Lesson 15: Why Were Logarithms Developed?

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 M3

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Student Outcomes . Students use logarithm tables to calculate products and quotients of multi-digit numbers without technology. . Students understand that logarithms were developed to speed up arithmetic calculations by reducing multiplication and division to the simpler operations of addition and subtraction. . Students solve logarithmic equations of the form log( ) = log( ) by equating = .

푋 푌 푋 푌 Lesson Notes This final lesson in Topic B includes two procedures that seem to be different but are closely related mathematically. First, students work with logarithm tables to see how applying logarithms simplified calculations in the days before computing machines and electronic technology. They also learn a bit of the history of how and why logarithms first appeared—a history often obscured when logarithmic functions are introduced as inverses of exponential functions. The last two pages of this document contain a base 10 table of logarithms that can be copied and distributed; such tables are also available on the web. Then, students learn to solve the final type of logarithmic equation, log( ) = log( ), where and are either real numbers or expressions that take on positive real values (A.SSE.A.2, F-LE.A.4). Using either technique requires that we know that the logarithm is a one-to-one function; that is, if log( ) = log푋( ), then 푌 = . Students푋 푌 do not yet have the vocabulary to be told this directly, but we do state it as fact in this lesson, and they will further explore the idea of one- to-one functions in Pre-Calculus. As with Lessons 10 and 12, this푋 lesson involves푌 only푋 base푌 10 logarithms, but the problem set does require that students do some work with logarithms base and base 2. Remind students to check for extraneous solutions when solving logarithmic equations. 푒

Classwork Discussion (4 minutes): How to Read a Table of Logarithms . For this lesson, we will pretend that we live in the time when logarithms were discovered, before there were calculators or computing machines. In this time, scientists, merchants, and sailors needed to make calculations for both astronomical observation and navigation. Logarithms made these calculations much easier, faster, and more accurate than calculation by hand. In fact, noted mathematician Pierre-Simon LaPlace (France, circa 1800) said that “[logarithms are an] admirable artifice which, by reducing to a few days the labour of many months, doubles the life of the astronomer, and spares him the errors and disgust inseparable from long calculations.” . A typical table of common logarithms, like the table at the end of this document, has many rows of numbers arranged in ten columns. The numbers in the table are decimals. In our table, they are given to four decimal places, and there are 90 rows of them (some tables of logarithms have 900 rows). Down the left-hand side of the table are the numbers from 1.0 to 9.9. Across the top of the table are the numbers from 0 to 9. To read the table, you locate the number whose logarithm you want using the numbers down the left of the table followed by the numbers across the top.

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. What does the number in the third row and second column represent (the entry for 1.21)?  The logarithm of 1.21, which is 0.0828. . The logarithm of numbers larger than 9.9 and smaller than 1.0 can also be found using this table. Suppose you want to find log(365). Is there any way we can rewrite this number to show a number between 1.0 and 9.9?  Rewrite 365 in scientific notation: 3.65 × 10 . . Can we simplify log(3.65 × 10 )? 2 MP.5 2 ( ) ( ) &  We can apply the formula for the logarithm of a product. Then, we have log 10 + log 3.65 = MP.6 2 + log(3.65). 2 . Now, all that is left is to find the value of log(3.65) using the table. What is the value of log(365)?  The table entry is 0.5623. That means log(365) 2 + 0.5623, so log(365) 2.5623. . How would you find log(0.365)? ≈ ≈  In scientific notation, 0.365 = 3.65 × 10 . So, once again you would find the row for 3.6 and the column for 5, and you would again find the− 1number 0.5623. But this time, you would have log(0.365) 1 + 0.5623, so log (0.365) 0.4377.

≈ − ≈

Example 1 (7 minutes) Scaffolding: Students will multiply multi-digit numbers without technology, and then use a table of . Students may need to be logarithms to find the same product using logarithms. reminded that if the logarithm is greater than . Find the product 3.42 × 2.47 without using a calculator. 1, a power of 10 greater  Using paper and pencil, and without any rounding, students should get than 1 is involved, and 8.4474. The point is to show how much time the multiplication of multi- only the decimal part of digit numbers can take. the number will be found in the table. . How could we use logarithms to find this product? . Struggling students should  If we take the logarithm of the product, we can rewrite the product as a attempt a simpler product sum of logarithms. such as 1.20 × 6.00 to . Rewrite the logarithm of the product as the sum of logarithms. illustrate the process. .  log(3.42 × 2.47) = log(3.42) + log(2.47) Advanced students may use larger or more precise . log(3.42) log(2.47) Use the table of logarithms to look up the values of and . numbers as a challenge.  According to the table, log(3.42) 0.5340, and log(2.47) 0.3927. To multiply a product such . Approximate the logarithm log(3.42 × 2.47). as 34.293 × 107.9821, ≈ ≈ students will have to  The approximate sum is employ scientific notation log(3.42 × 2.47) 0.5340 + 0.3927 and the property for the 0.9267. logarithm of a product. ≈ . What if there is more than one number that has a logarithm of 0.9267? ≈ Suppose that there are two numbers and that satisfy log( ) = 0.9267 and log( ) = 0.9267. Then, 10 . = , and 10 . = , so that = . This means that there is only one number that has the logarithm0 9267 0.9267. So,0 9267what is that number?푋 푌 푋 푌 푋 푌 푋 푌 . Can we find the exact number that has logarithm 0.9267 using the table?  The table says that log(8.44) 0.9263, and log(8.45) 0.9269.

≈ ≈

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. Which is closer?  log(8.45) 0.9267 . Since log(3.42 × 2.47) log (8.45), what can we conclude is an approximate value for 3.42 × 2.47? ≈  Since log(3.42 × 2.47) log (8.45), we know that 3.42 × 2.47 8.45. ≈ . Does this agree with the product you found when you did the calculation by hand? ≈ ≈  Yes, by hand we found that the product is 8.4474, which is approximately 8.45.

