Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM 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 will occur 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 demonstration of six basic properties of logarithms theoretically using the properties of exponents instead of numerical approximation as has been done in prior lessons. In the Problem Set, students will apply these properties to calculating logarithms, rewriting logarithmic expressions, and solving exponential equations base 10 (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 exercise. These exercises serve to remind students of the βmost important propertyβ of logarithms and prepare them for justifying the properties later in the lesson. Verify that students are breaking up the logarithm, using the property, and evaluating the logarithm at known values (e.g., 0.1, 10, 100). Opening Exercise Use the approximation π₯π¨π (π) β 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) β π + π. πππ β π. πππ 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) β βπ + π. ππππ β βπ. ππππ ο§ Ask students who are having trouble with any part of this exercise, βHow is the number in parentheses related to 2?β Follow that with the question, βSo how might you find its logarithm given that you know log(2)?β 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=2β 2 40 = 101 β 2 β 2 0.4 = 10β1 β 2 β 2 400 = 102 β 2 β 2 0.04 = 10β2 β 2 β 2 ο§ Advanced students may be challenged with a more general version of part (c): log(2π ). This can be explored by having students find log(25 ), log(26 ), etc. Lesson 12: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 160 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 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) β π β (π. ππππ) β π. ππππ Discussion (4 minutes) Discuss the properties of logarithms used in the Opening Exercises. ο§ In all three parts of the Opening Exercise, we used the important property log(π₯π¦) = log(π₯) + log(π¦). ο§ What are some other properties we used? οΊ ο§ We also used log(10) = 1 and log(0.1) = β1. It can be helpful to look further at properties of expressions involving logarithms. 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 β¦ ο§ When π₯ = 1? οΊ ο§ When π₯ = 0? οΊ ο§ πΏ=9 When π₯ = 10π ? οΊ ο§ The logarithm πΏ is not defined. There is no exponent of 10 that yields a value of 0. When π₯ = 109 ? οΊ ο§ πΏ=0 πΏ=π 3 When π₯ = β10? οΊ πΏ= 1 3 Exercises 1β6 (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. 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: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 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 1β6, explain why each statement below is a property of base ππ logarithms. 1. 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. Scaffolding: 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. Property 4: For any π > π, πππ₯π¨π (π) = π. Scaffolding: 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. Students in groups that struggle with Exercises 3β6 should be encouraged to check the property with numerical values for π, π₯, π, and π. The check may suggest an explanation. 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: Again, use an exponent rule as well as property π. 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 7β10 (8 minutes) These exercises bridge 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 as you see fit. 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 βtake the log of both sidesβ 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 apply to solve the equation. Any logarithm will work to solve the equations if applied properly, so students may find equivalent answers that appear to be different from those listed here. Lesson 12: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 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). a. π₯π¨π (πππ ππ ) π₯π¨π (π) + π π₯π¨π (π) + π π₯π¨π (π) b. π₯π¨π (βππ 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. πππβπ = π π+π ππ π₯π¨π (πππβπ ) = βπ₯π¨π (πππ+π ) π β π = β(π + π) ππ = π π=π c. πππππ = ππππβπ π₯π¨π (πππππ ) = π₯π¨π (ππππβπ ) ππ π₯π¨π (πππ) = (ππ β π) ππ = ππ β π π = βπ Lesson 12: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 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. πππ = ππ a. π₯π¨π (πππ ) = π₯π¨π (ππ ) π = π π₯π¨π (π) πππ b. π +π = ππ π +π π₯π¨π (πππ ) = π₯π¨π (ππ) π π + π = π₯π¨π (ππ) π = ±βπ₯π¨π (ππ) β π ππ = ππ c. 