Lesson 17: Graphing the Logarithm Function


 Patience Rodgers
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1 Lesson 17 Name Date Lesson 17: Graphing the Logarithm Function Exit Ticket Graph the function () = log () without using a calculator, and identify its key features. Lesson 17: Graphing the Logarithm Function 273
2 Lesson 17 Exit Ticket Sample Solutions Graph the function () = () without using a calculator, and identify its key features. Key features: The domain is (, ). The range is all real numbers. End behavior: As, (). As, (). Intercepts: intercept: There is one intercept at. intercept: The graph does not cross the axis. The graph passes through (, ). Lesson 17: Graphing the Logarithm Function 274
3 Lesson 17 Problem Set Sample Solutions For the Problem Set, students need graph paper. They should not use calculators or other graphing technology, except where specified in extension Problems 11 and 12. In Problems 2 and 3, students compare different representations of logarithmic functions. Problems 4 6 continue the reasoning from the lesson in which students observed logarithmic properties through the transformations of logarithmic graphs. Fluency problems 9 10 are a continuation of work done in Algebra I and have been placed in this lesson so that students recall concepts required in Lesson 19. Similar review problems occur in the next lesson. 1. The function () = () has function values in the table at right. a. Use the values in the table to sketch the graph of = () () b. What is the value of in () = ()? Explain how you know. Because the point (, ) is on the graph of = (), we know () =, so =. It follows that =. c. Identify the key features in the graph of = (). Because < <, the function values approach as, and the function values approach as. There is no intercept, and the intercept is. The domain of the function is (, ), the range is (, ), and the graph passes through (, ). Lesson 17: Graphing the Logarithm Function 275
4 Lesson Consider the logarithmic functions () = (), () = (), where is a positive real number, and. The graph of is given at right. a. Is >, or is <? Explain how you know. Since () =, and ()., the graph of lies below the graph of for. This means that is larger than, so we have >. (Note: The actual value of is.) b. Compare the domain and range of functions and. Functions and have the same domain, (, ), and the same range, (, ). c. Compare the intercepts and intercepts of and. Both and have an intercept at and no intercepts. d. Compare the end behavior of and. As, both () and (). 3. Consider the logarithmic functions () = () and () = (), where is a positive real number and. A table of approximate values of is given below. ().... a. Is >, or is <? Explain how you know. Since () =, and ()., the graph of lies above the graph of for. This means that is closer to than is, so we have <. (Note: The actual value of is.) b. Compare the domain and range of functions and. Functions and have the same domain, (, ), and the same range, (, ). Lesson 17: Graphing the Logarithm Function 276
5 Lesson 17 c. Compare the intercepts and intercepts of and. Both and have an intercept at and no intercepts. d. Compare the end behavior of and. As, both () and (). 4. On the same set of axes, sketch the functions () = () and () = ( ). a. Describe a transformation that takes the graph of to the graph of. The graph of is a vertical scaling of the graph of by a factor of. b. Use properties of logarithms to justify your observations in part (a). Using properties of logarithms, we know that () = ( )= () = (). Thus, the graph of is a vertical scaling of the graph of by a factor of. 5. On the same set of axes, sketch the functions () = () and () =. Lesson 17: Graphing the Logarithm Function 277
6 Lesson 17 a. Describe a transformation that takes the graph of to the graph of. The graph of is the graph of translated down by units. b. Use properties of logarithms to justify your observations in part (a). Using properties of logarithms, () = = () () =(). Thus, the graph of is a translation of the graph of down units. 6. On the same set of axes, sketch the functions () = () and () =. a. Describe a transformation that takes the graph of to the graph of. These two graphs coincide, so the identity transformation takes the graph of to the graph of. b. Use properties of logarithms to justify your observations in part (a). If =, then () =, so =. Then, =, so = (); thus, () = () for all >. 7. The figure below shows graphs of the functions () = (), () = (), and () = (). Lesson 17: Graphing the Logarithm Function 278