Discussion (2 minutes) In the above example, we showed that there was only one number that had logarithm 0.9267. This result generalizes to any number and any base of the logarithm—if log ( ) = log ( ), then = . We need to know this property both to use a logarithm table to look up values that produce푏 a certain푏 logarithmic value and to solve logarithmic equations later in the lesson. 푋 푌 푋 푌

If and are positive real numbers, or expressions that take on the value of positive real numbers, and log ( ) = log , then = . 푋 푌 푏 푏 푋 푌 푋 푌

Example 2 (4 minutes)

This example is a continuation of the first example, with the addition of scientific notation to further explain the power of logarithms. Because much of the reasoning was explained in Example 1, this should take much less time to work through. . Now, what if we needed to calculate (3.42 × 10 ) × (5.76 × 10 )? . Take the logarithm of this product, and find its approximate14 value12 using the logarithm table.  log (3.42 × 10 ) × (5.76 × 10 ) = log(3.42) + log(10 ) + log(5.76) + log(10 ) 14 12= log(3.42) + log(5.76) +1414 + 12 12 � 0�.5340 + 0.7604 + 26 27.2944 ≈ . Look up 0.2944 in the logarithm table. ≈  Since log(1.97) 0.2945, we can say that 0.2944 log(1.97). . How does that tell us which number has a logarithm approximately equal to 27.2944? ≈ ≈  log(1.97 × 10 ) = 27 + log(1.97), so log(1.97 × 10 ) 27.2944. . Finally, what is an approximate27 value of the product (3.42 × 1027 ) × (5.76 × 10 )? ≈  (3.42 × 10 ) × (5.76 × 10 ) 1.97 × 10 . 14 12 14 12 27 ≈ Example 3 (6 minutes)

. According to one estimate, the mass of the earth is roughly 5.28 × 10 kg, and the mass of the moon is about 7.35 × 10 kg. Without using a calculator but using the table of logarithms24 , find how many times greater the mass of earth22 is than the mass of the moon.

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. × .  Let be the ratio of the two masses. Then = = 10 . × 24 . 5 28 10 5 28 2 22 Taking푅 the logarithm of each side, 푅 7 35 10 7 35 ∙ 5.28 log( ) = log 10 7.35 2 푅 5.28� ∙ � = 2 + log 7.35 = 2 + log(�5.28)� log(7.35)

2 + 0.7226 0−.8663 ≈ 1.8563. − . Find 0.8563 in the table entries to estimate ≈.  In the table, 0.8563 is closest to log(7.18). 푅  So, log(71.8) 1.8563, and therefore, the mass of earth is approximately 71.8 times that of the moon. ≈ . Logarithms turn out to be very useful in dealing with especially large or especially small numbers. Scientific notation was probably developed as an attempt to do arithmetic using logarithms. How does it help to have those numbers expressed in scientific notation if we are going to use a logarithm table to perform multiplication or division?  Again, answers will differ, but students should at least recognize that scientific notation is helpful in working with very large or very small numbers. Using scientific notation, we can express each number as the product of a number between 1 and 10, and a power of 10. Taking the logarithm of the number allows us to use properties of logarithms base 10 to handle more easily any number where 0 < < 1 or 10. The logarithm of the number between 1 and 10 can be read from the table, and the exponent of the power of 10 can then be added to it. 푛 푛 푛 ≥ . Whenever we have a number of the form × 10 where is an integer and is a number between 1.0 and 9.9, the logarithm of this number will always be 푛+ log( ) and can be evaluated using a table of logarithms like the one included in this lesson. 푘 푛 푘 푛 푘

Discussion (4 minutes) Logarithms were devised by the Scottish mathematician John Napier (1550–1617) with the help of the English mathematician Henry Briggs (1561–1630) to simplify arithmetic computations with multi-digit numbers by turning multiplication and division into addition and subtraction. The basic idea is that while a sequence of powers like 2 , 2 , 2 , 2 , 2 , 2 , … is increasing multiplicatively, the sequence of its exponents is increasing additively. If numbers can0 be1 represented2 3 4 5 as the powers of a base, they can be multiplied by adding their exponents and divided by subtracting their exponents. Napier and Briggs published the first tables of what came to be called base 10 or common logarithms.

. It was Briggs’s idea to base the logarithms on the number 10. Why do you think he made that choice?  The number 10 is the base of our number system. So, taking 10 as the base of common logarithms makes hand calculations with logarithms easier. It’s really the same argument that makes scientific notation helpful: Powers of 10 are easy to use in calculations.

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Exercises 1–2 (12 minutes) Now that we know that if two logarithmic expressions with the same base are equal, then the arguments inside of the logarithms are equal, and we can solve a wider variety of logarithmic equations without invoking the definition each time. Due to the many logarithmic properties that the students now know, there are multiple approaches to solving these equations. Discuss different approaches with the students and their responses to Exercise 2.

Exercises 1–2

1. Solve the following equations. Remember to check for extraneous solutions because logarithms are only defined for positive real numbers. a. ( ) = ( ) ퟐ 퐥퐨퐠 풙 퐥퐨퐠 ퟒퟗ = = ퟐ or = 풙 ퟒퟗ Check: Both solutions are valid since and ( 풙 ) ퟕare both풙 −positiveퟕ numbers. ퟐ ퟐ The two solutions are and – . ퟕ −ퟕ Scaffolding: ퟕ ퟕ If the class seems to be struggling with the b. ( + ) + ( + ) = ( ) process to solve logarithmic equations, then 퐥퐨퐠 풙 ퟏ( +퐥퐨퐠)( 풙 ퟐ) = 퐥퐨퐠 (ퟕ풙 − ퟏퟕ ) either encourage them to create a graphic ( + )( ) = 퐥퐨퐠� 풙 ퟏ 풙 − ퟐ � 퐥퐨퐠 ퟕ풙 − ퟏퟕ organizer that summarizes the types of = problems and approaches that they should 풙 ퟏ 풙 − ퟐ ퟕ풙 − ퟏퟕ ퟐ + = use in each case, or hang one on the board 풙 − 풙 − ퟐ ퟕ풙 − ퟏퟕ ( ퟐ )( ) = for reference. A sample graphic organizer is 풙 − ퟖ풙 ퟏퟓ ퟎ included. 풙 − ퟓ 풙 − ퟑ= orퟎ = Check: Since + , 풙 ,ퟑ and 풙 ퟓ are all positive for either Rewrite problem in the form… = or = , both solutions are valid. 풙 ퟏ 풙 − ퟐ ퟕ풙 − ퟏퟕ log ( ) = log ( ) = log ( ) Then… 풙Thus,ퟑ the 풙solutionsퟓ to this equation are and 5. 푏 푏 푏 푌= 퐿 푌 = 푍 ퟑ 퐿 푏 푌 푌 푍 c. ( + ) = ( ) ퟐ 퐥퐨퐠 풙 ퟏ 퐥퐨퐠�풙 풙 − ퟐ � + = ( ) ퟐ = 풙 ퟏ 풙 풙 − ퟐ = ퟐ 풙 − ퟐ풙 ퟏ = −ퟐ풙 ퟏ 풙 − Check: Both + > and ( ) > , so the solution is valid. ퟐ ퟐ ퟏ ퟏ ퟏ ퟏ �− ퟐ� ퟏ ퟎ − ퟐ − ퟐ − ퟐ ퟎ − ퟐ Thus, is the only valid solution to this equation. ퟏ − ퟐ

