π (π) means ππ³ = π, then when π = π, π³ = π. b. π₯π¨π π (π) = π Because π³ = π₯π¨π π (π) means ππ³ = π, then when π = π, π³ = π. c. π₯π¨π π (ππ ) = π Because π³ = π₯π¨π π (π) means ππ³ = π, then when π = ππ , π³ = π. Lesson 13: Changing the Base 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 Scaffolding: By working in pairs, students should be able to reconstruct the arguments they used in Lesson 10. If they have trouble, they should be encouraged to use the definition and properties already justified. 177 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II d. ππ₯π¨π π(π) = π Because π³ = π₯π¨π π (π) means ππ³ = π, then π = ππ₯π¨π π(π). e. π₯π¨π π (π β π) = π₯π¨π π (π) + π₯π¨π π (π) By the rule ππ β ππ = ππ+π , ππ₯π¨π π(π) β ππ₯π¨π π(π) = ππ₯π¨π π(π)+π₯π¨π π(π) . By property 4, ππ₯π¨π π(π) β ππ₯π¨π π(π) = π β π. Therefore, π β π = ππ₯π¨π π(π)+π₯π¨π π(π). By property 4 again, π β π = ππ₯π¨π π(πβπ). So, the exponents must be equal, and π₯π¨π π (π β π) = π₯π¨π π (π) + π₯π¨π π (π). f. π₯π¨π π (ππ ) = π β π₯π¨π π (π) π By the rule (ππ )π = πππ , ππ π₯π¨π π(π) = (ππ₯π¨π π(π) ) . π By property 4, (ππ₯π¨π π(π) ) = ππ. π Therefore, ππ = πππ₯π¨π π (π). By property 4 again, ππ = ππ₯π¨π π(π ) . So, the exponents must be equal, and π₯π¨π π (ππ ) = π β π₯π¨π π (π). g. π π π₯π¨π π ( ) = −π₯π¨π π (π) By property 6, π₯π¨π π (ππ ) = π β π₯π¨π π (π). Let π = −π; then for π ≠ π, π₯π¨π π (π−π ) = (−π) β π₯π¨π π (π). π π Thus, π₯π¨π π ( ) = −π₯π¨π π (π). h. π π π₯π¨π π ( ) = π₯π¨π π (π) − π₯π¨π π (π) By property 5, π₯π¨π π (π β π) = π₯π¨π π (π) + π₯π¨π π (π). By property 7, for π ≠ π, π₯π¨π π (π−π ) = (−π) β π₯π¨π π (π). π π Therefore, π₯π¨π π ( ) = π₯π¨π π (π) − π₯π¨π π (π). Discussion (2 minutes) Define the natural logarithm in this Discussion. Because students often misinterpret the symbol ππ as the word in, take the time to emphasize that the notation is an L followed by an N, which comes from the French for natural logarithm: le logarithme naturel. ο§ Recall Euler’s number π from Lesson 5, which is an irrational number approximated by π ≈ 2.718β28 …. This number plays an important role in many parts of mathematics, and it is frequently used as the base of logarithms. Because exponential functions with base π are used to model growth and change of natural phenomena, a logarithm with base π is called a natural logarithm. The notation for the natural logarithm of a positive number π₯ is ln(π₯) = log π (π₯). ο§ What is the value of ln(1)? οΊ ln(1) = 0 Lesson 13: Changing the Base 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 178 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II ο§ What is the value of ln(π)? The value of ln(π 2 )? Of ln(π 3 )? ln(π) = 1, ln(π 2 ) = 2, and ln(π 3 ) = 3 οΊ ο§ Because scientists primarily use logarithms base 10 and base π, calculators often only have two logarithm buttons: LOG for calculating log(π₯) and LN for calculating ln(π₯). With the change of base formula, you can use either the common logarithm (base 10) or the natural logarithm (base π) to calculate the value of a logarithm with any allowable base π, so technically we only need one of those two buttons. However, each base has important uses, so most calculators are able to calculate logarithms in either base. Exercises 4–6 (5 minutes) Exercises 4–6 allow students to compare the values of ln(π₯) to the more familiar values of log(π₯) for a few values of