7 Lesson 17 a. Identify which graph corresponds to which function. Explain how you know. The top graph (in blue) is the graph of () = (), the middle graph (in green) is the graph of () = (), and the lower graph (in red) is the graph of () = (). We know this because the blue graph passes through the point (, ), the green graph passes through the point (, ), and the red graph passes through the point (, ). We also know that the higher the value of the base, the flatter the graph, so the graph of the function with the largest base,, must be the red graph on the bottom, and the graph of the function with the smallest base,, must be the blue graph on the top. b. Sketch the graph of () = () on the same axes. 8. The figure below shows graphs of the functions () = (), () = (), and () = (). a. Identify which graph corresponds to which function. Explain how you know. The top graph (in blue) is the graph of () = (), the middle graph (in red) is the graph of (), and the lower graph is the graph of () = () = (). We know this because the blue graph passes through the point (, ), the red graph passes through the point (, ), and the green graph passes through the point (, ). Lesson 17: Graphing the Logarithm Function 279
8 Lesson 17 b. Sketch the graph of () = () on the same axes. 9. For each function, find a formula for the function in terms of. Part (a) has been done for you. a. If () = +, find () = ( + ). () = ( + ) = ( + ) + ( + ) = + + b. If () = +, find () =. () = + c. If () = (), find () = when >. () = + () d. If () =, find () = ( ( + )). () = + e. If () =, find () = when. () = f. If () =, find () =. () = g. If () = (), find () = +. () = + h. If () = + +, find () = (()). () =(()) + () + Lesson 17: Graphing the Logarithm Function 280
9 Lesson For each of the functions and below, write an expression for (i) (), (ii) (), and (iii) () in terms of. Part (a) has been done for you. a. () =, () = + i. () = ( + ) = ( + ) ii. () = ( ) = + iii. () = ( ) = ( ) = b. () =, () = + i. () = ii. () = iii. () = c. () = +, () = i. () = ii. () = iii. () = + + d. () =, () = i. () = ii. iii. () = () = e. () =, () = i. () = = ii. () = ( ) = iii. () = Lesson 17: Graphing the Logarithm Function 281
10 Lesson 17 Extension: 11. Consider the functions () = () and () =. a. Use a calculator or other graphing utility to produce graphs of () = () and () = for. b. Compare the graph of the function () = () with the graph of the function () =. Describe the similarities and differences between the graphs. They are not the same, but they have a similar shape when. Both graphs pass through the points (, ) and (, ). Both functions appear to approach infinity slowly as. The graph of () = () lies below the graph of () = on the interval (, ), and the graph of appears to lie above the graph of on the interval (, ). The logarithm function is defined for >, and the radical function is defined for. Both functions appear to slowly approach infinity as. c. Is it always the case that () > for >? No, for <, () >. Between and, the graphs cross again, and we have > () for. Lesson 17: Graphing the Logarithm Function 282
11 Lesson Consider the functions () = () and () =. a. Use a calculator or other graphing utility to produce graphs of () = () and () = for. b. Compare the graph of the function () = () with the graph of the function () =. Describe the similarities and differences between the graphs. They are not the same, but they have a similar shape when. Both graphs pass through the points (, ) and (, ). Both functions appear to approach infinity slowly as. The graph of () = () lies below the graph of () = on the interval (, ), and the graph of appears to lie above the graph of on the interval (, ). The logarithm function is defined for >, and the radical function is defined for all real numbers. Both functions appear to approach infinity slowly as. c. Is it always the case that () > for >? No, if we extend the viewing window on the calculator, we see that the graphs cross again between and. Thus, () > for <, and () <for. Lesson 17: Graphing the Logarithm Function 283
12 Lesson 18 Name Date Lesson 18: Graphs of Exponential Functions and Logarithmic Functions Exit Ticket The graph of a logarithmic function () = log () is shown below. a. Explain how to find points on the graph of the function () =. b. Sketch the graph of the function () = on the same axes. Lesson 18: Graphs of Exponential Functions and Logarithmic Functions 291
13 Lesson 18 Exit Ticket Sample Solutions The graph of a logarithmic function () = () is shown below. a. Explain how to find points on the graph of the function () =. A point (, ) is on the graph of if the corresponding point (, ) is on the graph of. b. Sketch the graph of the function () = on the same axes. See graph above. Lesson 18: Graphs of Exponential Functions and Logarithmic Functions 292
14 Lesson 18 Problem Set Sample Solutions Problems 5 7 serve to review the process of computing () for given functions and in preparation for work with inverses of functions in Lesson Sketch the graphs of the functions () = and () = (). 2. Sketch the graphs of the functions () = and () = (). Lesson 18: Graphs of Exponential Functions and Logarithmic Functions 293
15 Lesson Sketch the graphs of the functions () = and () = on the same sheet of graph paper, and answer the following questions. a. Where do the two exponential graphs intersect? The graphs intersect at the point (, ). b. For which values of is <? If >, then <. c. For which values of is >? If <, then >. d. What happens to the values of the functions and as? As, both () and (). e. What are the domains of the two functions and? Both functions have domain (, ). Lesson 18: Graphs of Exponential Functions and Logarithmic Functions 294