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d. ( + ) + ( ) = ( )

퐥퐨퐠 풙 ퟒ 퐥퐨퐠 풙 − ퟏ 퐥퐨퐠 ퟑ풙( + )( ) = ( ) ( + )( ) = 퐥퐨퐠� 풙 ퟒ 풙 − ퟏ � 퐥퐨퐠 ퟑ풙 + = 풙 ퟒ 풙 − ퟏ ퟑ풙 ퟐ = 풙 ퟑ풙 − ퟒ ퟑ풙 ퟐ = or = 풙 − ퟒ ퟎ Check: Since ( ) is undefined when = , there풙 ퟐ is an풙 extraneous−ퟐ solution of = .

The only valid퐥퐨퐠 solutionퟑ풙 to this equation is 풙 . 55T −ퟐ 풙 −ퟐ

ퟐ e. ( ) ( ) = ( ) ퟐ 퐥퐨퐠 풙 − 풙 − 퐥퐨퐠 풙 − ퟐ 퐥퐨퐠( 풙 − ퟑ) + ( ) = ( ) ( )( ) = ( ퟐ ) 퐥퐨퐠 풙 − ퟐ 퐥퐨퐠 풙 − ퟑ 퐥퐨퐠 풙 − 풙 ( )( ) = ퟐ 퐥퐨퐠� 풙 − ퟐ 풙 − ퟑ � 퐥퐨퐠 풙 − 풙 + = ퟐ 풙 − ퟐ 풙 − ퟑ 풙 − 풙 ퟐ = ퟐ 풙 − ퟓ풙 ퟔ 풙 − 풙 ퟒ풙 = ퟔ ퟑ 풙 Check: When = , we have < , so ( ), ퟐ ( ), and ( ) are all undefined. So, ퟐ the solution = ퟑ is extraneous. 풙 ퟐ 풙 − ퟐ ퟎ 퐥퐨퐠 풙 − ퟐ 퐥퐨퐠 풙 − ퟑ 퐥퐨퐠 풙 − 풙 ퟑ There are no풙 validퟐ solutions to this equation.

f. ( ) + ( ) + ( + ) = ( )

퐥퐨퐠 풙 퐥퐨퐠 풙 − ퟏ 퐥퐨퐠 풙 ퟏ (ퟑ 퐥퐨퐠 풙)( + ) = ( ) ( ) = ( ퟑ) 퐥퐨퐠�풙 풙 − ퟏ 풙 ퟏ � 퐥퐨퐠 풙 ퟑ = ퟑ 퐥퐨퐠 풙 − 풙 퐥퐨퐠 풙 ퟑ = ퟑ 풙 − 풙 풙 Since ( ) is undefined, = is an extraneous solution.풙 ퟎ

There 퐥퐨퐠are noퟎ valid solutions풙 to thisퟎ equation.

g. ( ) = ( )

Two퐥퐨퐠 풙possible− ퟒ approaches−퐥퐨퐠 풙 − ퟐ to solving this equation are shown. ( ) + ( ) = ( ) = ( )( ) = ( ) ퟏ 퐥퐨퐠 풙 − ퟒ( 퐥퐨퐠)(풙 − ퟐ) = ퟎ 퐥퐨퐠 풙 − ퟒ = 퐥퐨퐠 � � 풙 − ퟐ 퐥퐨퐠� 풙 − ퟒ 풙 −+ퟐ � = 퐥퐨퐠 ퟏ ퟏ 풙 − ퟒ 풙 − ퟐ ퟏ ( )( 풙 − ퟒ) = ퟐ + = 풙 − ퟐ 풙ퟐ − ퟔ풙 ퟖ ퟏ + = = ± 풙 − ퟒ 풙 − ퟐ ퟏ 풙 − ퟔ풙 ퟕ ퟎ ퟐ + = 풙 − ퟔ풙 ퟖ ퟏ 풙 ퟑ √ퟐ ퟐ = ± 풙 − ퟔ풙 ퟕ ퟎ

Check: If = , then풙 <ퟑ , so√ퟐ ( ) is undefined. Thus is an extraneous solution.

The only valid풙 ퟑsolution− √ퟐ to this풙 equationퟐ 퐥퐨퐠 is 풙+− ퟐ . ퟑ − √ퟐ

ퟑ √ퟐ

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2. How do you know whether or not you need to use the definition of logarithm to solve an equation involving logarithms as we did in Lesson or whether you can use the methods of this lesson?

If the equation involves only logarithmicퟏퟓ expressions, then it can be reorganized to be of the form ( ) = ( ) and then solved by equating = . If there are constants involved, then the equation can be solved by the definition. 퐥퐨퐠 푿 퐥퐨퐠 풀 푿 풀

Closing (2 minutes) Ask students the following questions and after coming to a consensus, have students record the answers in their notebooks. . How do we use a table of logarithms to compute a product of two numbers and ?  We look up approximations to log( ) and log( ) in the table, add those logarithms, and then look up 푥 푦 the sum in the table to extract the approximate product. 푥 푦 . Does this process provide an exact answer? Explain how you know.  It is only an approximation because the table only allows us to look up to two decimal places and log( ) to decimal places. 푥 . How do we solve an equation in which every term contains a logarithm? 푥 푓표푢푟  We rearrange the terms to get an equation of the form log( ) = log ( ), then equate = , and solve from there. 푋 푌 푋 푌 . How does that differ from solving an equation that contains constant terms?  If an equation has constant terms, then we rearrange the equation to the form log( ) = , apply the definition of the logarithm, and solve from there. 푋 푐

Lesson Summary

A table of base logarithms can be used to simplify multiplication of multi-digit numbers:

1. To computeퟏퟎ × for positive real numbers and , look up the values ( ) and ( ) in the logarithm table. 푨 푩 푨 푩 퐥퐨퐠 푨 퐥퐨퐠 푩 2. Add ( ) and ( ). The sum can be written as + , where is an integer and < is the decimal part. 퐥퐨퐠 푨 퐥퐨퐠 푩 풌 풅 풌 ퟎ ≤ 풅 ퟏ 3. Look back at the table and find the entry closest to the decimal part, . × . 4. The product of that entry and is an approximation to 풅 풌 ퟏퟎ 푨 푩 A similar process simplifies division of multi-digit numbers:

1. To compute ÷ for positive and , look up the values ( ) and ( ) in the logarithm table. 2. Calculate ( ) ( ). The difference can be written as + , where is an integer and < is 푨 푩 푨 푩 퐥퐨퐠 푨 퐥퐨퐠 푩 the decimal part. 퐥퐨퐠 푨 − 퐥퐨퐠 푩 풌 풅 풌 ퟎ ≤ 풅 ퟏ 3. Look back at the table, and find the entry closest to the decimal part, . ÷ . 4. The product of that entry and is an approximation to 풅 풌 ퟏퟎ 푨 푩 For any positive values and , if ( ) = ( ), we can conclude that = . This property is the essence of

how a logarithm table works, and it allows풃 us to solve풃 equations with logarithmic expressions on both sides of the 푿 풀 퐥퐨퐠 푿 퐥퐨퐠 풀 푿 풀

Exit Ticket (4 minutes)

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Name Date

Lesson 15: Why Were Logarithms Developed?

Exit Ticket

The surface area of Jupiter is 6.14 × 10 km , and the surface area of the Earth is 5.10 × 10 km². Without using a calculator but using the table of logarithms10 , find2 how many times greater the surface area of Jupiter8 is than the surface area of Earth.

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Exit Ticket Sample Solutions

The surface area of Jupiter is . × , and the surface area of the Earth is . × ². Without using a calculator but using the table of logarithmsퟏퟎ , findퟐ how many times greater the surface area of Jupiterퟖ is than the surface area of Earth. ퟔ ퟏퟒ ퟏퟎ 퐤퐦 ퟓ ퟏퟎ ퟏퟎ 퐤퐦

. × . Let be the ratio of the two surface areas. Then, = = . . × ퟏퟎ . ퟔ ퟏퟒ ퟏퟎ ퟔ ퟏퟒ ퟐ 푹 푹 ퟖ ∙ ퟏퟎ Taking the logarithm of each side, ퟓ ퟏퟎ ퟏퟎ ퟓ ퟏퟎ . ( ) = . ퟐ ퟔ ퟏퟒ . 퐥퐨퐠 푹 = 퐥퐨퐠+� ∙ ퟏퟎ � ퟓ ퟏퟎ . ퟔ ퟏퟒ = ퟐ + 퐥퐨퐠(� . )� ( . ) ퟓ ퟏퟎ + . . ퟐ 퐥퐨퐠 ퟔ ퟏퟒ − 퐥퐨퐠 ퟓ ퟏퟎ . . ≈ ퟐ ퟎ ퟕퟖퟖퟐ − ퟎ ퟕퟎퟕퟔ Find . in the table entries to estimate .≈ ퟐ ퟎퟖퟎퟔ

Look ퟎupퟎퟖퟎퟔ . , which is closest to ( . 푹). Note that + . ( ) + ( . ), so

( ퟎ) ퟎퟖퟎퟔ. . Therefore, the퐥퐨퐠 surfaceퟏ ퟐퟎ area of Jupiterퟐ is approximatelyퟎ ퟎퟖퟎퟔ ≈ 퐥퐨퐠 ퟏퟎퟎ times퐥퐨퐠 thatퟏ oퟐퟎf Earth. 퐥퐨퐠 ퟏퟐퟎ ≈ ퟐ ퟎퟖퟎퟔ ퟏퟐퟎ

Problem Set Sample Solutions These problems give students additional practice using base 10 logarithms to perform arithmetic calculations and solve equations.

1. Use the table of logarithms to approximate solutions to the following logarithmic equations. a. ( ) = .

퐥퐨퐠In the풙 tableퟎ, ퟓퟎퟒퟒ. is closest to ( . ), so ( ) ( . ). Therefore, . . ퟎ ퟓퟎퟒퟒ 퐥퐨퐠 ퟑ ퟏퟗ 퐥퐨퐠 풙 ≈ 퐥퐨퐠 ퟑ ퟏퟗ 풙 ≈ ퟑ ퟏퟗ b. ( ) = . [Hint: Begin by writing . as [( . ) + ] .]

퐥퐨퐠 풙 −ퟎ ퟓퟎퟒퟒ ( −)ퟎ=ퟓퟎퟒퟒ[( . −ퟎ) +ퟓퟎퟒퟒ] ퟏ − ퟏ = . 풍풐품 풙 −ퟎ ퟓퟎퟒퟒ ퟏ − ퟏ

In the table, . is closest to ( . ), so ( ) ( . ) 50T ퟎ ퟒퟗퟓퟔ − ퟏ ퟎ ퟒퟗퟓퟔ 풍풐품 ퟑ ퟏퟑ 풍풐품( . 풙 )≈ 풍풐품(ퟑ ퟏퟑ) − ퟏ . ≈ 풍풐품 ퟑ ퟏퟑ − 풍풐품 ퟏퟎ ퟑ ퟏퟑ ≈ 풍풐품(� . �). ퟏퟎ Therefore, . . ≈ 풍풐품 ퟎ ퟑퟏퟑ

Alternatively,풙 ≈ ퟎ .ퟑퟏퟑ is the opposite of . , so is the reciprocal of the answer in part (a). Thus, . . . −ퟏ −ퟎ ퟓퟎퟒퟒ ퟎ ퟓퟎퟒퟒ 풙 풙 ≈ ퟑ ퟏퟗ ≈ ퟎ ퟑퟏퟑ

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c. ( ) = .

퐥퐨퐠 풙 ퟑퟓ ퟓퟎퟒퟒ ( ) = + . = ( ) + . 퐥퐨퐠 풙 ퟑퟓ ퟎ ퟓퟎퟒퟒ ( ퟑퟓ) + ( . ) 퐥퐨퐠 ퟏퟎ ퟎ ퟓퟎퟒퟒ ( . ퟑퟓ × ) ≈ 퐥퐨퐠 ퟏퟎ 풍풐품 ퟑ ퟏퟗ ퟑퟓ Therefore, . × . ≈ 퐥퐨퐠 ퟑ ퟏퟗ ퟏퟎ ퟑퟓ 풙 ≈ ퟑ ퟏퟗ ퟏퟎ d. ( ) = .