π₯ and to conclude that for any π₯ ≥ 1, log(π₯) ≤ ln(π₯). Students need a calculator with an LN key. They should work in pairs on these exercises, with one student using the calculator and the other recording the result. They should share their results for Exercise 4 in a class discussion before moving on. Scaffolding: 4. Use the LN and LOG keys on your calculator to find the value of each logarithm to four decimal places. a. b. c. d. e. π₯π§(π) π₯π§(π) π₯π§(ππ) π₯π§(ππ) π₯π§(πππ) π. ππππ π. ππππ π. ππππ π. ππππ π. ππππ π₯π¨π (π) π₯π¨π (π) π₯π¨π (ππ) π₯π¨π (ππ) π₯π¨π (πππ) π. ππππ π. ππππ π. ππππ π. ππππ π. ππππ Students who are not familiar with the LN key on the calculator can check how it works by evaluating the following expressions: ln(1), ln(π), Make a conjecture that compares values of π₯π¨π (π) to π₯π§(π) for π ≥ π. 5. ln(π 3 ). It appears that for π ≥ π, π₯π¨π (π) ≤ π₯π§(π). 6. Justify your conjecture in Exercise 5 using the change of base formula. By the change of base formula, π₯π¨π (π) = π₯π§(π) . Then π₯π§(ππ) ⋅ π₯π¨π (π) = π₯π§(π). Since π₯π§(ππ) π₯π§(ππ) ≈ π. π, π₯π¨π (π) ≤ π₯π§(ππ) ⋅ π₯π¨π (π), and thus π₯π¨π (π) ≤ π₯π§(π). Example 2 (3 minutes) This example introduces more complicated expressions involving logarithms and showcases the power of logarithms in rearranging logarithmic expressions. Students have done similar exercises in their homework in prior lessons for base10 logarithms, so this example and the following exercises demonstrate how the same procedures apply to natural logarithms. Remind students that Exercise 3 established that the logarithm properties developed for base-10 logarithms apply for logarithms of any base, including base π. ο§ 1 Write as an expression containing only one logarithm: ln(π 2 ) + ln ( 2) − ln(√π). π οΊ ο§ 1 1 2 ln(π 2 ) + ln ( 2) − ln(√π) = 2 ln(π) − 2 ln(π) − β ln(π) = − π Therefore, ln(π 2) 1 1 ln(π) 2 1 2 + ln ( 2) − ln(√π) = − ln(π). Lesson 13: π Changing the Base 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 179 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Exercises 7–8 (6 minutes) Exercise 7 follows Example 2 by asking students to simplify more complicated logarithmic expressions. In Exercise 7, students condense a sum of logarithmic expressions to an expression containing only one logarithm, while in Exercise 8, students take a single complicated logarithm and break it up into simpler parts. Students should work in pairs on these exercises, sharing their results in a class discussion before the Closing. 7. Write as a single logarithm. a. 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) π = π₯π§(π β ππ β π) = π₯π§(πππ) = π₯π§(ππ ) = π π₯π§(π) Any of the last three expressions is an acceptable final answer. b. π 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§ (πππ ) − π₯π§(π) π = π₯π§(π) + π₯π§(π) − π₯π§(π) = π₯π§(π β π) − π₯π§(π) ππ = π₯π§ ( ) π = π₯π§(ππ) π π Therefore, π₯π§(π) + π₯π§(ππ) − π₯π§(π) = π₯π§(ππ). 