16 Lesson Use the information from Problem 3 together with the relationship between graphs of exponential and logarithmic functions to sketch the graphs of the functions () = () and () = () on the same sheet of graph paper. Then, answer the following questions. a. Where do the two logarithmic graphs intersect? The graphs intersect at the point (, ). b. For which values of is () <()? When <, we have () <(). c. For which values of is () >()? When >, we have () >(). d. What happens to the values of the functions and as? As, both () and (). e. What are the domains of the two functions and? Both functions have domain (, ). Lesson 18: Graphs of Exponential Functions and Logarithmic Functions 295
17 Lesson For each function, find a formula for the function in terms of. a. If () =, find () = + (). () = b. If () = +, find () = ( + ) (). () = + c. If () = + + +, find () = () = + () + (). d. If () = + + +, find () = () = + () (). 6. In Problem 5, parts (c) and (d), list at least two aspects about the formulas you found as they relate to the function () = The formula for 1(c) is all of the even power terms of. The formula for 1(d) is all of the odd power terms of. The sum of the two functions gives back again; that is, ()() + ()() =. 7. For each of the functions and below, write an expression for (i) (), (ii) (), and (iii) () in terms of. a. () =, () = i. () = ii. () = iii. () = b. () =, () = + for two numbers and, when is not or i. () = ii. () = iii. () =, which is equivalent to () = () + c. () = + +, () =, when is not or i. () = ii. iii. () = () = d. () =, () = () i. () = ii. () = iii. () = Lesson 18: Graphs of Exponential Functions and Logarithmic Functions 296
18 Lesson 18 e. () = (), () = i. () = ii. () = iii. () = (()) f. () =, () = i. () = ii. () = iii. () = Lesson 18: Graphs of Exponential Functions and Logarithmic Functions 297
19 Lesson 19 Name Date Lesson 19: The Inverse Relationship Between Logarithmic and Exponential Functions Exit Ticket 1. The graph of a function is shown below. Sketch the graph of its inverse function on the same axes. 2. Explain how you made your sketch. 3. The function graphed above is the function () = log () +2 for >0. Find a formula for the inverse of this function. Lesson 19: The Inverse Relationship Between Logarithmic and Exponential Functions 308
20 Lesson 19 Exit Ticket Sample Solutions 1. The graph of a function is shown below. Sketch the graph of its inverse function on the same axes. 2. Explain how you made your sketch. Answers will vary. Example: I drew the line given by = and reflected the graph of across it. 3. The graph of the function above is the function () = () + for >. Find a formula for the inverse of this function. = () + = () + = () () = = () = Problem Set Sample Solutions 1. For each function below, find two functions and such that () = (). (There are many correct answers.) a. () = ( + ) Possible answer: () =, () = + b. () = Possible answer: () =, () = c. () = Possible answer: () =, () = Lesson 19: The Inverse Relationship Between Logarithmic and Exponential Functions 309
21 Lesson 19 d. () = Possible answer: () =, () = e. () = ( + ) + ( + ) Possible answer: () = +, () = + f. () = ( + ) Possible answer: () =, () = + g. () = ( + ) Possible answer: () = (), () = + h. () = ( + ) Possible answer: () = (), () = + i. () = (()) Possible answer: () = (), () = () 2. Let be the function that assigns to each student in your class his or her biological mother. a. Use the definition of function to explain why is a function. The function has a welldefined domain (students in the class) and range (their mothers), and each student is assigned one and only one biological mother. b. In order for to have an inverse, what condition must be true about the students in your class? If a mother has several children in the same classroom, then there would be no way to define an inverse function that picks one and only one student for each mother. The condition that must be true is that there are no siblings in the class. c. If we enlarged the domain to include all students in your school, would this larger domain function have an inverse? Probably not. Most schools have several students who are siblings. 3. The table below shows a partially filledout set of inputoutput pairs for two functions and that have the same finite domain of {,,,,,,,, }. ().... ().... a. Complete the table so that is invertible but is definitely not invertible. Answers will vary. For, all output values should be different. For, at least two output values for two different inputs should be the same number. Lesson 19: The Inverse Relationship Between Logarithmic and Exponential Functions 310