퐥퐨퐠 풙 ퟒ ퟗퟐퟎퟏ ( ) = + . = ( ) + . 퐥퐨퐠 풙 ퟒ ퟎ ퟗퟐퟎퟏ ( ퟒ) + ( . ) 퐥퐨퐠 ퟏퟎ ퟎ ퟗퟐퟎퟏ ( . ퟒ × ) ≈ 퐥퐨퐠 ퟏퟎ 풍풐품 ퟖ ퟑퟐ ퟒ Therefore, , . ≈ 퐥퐨퐠 ퟖ ퟑퟐ ퟏퟎ

풙 ≈ ퟖퟑ ퟐퟎퟎ 2. Use logarithms and the logarithm table to evaluate each expression. a. .

√ퟐ ퟑퟑ ( . ) = ( . ) = ( . ) = . ퟏ ퟐ ퟏ ퟏ 퐥퐨퐠 � ퟐ ퟑퟑ � 퐥퐨퐠 ퟐ ퟑퟑ ퟎ ퟑퟔퟕퟒ ퟎ ퟏퟖퟑퟕ Thus, . = . , and locatingퟐ . in ퟐthe logarithm table gives a value approximately . . Therefore, . . . 퐥퐨퐠�√ퟐ ퟑퟑ� ퟎ ퟏퟖퟑퟕ ퟎ ퟏퟖퟑퟕ ퟏ ퟓퟑ √ퟐ ퟑퟑ ≈ ퟏ ퟓퟑ b. , ,

ퟏퟑ ퟓퟎퟎ ∙ ퟑ ퟔퟎퟎ ( . . ) = ( . ) + ( . ) + + ퟒ ퟑ . + . + 풍풐품 ퟏ ퟑퟓ ∙ ퟏퟎ ∙ ퟑ ퟔ ∙ ퟏퟎ 풍풐품 ퟏ ퟑퟓ 풍풐품 ퟑ ퟔ ퟒ ퟑ . ≈ ퟎ ퟏퟑퟎퟑ ퟎ ퟓퟓퟔퟑ ퟕ Thus, ( , , ) . . Locating .≈ ퟕ ퟔퟖퟔퟔ in the logarithm table gives a value of . . Therefore, the product is approximately . × . 퐥퐨퐠 ퟏퟑ ퟓퟎퟎ ∙ ퟑ ퟔퟎퟎ ≈ ퟕ ퟔퟖퟔퟔ ퟎ ퟕퟔퟖퟔퟔ ퟒ ퟖퟔ ퟒ ퟖퟔ ퟏퟎ . × c. . × ퟗ ퟕ ퟐ ퟏퟎ ퟓ ퟏ ퟑ ퟏퟎ( . ) + ( ) ( . ) ( ) . + . . ퟗ ퟓ Locating퐥퐨퐠 ퟕ .ퟐ 퐥퐨퐠 in theퟏퟎ logarithm− 퐥퐨퐠 ퟏ tableퟑ − gives퐥퐨퐠 ퟏퟎ . ≈. Soퟎ ퟖퟓퟕퟑthe quotientퟗ − ퟎis ퟏퟏퟑퟗapproximately− ퟓ ≈ ퟒ ퟕퟒퟑퟒ. × . ퟒ ퟎ ퟕퟒퟑퟒ ퟓ ퟓퟒ ퟓ ퟓퟒ ퟏퟎ 3. Solve for : ( ) + ( ) = ( ).

풙 퐥퐨퐠 ퟑ ퟐ퐥퐨퐠 풙 퐥퐨퐠 ퟐퟕ ( ) + ( ) = ( ) ( ) + ( ) = ( ) 퐥퐨퐠 ퟑ ퟐ 퐥퐨퐠 풙 퐥퐨퐠 ퟐퟕ ( ퟐ ) = ( ) 퐥퐨퐠 ퟑ 퐥퐨퐠 풙 퐥퐨퐠 ퟐퟕ ퟐ = 퐥퐨퐠 ퟑ풙 퐥퐨퐠 ퟐퟕ ퟐ = ퟑ풙 ퟐퟕ ퟐ = 풙 ퟗ 풙 ퟑ

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4. ( ) + ( + ) = ( )

퐥퐨퐠 ퟑ풙 퐥퐨퐠 풙 ퟒ 퐥퐨퐠 ퟏퟓ ( + ) = ( ) ퟐ + = 퐥퐨퐠 ퟑ풙 ퟏퟐ풙 퐥퐨퐠 ퟏퟓ + ퟐ = ퟑ풙 ퟏퟐ풙 ퟏퟓ + ퟐ = ퟑ풙 ퟏퟐ풙 − ퟏퟓ ퟎ (ퟐ + ) ( + ) = ퟑ풙 ퟏퟓ풙 − ퟑ풙 − ퟏퟓ ퟎ ( )( + ) = ퟑ풙 풙 ퟓ − ퟑ 풙 ퟓ ퟎ Thus, and – solve the quadratic equation, butퟑ 풙 − isퟑ an풙 extraneousퟓ ퟎ solution. Hence, Is the only solution.

ퟏ ퟓ −ퟓ ퟏ 5. Solve for . ( ) = ( + ) + ( ) a. 풙 퐥퐨퐠 (풙 ) =퐥퐨퐠 (풚 +풛 ) +퐥퐨퐠 풚( − 풛 ) ( ) = ( + )( ) 풍풐품 풙 풍풐품 풚 풛 풍풐품 풚 − 풛 ( ) = ( ) 풍풐품 풙 풍풐품� 풚 풛 풚 − 풛 � = ퟐ ퟐ 풍풐품 풙 풍풐품 풚 − 풛 ퟐ ퟐ 풙 풚 − 풛

b. ( ) = ( ) + ( ) + ( )– ( )

퐥퐨퐠(풙) = �퐥퐨퐠(풚) + 퐥퐨퐠(풛)� + �퐥퐨퐠( 풚)– 퐥퐨퐠( 풛) � ( ) = ( ) 퐥퐨퐠 풙 �퐥퐨퐠 풚 퐥퐨퐠 풛 � �퐥퐨퐠 풚 퐥퐨퐠 풛 � ( ) = ( ) 퐥퐨퐠 풙 ퟐ 퐥퐨퐠 풚 = ퟐ 퐥퐨퐠 풙 퐥퐨퐠 풚 ퟐ 풙 풚

6. If and are positive real numbers, and ( ) = + ( ), express in terms of .

Since풙 풚( ) = + ( ), we see that퐥퐨퐠 풚( ) =ퟏ 퐥퐨퐠( 풙 ). Then 풚= . 풙

퐥퐨퐠 ퟏퟎ풙 ퟏ 퐥퐨퐠 풙 퐥퐨퐠 풚 퐥퐨퐠 ퟏퟎ풙 풚 ퟏퟎ풙 7. If , , and are positive real numbers, and ( ) ( ) = ( ) ( ), express in terms of and .