8. Write each expression as a sum or difference of constants and logarithms of simpler terms. a. π₯π§ ( √πππ ππ ) √πππ π₯π§ ( π ) = π₯π§(√π) + π₯π§ (√ππ ) − π₯π§(ππ ) π π π = π₯π§(π) + π₯π§(π) − π π π π b. (π+π) ) ππ +ππ π₯π§ ( π₯π§ ( (π + π)π ) = π₯π§(π + π)π − π₯π§(ππ + ππ ) ππ + ππ = π π₯π§(π + π) − π₯π§(ππ + ππ ) The point of this simplification is that neither of these terms can be simplified further. Lesson 13: Changing the Base 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 180 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Closing (4 minutes) Have students summarize the lesson by discussing the following questions and coming to a consensus before students record the answers in their notebooks: ο§ What is the definition of the logarithm base π? οΊ ο§ What does ln(π₯) represent? οΊ ο§ If there exist numbers π, πΏ, and π₯ so that π πΏ = π₯, then πΏ = log π (π₯). The notation ln(π₯) represents the logarithm of π₯ base π; that is, ln(π₯) = log π (π₯). How can we use a calculator to approximate a logarithm to a base other than 10 or π? οΊ Use the change of base formula to convert a logarithm with base π to one with base 10 or base π; then, use the appropriate calculator function. Lesson Summary We have established a formula for changing the base of logarithms from π to π: π₯π¨π π (π) = π₯π¨π π (π) . π₯π¨π π (π) In particular, the formula allows us to change logarithms base π to common or natural logarithms, which are the only two kinds of logarithms that most calculators compute: π₯π¨π π (π) = π₯π¨π (π) π₯π§(π) = . π₯π¨π (π) π₯π§(π) We have also established the following properties for base π logarithms. If π, π, π, and π are all positive real numbers with π ≠ π and π ≠ π and π is any real number, then: 1. π₯π¨π π (π) = π 2. π₯π¨π π (π) = π 3. π₯π¨π π (ππ ) = π 4. ππ₯π¨π π(π) = π 5. π₯π¨π π (π β π) = π₯π¨π π (π) + π₯π¨π π (π) 6. π₯π¨π π (ππ ) = π β π₯π¨π π (π) 7. π₯π¨π π ( ) = −π₯π¨π π (π) 8. π₯π¨π π ( ) = π₯π¨π π (π) − π₯π¨π π (π). π π π π Exit Ticket (5 minutes) Lesson 13: Changing the Base 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 181 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Name Date Lesson 13: Changing the Base Exit Ticket 1. Are there any properties that hold for base-10 logarithms that would not be valid for the logarithm base π? Why? Are there any properties that hold for base-10 logarithms that would not be valid for some positive base π, such that π ≠ 1? 2. Write each logarithm as an equivalent expression involving only logarithms base 10. 3. a. log 3 (25) b. log100 (π₯ 2 ) Rewrite each expression as an equivalent expression containing only one logarithm. a. 3 ln(π + π) − 2 ln(π) − 7 ln(π) π₯ π¦ b. ln(π₯π¦) − ln ( ) Lesson 13: Changing the Base 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 182 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Exit Ticket Sample Solutions 1. Are there any properties that hold for base-ππ logarithms that would not be valid for the logarithm base π? Why? Are there any properties that hold for base-ππ logarithms that would not be valid for some positive base π, such that π ≠ π? No. Any property that is true for a base-ππ logarithm will be true for a base-π logarithm. The only difference between a common logarithm and a natural logarithm is a scale change because π₯π¨π (π) = π₯π§(π) = π₯π§(π) , and π₯π§(ππ) π₯π¨π (π) . π₯π¨π (π) Since π₯π¨π π (π) = π₯π¨π (π) , we would only encounter a problem if π₯π¨π (π) = π, but this only happens when π = π, and π π₯π¨π (π) is not a valid base for logarithms. 2. Write each logarithm as an equivalent expression involving only logarithms base ππ. a. π₯π¨π π (ππ) π₯π¨π π (ππ) = b. π₯π¨π (ππ) π₯π¨π (π) π₯π¨π πππ (ππ ) π₯π¨π πππ (ππ ) = π₯π¨π (ππ ) π₯π¨π (πππ) π π₯π¨π (π) π = π₯π¨π (π) = 3. Rewrite each expression as an equivalent expression containing only one logarithm. a. 