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 1β6, we have established that the property holds, so we can use these properties in our future work with logarithms. The Lesson Summary might be posted in the classroom for at least the rest of the module. The Exit Ticket asks the students to show that properties 7 and 8 hold. 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: 1. 2. π π π π₯π¨π ( ) = π₯π¨π (π) β π₯π¨π (π) π π₯π¨π ( ) = βπ₯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: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 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 1. State as many of the six properties of logarithms as you can. 2. Use the properties of logarithms to show that log ( ) = βlog(π₯) for all π₯ > 0. 3. Use the properties of logarithms to show that log ( ) = log(π₯) β log(π¦) for π₯ > 0 and π¦ > 0. 1 π₯ π₯ π¦ Lesson 12: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 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 1. State as many of the six properties of logarithms as you can. π₯π¨π (π) = π π₯π¨π (ππ) = π π₯π¨π (πππ ) = π πππ₯π¨π (π) = π π₯π¨π (π β π) = π₯π¨π (π) + π₯π¨π (π) π₯π¨π (ππ ) = π β π₯π¨π (π) 2. π π Use the properties of logarithms to show that π₯π¨π ( ) = βπ₯π¨π (π) for π > π. By property 6, π₯π¨π (ππ ) = π β π₯π¨π (π). π π Let π = βπ, then for π > π, π₯π¨π (πβπ ) = (βπ) β π₯π¨π (π), which is equivalent to π₯π¨π ( ) = βπ₯π¨π (π). π π Thus, for any π > π, π₯π¨π ( ) = βπ₯π¨π (π). 3. π π Use the properties of logarithms to show that π₯π¨π ( ) = π₯π¨π (π) β 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 2 above, for π > π, π₯π¨π (πβπ ) = (βπ) β π₯π¨π (π). Therefore, π π π₯π¨π ( ) = π₯π¨π (π) + π₯π¨π ( ) π π = π₯π¨π (π) + (βπ)π₯π¨π (π) = π₯π¨π (π) β π₯π¨π (π). π π Thus, for any π > π and π > π, π₯π¨π ( ) = π₯π¨π (π) β π₯π¨π (π). Problem Set Sample Solutions Problems 1β7 give students an opportunity to practice using the properties they have established in this lesson, and in the remaining problems, students apply base 10 logarithms to solve simple exponential equations. 1. Use the approximate logarithm values below to estimate 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) β π. ππππ. Lesson 12: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 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) β 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) β 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 ( ) = βπ₯π¨π (π) β βπ. ππππ. π f. π π π₯π¨π ( ) Using property 8, π π₯π¨π ( ) = π₯π¨π (π) β π₯π¨π (π) β βπ. πππ. π g. π π₯π¨π (βπ) Using property 6, π π π₯π¨π (βπ) = π₯π¨π (ππ ) = 2. π π₯π¨π (π) β 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 πβπ c. π₯π¨π (πΏπ ) d. ππ e. π₯π¨π (ππ) π π₯π¨π (βπ) π π π f. π π₯π¨π ( β ) π π₯π¨π (πΏππ ππ ) π + ππ + ππ πβπ π Lesson 12: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 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 ( ) β π₯π¨π ( ) π π₯π¨π ( ) π c. π π π₯π¨π (ππ) + π₯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 (βπ) + π₯π¨π ( ) + π π₯π¨π (π) π π π₯π¨π (π± π ) b. π π₯π¨π ( πβπ) + π₯π¨π (βππ ) π₯π¨π (π) c. π π π₯π¨π (π) + π π₯π¨π (π) β π₯π¨π (π) π₯π¨π ( d. π π πππ βπ (π₯π¨π (π) β π π₯π¨π (π) + π₯π¨π (π)) π π₯π¨π ( β e. ) ππ ) ππ π(π₯π¨π (π) β π₯π¨π (ππ)) + π(π₯π¨π (π) β 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. Use properties of logarithms to rewrite the following expressions in an equivalent form containing only π₯π¨π (π), π₯π¨π (π), π₯π¨π (π), and numbers. a. πππ ππ ) βπ π₯π¨π ( π₯π¨π (π) + π π₯π¨π (π) + π π₯π¨π (π) β π π₯π¨π (π) π π b. π₯π¨π ( ππ βπππ ππ π ) π π π₯π¨π (ππ) β π₯π¨π (π) + π₯π¨π (π) β π₯π¨π (π) π π Lesson 12: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 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 c. πππππ ) ππ π₯π¨π ( π + π π₯π¨π (π) β π π₯π¨π (π) d. ππ ππ ) πππ π₯π¨π (β π (π π₯π¨π (π) + π π₯π¨π (π) β π β π₯π¨π (π)) π e. π₯π¨π ( π ) ππππ π βπ β π π₯π¨π (π) β π₯π¨π (π) 6. π π Express π₯π¨π ( β π π π ) + (π₯π¨π ( ) β 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 ( β ) + (π₯π¨π ( ) β π₯π¨π ( )) = π₯π¨π ( ) + π₯π¨π ( ) β π₯π¨π ( ) π π+π π π+π π(π + π) π π+π = βπ₯π¨π (π(π + π)) β π₯π¨π (π) + π₯π¨π (π + π) = βπ₯π¨π (π) β π₯π¨π (π + π) β π₯π¨π (π) + π₯π¨π (π + π) = βπ π₯π¨π (π) 7. Show that π₯π¨π (π + βππ β π) + π₯π¨π (π β βππ β π) = π for π β₯ π. π₯π¨π (π + βππ β π) + π₯π¨π (π β βππ β π) = π₯π¨π ((π + βππ β π) (π β βππ β π)) π = π₯π¨π (ππ β (βππ β π) ) = π₯π¨π (ππ β ππ + π) = π₯π¨π (π) =π 8. If ππ = πππ.ππ, 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 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 = ±βπ₯π¨π (πππ) Lesson 12: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 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 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 β π₯π¨π (ππ) = π₯π¨π (πππ β π) ππ β 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 β π₯π¨π (π) = π₯π¨π (πππ) + π₯π¨π (π) π + π₯π¨π (π) ππ = π₯π¨π (π) π= e. π + π₯π¨π (π) π π₯π¨π (π) ππ± = πβππ±+π π₯π¨π (ππ ) = π₯π¨π (πβππ+π ) π π₯π¨π (π) = (βππ + 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) b. πππ = ππ π = π₯π¨π (ππ) c. πππ = πππ π³ = π₯π¨π (πππ) Lesson 12: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 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 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 (ππ) = 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 a constant difference of 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, and ππππ is less than that, 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 ( ) = ( ) = f. π ππ and π πππ , so the solution is between π and π. π π π π π π , and is between and , so the solution is between π and π. π π π π πππ = ππππ πππ = ππππ. Since ππππ is less than ππππ 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 ππππth. a. πππ = ππ π₯π¨π (πππ ) = π₯π¨π (ππ) π π₯π¨π (ππ) = π₯π¨π (ππ) π₯π¨π (ππ) π= π₯π¨π (ππ) π β π. πππ b. πππ = ππ π= π₯π¨π (ππ) π₯π¨π (ππ) π β π. πππ Lesson 12: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 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 c. ππππ = ππππ π= π₯π¨π (ππππ) π₯π¨π (πππ) π β π. πππ d. ( π π ) = π. ππ ππ π π=β π₯π¨π ( π ) ππ π β π. πππ e. π π π ( ) = π π π π₯π¨π ( ) π π π₯π¨π ( ) π π β π. πππ π= f. πππ = ππππ π= π₯π¨π (ππππ) π₯π¨π (ππ) π β π. πππ 13. Show that the value of π that satisfies 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 β πππ . Thus, π = π₯π¨π (π) + π is a solution to the equations πππ = π β 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= b. π₯π¨π (π) π₯π¨π (π) πππβπ = ππ π₯π¨π (πππβπ ) = π₯π¨π (ππ) π = π + π₯π¨π (ππ) Lesson 12: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 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 c. πππ + πππ+π = ππ πππ (π + ππ) = ππ πππ = π π=π d. π β ππ = ππ βππ = π ππ = βπ 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¨) β 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¨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 will be 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 + ). 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) Lesson 12: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 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 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 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: Date: Properties of Logarithms 7/29/17 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 174 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
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