풙 풚 풛 (퐥퐨퐠) 풙 − (퐥퐨퐠) =풚 (퐥퐨퐠) 풚 − (퐥퐨퐠) 풛 풚 풙 풛 ( ) + ( ) = ( ) 퐥퐨퐠 풙 − 퐥퐨퐠 풚 퐥퐨퐠 풚 − 퐥퐨퐠 풛 ( ) = ( ) 퐥퐨퐠 풙 퐥퐨퐠 풛 ퟐ 퐥퐨퐠 풚 = ퟐ 퐥퐨퐠 풙풛 퐥퐨퐠 풚 = ퟐ 풙풛 풚 풚 √풙풛

8. If and are positive real numbers and ( ) = ( + ) ( ) , express in terms of .

풙 풚 퐥퐨퐠( )풙= 풚�퐥퐨퐠( +풚 )ퟏ − 퐥퐨퐠( )풚 � 풙 풚 + 퐥퐨퐠(풙) = 풚�퐥퐨퐠 풚 ퟏ − 퐥퐨퐠 풚 � 풚 ퟏ 퐥퐨퐠 풙 풚 �퐥퐨퐠 � �� +풚 ( ) = 풚 풚 ퟏ 퐥퐨퐠 풙 퐥퐨퐠+�� 풚 = 풚 풚 ퟏ 풙 � � 풚

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9. If and are positive real numbers and ( ) = + ( ), express in terms of .

Since풙 풚( ) = + ( ) = +퐥퐨퐠 풚 ( )ퟑ, weퟐ see퐥퐨퐠 that풙 ( ) =풚 ( 풙). Then = . ퟐ ퟐ ퟐ ퟐ 퐥퐨퐠 ퟏퟎퟎퟎ풙 ퟑ 퐥퐨퐠 풙 ퟑ ퟐ 퐥퐨퐠 풙 퐥퐨퐠 풚 퐥퐨퐠 ퟏퟎퟎퟎ풙 풚 ퟏퟎퟎퟎ풙 10. If , , and are positive real numbers and ( ) = ( ) + ( ) , express in terms of and .

Since풙 풚 풛 = ( ) + ( ) , we퐥퐨퐠 see풛 that퐥퐨퐠 풚( ) =ퟐ 퐥퐨퐠 풙 − ퟏ, so = 풛 . 풙 풚 ퟐ ퟐ ퟐ 풙 풚 풙 풚 풙 풚 퐥퐨퐠 � ퟏퟎ � 퐥퐨퐠 풚 ퟐ 퐥퐨퐠 풙 − ퟏ 퐥퐨퐠 풛 퐥퐨퐠 � ퟏퟎ � 풛 ퟏퟎ 11. Solve the following equations. a. ( ) ( ) = ( )

퐥퐧 ퟏퟎ − 퐥퐧 ퟕ − 풙 퐥퐧 풙 = ( ) ퟏퟎ 퐥퐧 � � 퐥퐧 풙 ퟕ − 풙 = ퟏퟎ = ( 풙 ) ퟕ − 풙 + = ퟏퟎ 풙 ퟕ − 풙 ( ퟐ )( ) = 풙 − ퟕ풙 ퟏퟎ ퟎ = or = 풙 − ퟓ 풙 − ퟐ ퟎ Check: If = or = , then the expressions 풙and ퟐ 풙 areퟓ positive. Thus, both and are valid solutions to this equation. 풙 ퟐ 풙 ퟓ 풙 ퟕ − 풙

ퟐ ퟓ b. ( + ) + ( ) = ( ) ( + )( ) = ( ) 퐥퐧 풙 ퟐ 퐥퐧 풙 − ퟐ 퐥퐧 ퟗ풙 − ퟐퟒ = 퐥퐧� 풙 ퟐ 풙 − ퟐ � 퐥퐧 ퟗ풙 − ퟐퟒ ퟐ + = 풙 − ퟒ ퟗ풙 − ퟐퟒ ( ퟐ )( ) = 풙 − ퟗ풙 ퟐퟎ ퟎ = or = 풙 − ퟒ 풙 − ퟓ ퟎ Check: If = or = , then the expressions 풙+ ,ퟒ 풙, andퟓ are all positive. Thus, both and are valid solutions to this equation. 풙 ퟒ 풙 ퟓ 풙 ퟐ 풙 − ퟐ ퟗ풙 − ퟐퟒ ퟒ ퟓ c. ( + ) + ( ) = ( )

퐥퐧 풙 ퟐ 퐥퐧 풙 − ퟐ 퐥퐧 −ퟐ풙 − ퟏ ( + )( ) = ( ) ( ) = ( ) 퐥퐧� 풙 ퟐ 풙 − ퟐ � 퐥퐧 −ퟐ풙 − ퟏ ퟐ = 퐥퐧 풙 − ퟒ 퐥퐧 −ퟐ풙 − ퟏ ퟐ + = 풙 − ퟒ −ퟐ풙 − ퟏ ( ퟐ+ )( ) = 풙 ퟐ풙 − ퟑ ퟎ = or = 풙 ퟑ 풙 − ퟏ ퟎ So, = or = , but = makes the input풙 to− bothퟑ logarithms풙 ퟏ on the left-hand side negative, and = makes the input to the second and third logarithms negative. Thus, there are no solutions to the original equation.풙 − ퟑ 풙 ퟏ 풙 −ퟑ 풙 ퟏ

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12. Suppose the formula = ( + ) gives the population of a city growing at an annual percent rate where is the population years ago. 풕 푷 푷ퟎ ퟏ 풓 푷 풓 푷ퟎ a. Find the time풕 is takes this population to double. Let = ; then풕 ,

푷 =ퟐ푷ퟎ ( + ) = ( + ) 풕 ퟐ푷ퟎ 푷ퟎ ퟏ 풓 ( ) = ( +풕 ) ퟐ ퟏ 풓 ( ) = ( + 풕) 퐥퐨퐠 ퟐ 퐥퐨퐠 ퟏ 풓 ( ) 퐥퐨퐠 ퟐ = 풕 퐥퐨퐠 ퟏ 풓. ( + ) 퐥퐨퐠 ퟐ 풕 퐥퐨퐠 ퟏ 풓

b. Use the structure of the expression to explain why populations with lower growth rates take a longer time to double.