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§(ππ )) = π₯π§((π + π)π ) − π₯π§(ππ ππ ) (π + π)π = π₯π§ ( π π ) π π b. π π π₯π§(ππ) − 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§(ππ ) π π Therefore, π₯π§(ππ) − π₯π§ ( ) is equivalent to both π π₯π§(π) and π₯π§(ππ ). Lesson 13: Changing the Base 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 183 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 13 M3 ALGEBRA II Problem Set Sample Solutions 1. Evaluate each of the following logarithmic expressions, approximating to four decimal places if necessary. Use the LN or LOG key on your calculator rather than a table. a. π₯π¨π π (ππ) π₯π¨π π (ππ) = π₯π¨π (ππ) π₯π¨π (π) = π₯π¨π (ππ ) π₯π¨π (ππ ) π β π₯π¨π (π) π β π₯π¨π (π) π = π = π π Therefore, π₯π¨π π (ππ) = . b. π₯π¨π π (ππ) π₯π¨π π (ππ) = π₯π¨π (ππ) π₯π¨π (π) ≈ π. ππππ Therefore, π₯π¨π π (ππ) ≈ π. ππππ. c. 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. ππππ Therefore, π₯π¨π π (π) + π₯π¨π π (π) ≈ π. ππππ. 2. Use logarithmic properties and the fact that π₯π§(π) ≈ π. ππ and π₯π§(π) ≈ π. ππ to approximate the value of each of the following logarithmic expressions. Do not use a calculator. a. π₯π§(ππ ) π₯π§(ππ ) = π π₯π§(π) =π Therefore, π₯π§(ππ ) = π. b. π₯π§(π) π₯π§(π) = π₯π§(π) + π₯π§(π) ≈ π. ππ + π. ππ ≈ π. ππ Therefore, π₯π§(π) ≈ π. ππ. Lesson 13: Changing the Base 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 184 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II c. π₯π§(πππ) π₯π§(πππ) = π₯π§(π ⋅ ππ) = π₯π§(π) + π₯π§(ππ) ≈ π π₯π§(π) + π π₯π§(π) ≈ π. ππ + π. ππ ≈ π. ππ Therefore, π₯π§(πππ) ≈ π. ππ. d. π π π₯π§ ( ) π π₯π§ ( ) = π₯π§(π) − π₯π§(π) π = π₯π§(ππ ) − π₯π§(π) ≈ π(π. ππ) − π. ππ ≈ π. ππ π π Therefore, π₯π§ ( ) ≈ π. ππ. 3. Compare the values of π₯π¨π π (ππ) and π₯π¨π π ( π π ) without using a calculator. ππ Using the change of base formula, π₯π¨π π (ππ) π π₯π¨π π ( ) π π₯π¨π π (ππ) = −π = −π₯π¨π π (ππ) π = π₯π¨π π ( ) . ππ π₯π¨π π (ππ) = π Thus, π₯π¨π π (ππ) = π₯π¨π π ( π 4. π ). ππ Show that for any positive numbers π and π with π ≠ π and π ≠ π, π₯π¨π π (π) β π₯π¨π π (π) = π. Using the change of base formula, π₯π¨π π (π) = π₯π¨π π (π) π = . π₯π¨π π (π) π₯π¨π π (π) Thus, π₯π¨π π (π) β π₯π¨π π (π) = 5. π β π₯π¨π π (π) = π. π₯π¨π π(π) Express π in terms of π, π, and π if π₯π§(π) − π₯π§(π) = ππ. π₯π§(π) − π₯π§(π) = ππ π π₯π§ ( ) = ππ π π = πππ π π = π πππ Lesson 13: Changing the Base 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 185 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II 6. Rewrite each expression in an equivalent form that only contains one base-ππ logarithm. a. π₯π¨π π(πππ) π π₯π¨π π (πππ) = b. 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₯π¨π (π) π₯π¨π (π) π₯π¨π (π) π₯π¨π (π) π₯π¨π (π) π ), for positive real values of π ≠ π ππ π π₯π¨π ( ) π ππ = − π π₯π¨π π ( ) = ππ π₯π¨π (π) π₯π¨π (π) c. π₯π¨π π(πππππ) π π₯π¨π π (πππππ) = d. ) π = π π₯π¨π (π )+π₯π¨π (ππ 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¨π π(π. ππ) π₯π¨π π (π. ππ) = 7. π π₯π¨π (π ⋅ππ π₯π¨π (π) ππ π₯π¨π (πππ ) π₯π¨π (ππ)−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) Write each number in terms of natural logarithms, and then use the properties of logarithms to show that it is a rational number. a. π₯π¨π π(√ππ) π π π π₯π§(√ππ) π₯π§ (π ) π π₯π§(π) π = = = π₯π§(π) π₯π§(ππ ) π π₯π§(π) π b. π₯π¨π π(ππ) π₯π§(ππ) π₯π§(ππ ) π = = π₯π§(π) π₯π§(ππ ) π