If is a decimal between and , then the denominator will be a number between and ( ). Thus, the value of will be large for small values of and getting closer to as increases. 풓 ퟎ ퟏ ퟎ 퐥퐨퐠 ퟐ 풕 풓 ퟏ 풓 c. Use the structure of the expression to explain why the only way to double the population in one year is if there is a percent growth rate. = ( ) = ( + ) For the populationퟏퟎퟎ to double, we need to have . This happens if , and then we have = + and = . 풕 ퟏ 퐥퐨퐠 ퟐ 퐥퐨퐠 ퟏ 풓

ퟐ ퟏ 풓 풓 ퟏ 13. If > , + > , > , and ( ) = ( + ) + ( ), find in terms of and .

Applying풙 ퟎ 풂properties풃 ퟎ of풂 logarithms,풃 퐥퐨퐠 we풙 have퐥퐨퐠 풂 풃 퐥퐨퐠 풂 − 풃 풙 풂 풃 ( ) = ( + ) + ( ) = ( + )( ) 퐥퐨퐠 풙 퐥퐨퐠 풂 풃 퐥퐨퐠 풂 − 풃 = ( ). 퐥퐨퐠� 풂 풃 풂 − 풃 � ퟐ ퟐ So, = . 퐥퐨퐠 풂 − 풃 ퟐ ퟐ 풙 풂 − 풃 14. Jenn claims that because ( ) + ( ) + ( ) = ( ), then ( ) + ( ) + ( ) = ( ). a. Is she correct? Explain how you know. 퐥퐨퐠 ퟏ 퐥퐨퐠 ퟐ 퐥퐨퐠 ퟑ 퐥퐨퐠 ퟔ 퐥퐨퐠 ퟐ 퐥퐨퐠 ퟑ 퐥퐨퐠 ퟒ 퐥퐨퐠 ퟗ MP.3 Jenn is not correct. Even though ( ) + ( ) + ( ) = ( ) = ( ), the logarithm properties give ( ) + ( ) + ( ) = ( ) = ( ). Since < , we know that ( ) < , and since > , we퐥퐨퐠 knowퟏ that퐥퐨퐠 ퟐ ( )퐥퐨퐠> ퟑ, so clearly퐥퐨퐠 ퟏ ∙ ퟐ(∙ ퟑ) 퐥퐨퐠( ퟔ ). 퐥퐨퐠 ퟐ 퐥퐨퐠 ퟑ 퐥퐨퐠 ퟒ 퐥퐨퐠 ퟐ ∙ ퟑ ∙ ퟒ 퐥퐨퐠 ퟐퟒ ퟗ ퟏퟎ 퐥 퐨퐠 ퟗ ퟏ ퟐퟒ ퟏퟎ 퐥퐨퐠 ퟐퟒ ퟏ 퐥퐨퐠 ퟗ ≠ 퐥퐨퐠 ퟐퟒ

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b. If ( ) + ( ) + ( ) = ( + + ), express in terms of and . Explain how this result relates to your answer to part (a). 퐥퐨퐠 풂 퐥퐨퐠 풃 퐥퐨퐠 풄 퐥퐨퐠 풂 풃 풄 풄 풂 풃 ( ) + ( ) + ( ) = ( + + ) ( ) = ( + + ) 퐥퐨퐠 풂 퐥퐨퐠 풃 퐥퐨퐠 풄 퐥퐨퐠 풂 풃 풄 = + + 퐥퐨퐠 풂풃풄 퐥퐨퐠 풂 풃 풄 = + MP.3 풂풃풄 풂 풃 풄 ( ) = + 풂풃풄 − 풄 풂 풃 + 풄 풂풃 − ퟏ = 풂 풃 풂 풃 풄 풂풃 − ퟏ + If ( ) + ( ) + ( ) were equal to ( ), then we would have = . However, = = ퟐ ퟑ ퟐ+ퟑ ퟓ 퐥퐨퐠 ,ퟐ so we퐥퐨퐠 knowퟑ that퐥퐨퐠 ퟒ ( ) + ( ) +퐥퐨퐠 (ퟗ ) ( ). ퟒ ퟐ∙ퟑ−ퟏ ퟐ∙ퟑ−ퟏ ퟓ ퟏ ≠ ퟒ 퐥퐨퐠 ퟐ 퐥퐨퐠 ퟑ 퐥퐨퐠 ퟒ ≠ 퐥퐨퐠 ퟗ

c. Find other values of , , and so that ( ) + ( ) + ( ) = ( + + ).

Many answers are possible;풂 풃 in풄 fact, any퐥퐨퐠 positive풂 values퐥퐨퐠 풃 of 퐥퐨퐠 and풄 where퐥퐨퐠 풂 풃 will풄 produce so ( ) + ( ) + ( ) = ( + + ) = = = that . One such풂 answer풃 is 풂풃 ≠, ퟏ , and 풄 . ퟏ 퐥퐨퐠 풂 퐥퐨퐠 풃 퐥퐨퐠 풄 퐥퐨퐠 풂 풃 풄 풂 ퟑ 풃 ퟕ 풄 ퟐ

15. In Problem 7 of Lesson 12, you showed that for , + + = . It follows that + = . What does this tell us aboutퟐ the relationship ퟐbetween the expressions 풙 ≥ ퟏ 퐥퐨퐠�풙 √풙 − ퟏ� 퐥퐨퐠�풙 − √풙 − ퟏ� ퟎ + ퟐ and ? ퟐ 퐥퐨퐠�풙 √풙 − ퟏ� −퐥퐨퐠�풙 − √풙 − ퟏ� ퟐ ퟐ 풙 √풙 − ퟏ 풙 − √풙 − ퟏ Since we know + = , and log = , we know ퟐ ퟐ ퟐ ퟏ 퐥퐨퐠�풙 √풙 − ퟏ� − 퐥퐨퐠�풙 − √풙 − ퟏ� − ⁡�풙 − √풙 − ퟏ� 풍풐품 � � ퟐ � that + = . Then, + = . We can verify that풙− these풙 −ퟏ expressions ퟐ ퟏ ퟐ ퟏ are reciproca퐥퐨퐠�풙 ls√ 풙by −multiplyingퟏ� 퐥퐨퐠 them� � together:ퟐ � 풙 √풙 − ퟏ � ퟐ 풙− 풙 −ퟏ 풙− 풙 −ퟏ + = + ퟐ ퟐ ퟐ ퟐ ퟐ ퟐ ퟐ �풙 �풙 − ퟏ� �풙 − �풙 − ퟏ� = 풙 풙( �풙 −)ퟏ − 풙�풙 − ퟏ − ��풙 − ퟏ� = ퟐ. ퟐ 풙 − 풙 − ퟏ ퟏ

16. Use the change of base formula to solve the following equations. a. ( ) = ( + ) ퟐ ퟏퟎퟎ ( + ) 퐥퐨퐠 풙 퐥퐨퐠 풙 − ퟐ풙 ퟔ ( ) = ퟐ ( ) 퐥퐨퐠 풙 − ퟐ풙 ퟔ 퐥퐨퐠 풙 ( ) = 퐥퐨퐠( ퟏퟎퟎ + ) ퟏ ퟐ 퐥퐨퐠(풙) = 퐥퐨퐠( 풙 − ퟐ풙+ )ퟔ ퟐ ( ) = ( ퟐ + ) ퟐ 퐥퐨퐠 풙 퐥퐨퐠 풙 − ퟐ풙 ퟔ ퟐ = ퟐ + 퐥퐨퐠 풙 퐥퐨퐠 풙 − ퟐ풙 ퟔ ퟐ = ퟐ 풙 풙 − ퟐ풙 ퟔ = ퟐ풙 ퟔ Since both sides of the equation are defined풙 for ퟑ = , the only solution to this equation is .

풙 ퟑ ퟑ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 M3

ALGEBRA II

b. ( ) = ( ) ( ) 퐥퐨퐠 풙 − ퟐ 퐥퐨퐠ퟏퟎퟎ ퟏퟒ − 풙 ( ) = ( ) 퐥퐨퐠 ퟏퟒ − 풙 퐥퐨퐠 풙 − ퟐ ( ) = 퐥퐨퐠 (ퟏퟎퟎ ) ퟏ 퐥퐨퐠(풙 − ퟐ) = 퐥퐨퐠( ퟏퟒ −)풙 ퟐ (( ) ) = ( ) ퟐ 퐥퐨퐠 풙 − ퟐ 퐥퐨퐠 ퟏퟒ − 풙 ( )ퟐ = 퐥퐨퐠 풙 − ퟐ 퐥퐨퐠 ퟏퟒ − 풙 + ퟐ = 풙 − ퟐ ퟏퟒ − 풙 ퟐ = 풙 − ퟒ풙 ퟒ ퟏퟒ − 풙 ( ퟐ )( + ) = 풙 − ퟑ풙 − ퟏퟎ ퟎ Thus, either = or = . Since the left side of the equation is undefined when = , but both sides 풙 − ퟓ 풙 ퟐ ퟎ are defined for = , the only solution to the equation is . 풙 ퟓ 풙 −ퟐ 풙 −ퟐ 풙 ퟓ ퟓ c. ( + ) = ( + + ) ퟐ ( + ) = ( + + ) 퐥퐨퐠ퟐ 풙 ퟏ 퐥퐨퐠ퟒ 풙 ퟑ풙 ퟒ ퟐ ퟐ ퟒ( + + ) 퐥퐨퐠 (풙 + ퟏ) = 퐥퐨퐠 풙 ퟑ풙 ퟒ ퟐ ( ) 퐥퐨퐠ퟐ 풙 ퟑ풙 ퟒ 퐥퐨퐠ퟐ(풙 + ퟏ) = ( + + ) 퐥퐨퐠ퟐ ퟒ (( + ) ) = ( ퟐ + + ) ퟐ 퐥퐨퐠ퟐ 풙 ퟏ 퐥퐨퐠ퟐ 풙 ퟑ풙 ퟒ ( + )ퟐ = + ퟐ + 퐥퐨퐠ퟐ 풙 ퟏ 퐥퐨퐠ퟐ 풙 ퟑ풙 ퟒ + + ퟐ = ퟐ + + 풙 ퟏ 풙 ퟑ풙 ퟒ ퟐ + = ퟐ + 풙 ퟐ풙 ퟏ 풙 ퟑ풙 ퟒ = ퟐ풙 ퟏ ퟑ풙 ퟒ Since the left side of the equation is undefined for =풙 −, thereퟑ is no solution to this equation. 풙 −ퟑ d. ( ) = ( + ) ퟑ ퟐ 퐥퐨퐠ퟐ 풙 − ퟏ 퐥퐨퐠ퟖ 풙 − ퟐ풙 − ퟐ풙 ퟓ ( + ) ( ) = ퟑ (ퟐ ) 풍풐품ퟐ 풙 − ퟐ풙 − ퟐ풙 ퟓ 풍풐품ퟐ(풙 − ퟏ) = ( + ) 풍풐품ퟐ ퟖ (( ) ) = ( ퟑ ퟐ + ) ퟑ 풍풐품ퟐ 풙 − ퟏ 풍풐품ퟐ 풙 − ퟐ풙 − ퟐ풙 ퟓ ( )ퟑ = ퟑ ퟐ + 풍풐품ퟐ 풙 − ퟏ 풍풐품ퟐ 풙 − ퟐ풙 − ퟐ풙 ퟓ + ퟑ = ퟑ ퟐ + 풙 − ퟏ 풙 − ퟐ풙 − ퟐ풙 ퟓ ퟑ ퟐ + = ퟑ ퟐ 풙 − ퟑ풙 ퟑ풙 − ퟏ 풙 − ퟐ풙 − ퟐ풙 ퟓ ( ퟐ )( ) = 풙 − ퟓ풙 ퟔ ퟎ Since both sides of the equation are풙 defined− ퟑ 풙 for− ퟐ = ퟎ and = , and are both valid solutions to this equation. 풙 ퟑ 풙 ퟐ ퟐ ퟑ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 M3

ALGEBRA II

+ ( ) 17. Solve the following equation: ( ) = . � � ퟐ 풍풏 ퟑ 풍풏 풙 � � Rewrite the left-hand side using퐥퐨퐠 theퟗ change풙 -of풍풏-baseퟏퟎ formula: ( ) ( ) = ( ) 풍풏 ퟗ풙 풍풐품 ퟗ풙 ( ) + ( ) = 풍풏 ퟏퟎ ( ) 풍풏 ퟗ 풍풏 풙 ( ) + ( ) = 풍풏 ퟏퟎ . ퟐ ( ) 풍풏 ퟑ 풍풏 풙 Thus, the equation is true for all > . 풍풏 ퟏퟎ

풙 ퟎ

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ALGEBRA II

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 M3

ALGEBRA II

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