c. π π π₯π¨π π ( ) π π₯π§ ( ) π₯π§(π−π ) π π = =− π₯π§(π) π₯π§(ππ ) π 8. Write each expression as an equivalent expression with a single logarithm. Assume π, π, and π are positive real numbers. a. π₯π§(π) + π π₯π§(π) − π π₯π§(π) πππ π₯π§ ( π ) π Lesson 13: Changing the Base 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 186 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 13 M3 ALGEBRA II b. π π (π₯π§(π + π) − π₯π§(π³)) π+π π₯π§ (√ ) π c. (π + π) + π₯π§(π) (π + π) π₯π§(π) + π₯π§(π) = π₯π§(ππ+π ) + π₯π§(π) = π₯π§(ππ+π β π) 9. Rewrite each expression as sums and differences in terms of π₯π§(π), π₯π§(π), and π₯π§(π). a. π₯π§(ππππ ) π₯π§(π) + π₯π§(π) + π π₯π§(π) b. ππ ) πππ π₯π§ ( π − π₯π§(π) − π₯π§(π) − π₯π§(π) c. π π π₯π§ (√ ) π (π₯π§(π) − π₯π§(π)) π 10. Use base-π logarithms to rewrite each exponential equation as a logarithmic equation, and solve the resulting equation. Use the change of base formula to convert to a base-ππ logarithm that can be evaluated on a calculator. Give each answer to π decimal places. If an equation has no solution, explain why. a. πππ = ππ ππ = π₯π¨π π(ππ) π π = π₯π¨π π(ππ) π π₯π¨π (ππ) π= π π₯π¨π (π) π ≈ π. ππππ b. ππ = ππ β 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) π ≈ π. ππππ Lesson 13: Changing the Base 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 187 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II c. ππ+π − ππ = ππ ππ (ππ − π) = ππ ππ ππ = ππ ππ π = π₯π¨π π ( ) ππ ππ π₯π¨π ( ) ππ π= π₯π¨π (π) π₯π¨π (ππ) − π₯π¨π (ππ) π= π₯π¨π (π) π ≈ −π. ππππ d. π ππ = π. ππ ππ = π₯π¨π π (π. ππ) π₯π¨π (π. ππ) ππ = π₯π¨π (π) ππ ≈ −π. ππππ This equation has no real solution because ππ cannot be negative for any real number π. 11. In Lesson 6, you discovered that π₯π¨π (π β πππ ) = π + π₯π¨π (π) by looking at a table of logarithms. Use the properties of logarithms to justify this property for an arbitrary base π > π with π ≠ π. That is, show that π₯π¨π π (π β ππ ) = π + π₯π¨π π (π). π₯π¨π π (π β ππ ) = π₯π¨π π (π) + π₯π¨π π (ππ ) = π + π₯π¨π π (π) 12. Larissa argued that since π₯π¨π π (π) = π and π₯π¨π π (π) = π, then it must be true that π₯π¨π π (π) = π. π. Is she correct? Explain how you know. MP.3 π Larissa is not correct. If π₯π¨π π (π) = π. π, then ππ.π = π, so π = ππ = π√π. Since π ≠ π√π, Larissa’s calculation is not correct. 13. Extension: Suppose that there is some positive number π so that π₯π¨π π (π) = π. ππ π₯π¨π π (π) = π. ππ π₯π¨π π (π) = π. ππ. a. Use the given values of π₯π¨π π (π), π₯π¨π π (π), and π₯π¨π π (π) to evaluate the following logarithms. i. 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. ππ = π. ππ Lesson 13: Changing the Base 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 188 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II ii. π₯π¨π π (π) π₯π¨π π (π) = π₯π¨π π (ππ ) = π ⋅ π₯π¨π π (π) = π ⋅ π. ππ = π. ππ iii. π₯π¨π π (ππ) π₯π¨π π (ππ) = π₯π¨π π(π ⋅ π) = π₯π¨π π (π) + π₯π¨π π (π) = π. ππ + π. ππ = π. ππ iv. π₯π¨π π (πππ) π₯π¨π π (πππ) = π₯π¨π π(π ⋅ πππ) = π₯π¨π π (π) + π₯π¨π π (πππ) = π. ππ + π π₯π¨π π (ππ) = π. ππ + π(π. ππ) = π. ππ + π. ππ = π. ππ b. Use the change of base formula to convert π₯π¨π π (ππ) to base ππ, and solve for π. Give your answer to four decimal places. From part (iii) above, π₯π¨π π (ππ) = π. ππ. Then, π. ππ = π₯π¨π π (ππ) π₯π¨π ππ (ππ) π. ππ = π₯π¨π ππ (π) π π. ππ = π₯π¨π ππ (π) π = π₯π¨π ππ (π) π. ππ π π = πππ.ππ π ≈ π. ππππ. 14. Use a logarithm with an appropriate base to solve the following exponential equations. a. πππ = ππ π₯π¨π π (πππ ) = π₯π¨π π(ππ) ππ = π π π= π Lesson 13: Changing the Base 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 189 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II b. ππ+π = πππ π₯π¨π π (ππ+π ) = π₯π¨π π(πππ ) π + π = ππ ⋅ π₯π¨π π(π) π + π = ππ ⋅ π ππ = π π π= π c. πππ−π = πππ+π π₯π¨π π (πππ−π ) = π₯π¨π π (πππ+π ) (ππ − π) π₯π¨π π (π) = (π + π) π₯π¨π π (ππ) ππ − π = π(π + π) ππ − π = ππ + π π=π d. π π πππ = ( ) ππ π ππ π₯π¨π π (πππ ) = π₯π¨π π (( ) ) π π ππ π₯π¨π π (π) = ππ π₯π¨π π ( ) π ππ = ππ(−π) ππ = π π=π e. ππ.ππ+π = πππ π₯π¨π π (ππ.ππ+π ) = π₯π¨π π (πππ) (π. ππ + π)π₯π¨π π (π) = π₯π¨π π (ππ ) π. ππ + π = π π. ππ = π π=π 15. Solve each exponential equation. a. πππ = ππ b. π=π c. e. π=π πππ = πππ π= d. π π πππ−π = πππ = πππ+π πππ−π = πππ π= π π f. π π= π Lesson 13: π π πππ ππ−π =π π = −π Changing the Base 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 190 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II g. π πππ−π ππ = ππ h. =π π= π=π i. π = ππ π= k. ππ+π = ππ j. π₯π§(π) π₯π§(π) π = −π + πππ+π = ππ π= o. π= πππ−π = ππ+π π₯π¨π (ππ) π₯π¨π (π) ππ+π = ππ−π l. π = −π + π₯π¨π (ππ) m. π₯π§(ππ) π₯π§(π) π₯π¨π (π) − π₯π¨π (π) π₯π¨π (π) + π₯π¨π (π) πππ = π n. 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 = π + π₯π§(π) 16. In Problem 9(e) of Lesson 12, you solved the equation ππ = π−ππ+π using the logarithm base ππ. a. Solve ππ = π−ππ+π using the logarithm base π. π₯π¨π π (ππ ) = π₯π¨π π (π−ππ+π ) π = (−ππ + π)π₯π¨π π (π) π = −ππ 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(π + π π₯π¨π π (π)) = π π₯π¨π π (π) π= b. π π₯π¨π π (π) π + π π₯π¨π π (π) Apply the change of base formula to show that your answer to part (a) agrees with your answer to Problem 9(e) of Lesson 12. Changing from base π to base ππ, we see that π₯π¨π π (π) = π₯π¨π (π) . π₯π¨π (π) Then, 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) which was the answer from Problem 9(e) of Lesson 12. Lesson 13: Changing the Base 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 191 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 13 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II c. Solve ππ = π−ππ+π using the logarithm base π. π₯π¨π π (ππ ) = π₯π¨π π (π−ππ+π ) π π₯π¨π π (π) = −ππ + π ππ + π π₯π¨π π (π) = π π(π + π₯π¨π π (π)) = π π= d. π π + π₯π¨π π (π) Apply the change of base formula to show that your answer to part (c) also agrees with your answer to Problem 9(e) of Lesson 12. Changing from base π to base ππ, we see that π₯π¨π π (π) = π₯π¨π (π) . π₯π¨π (π) Then, 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+ which was the answer from Problem 9(e) of Lesson 12. 17. Pearl solved the equation ππ = ππ as follows: π₯π¨π (ππ ) = π₯π¨π (ππ) π π₯π¨π (π) = π π= π . π₯π¨π (π) Jess solved the equation ππ = ππ as follows: π₯π¨π π (ππ ) = π₯π¨π π (ππ) π π₯π¨π π (π) = π₯π¨π π (ππ) π = π₯π¨π π (ππ). Is Pearl correct? Is Jess correct? Explain how you know. Both Pearl and Jess are correct. If we take Jess’s solution and apply the change of base formula, we have π = π₯π¨π π (ππ) π₯π¨π (ππ) = π₯π¨π (π) π = . π₯π¨π (π) MP.3 Thus, the two solutions are equivalent, and both students are correct. Lesson 13: Changing the Base 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 192 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