22 Lesson 19 b. Graph both functions and use their graphs to explain why is invertible and is not. Answers will vary. The graph of has one unique output for every input, so it is possible to undo and map each of its outputs to a unique input. The graph of has at least two input values that map to the same output value. Hence, there is no way to map that output value back to a unique multiple of. Hence, cannot have an inverse function because such a correspondence is not a function. 4. Find the inverse of each of the following functions. In each case, indicate the domain and range of both the original function and its inverse. a. () = = = + = The inverse function is () = +. Both functions and have a domain and range of all real numbers. b. () = + + = = + = + =( + ) + = The inverse function is () = +. Domain of and range of : all real numbers with Range of and domain of : all real numbers with c. () = = () = () + = The inverse function is () = () +. Domain of and range of : all real numbers Range of and domain of : all real numbers with > Lesson 19: The Inverse Relationship Between Logarithmic and Exponential Functions 311
23 Lesson 19 d. () = = () = = () = ( ()) The inverse function is () = (). Domain of and range of : all real numbers Range of and domain of : all real numbers with > e. () = ( + ) = ( + ) = ( + ) = + ( )= The inverse function is () =. Domain of and range of : all real numbers with > Range of and domain of : all real numbers f. () = + ( + ) = + ( + ) = ( + ) = + = ( ) = The inverse function is () =( ). Domain of and range of : all real numbers with > Range of and domain of : all real numbers Lesson 19: The Inverse Relationship Between Logarithmic and Exponential Functions 312
24 Lesson 19 g. () = + = + = () ( + ) = ( + ) = ( + ) = + ( ) = The inverse function is () = ( ). Domain of and range of : all real numbers with > Range of and domain of : all real numbers h. () = () ( + ) = () ( + ) = + = + + = = ( ) = = The inverse function is () =. Domain of and range of : all real numbers with > Range of and domain of : all real numbers < i. () = + = + + = = () = = = () = = () The inverse function is () =. Domain of and range of : all real numbers Range of and domain of : all real numbers, < < Lesson 19: The Inverse Relationship Between Logarithmic and Exponential Functions 313
25 Lesson Even though there are no real principal square roots for negative numbers, principal cube roots do exist for negative numbers: is the real number since =. Use the identities = and = for any real number to find the inverse of each of the functions below. In each case, indicate the domain and range of both the original function and its inverse. a. () = for any real number. = = = = = ( ) () = ( ) Domain of and range of : all real numbers Range of and domain of : all real numbers b. () = for any real number. = = = = + = ( + ) () = ( + ) Domain of and range of : all real numbers Range of and domain of : all real numbers c. () = () + for any real number. = () + = () + = () = = = + () = + Domain of and range of : all real numbers Range of and domain of : all real numbers Lesson 19: The Inverse Relationship Between Logarithmic and Exponential Functions 314
26 Lesson Suppose that the inverse of a function is the function itself. For example, the inverse of the function () = (for ) is just itself again, () = (for ). What symmetry must the graphs of all such functions have? (Hint: Study the graph of Exercise 5 in the lesson.) All graphs of functions that are selfinverses are symmetric with respect to the diagonal line given by the equation =. That is, a reflection across the line given by = takes the graph back to itself. 7. There are two primary scales for measuring daily temperature: degrees Celsius and degrees Fahrenheit. The United States uses the Fahrenheit scale, and many other countries use the Celsius scale. When traveling abroad you often need to convert between these two temperature scales. Let be the function that inputs a temperature measure in degrees Celsius, denoted by, and outputs the corresponding temperature measure in degrees Fahrenheit, denoted by. a. Assuming that is linear, we can use two points on the graph of to determine a formula for. In degrees Celsius, the freezing point of water is, and its boiling point is. In degrees Fahrenheit, the freezing point of water is, and its boiling point is. Use this information to find a formula for the function. (Hint: Plot the points and draw the graph of first, keeping careful track of the meaning of values on the axis and axis.) () = + b. If the temperature in Paris is, what is the temperature in degrees Fahrenheit? Since () =, it is in Paris. c. Find the inverse of the function and explain its meaning in terms of degrees Fahrenheit and degrees Celsius. The inverse of is () = (). Given the measure of a temperature reported in degrees Fahrenheit, the function converts that measure to degrees Celsius. d. The graphs of and its inverse are two lines that intersect in one point. What is that point? What is its significance in terms of degrees Celsius and degrees Fahrenheit? The point is (, ). This means that is the same temperature as. Extension: Use the fact that, for >, the functions () = and () = () are increasing to solve the following problems. Recall that an increasing function has the property that if both and are in the domain of and <, then () < (). 8. For which values of is <,,? <,, <,, = (,, ) 9. For which values of is () <,,? () <,, <,, Lesson 19: The Inverse Relationship Between Logarithmic and Exponential Functions 315
27 Lesson 20 Name Date Lesson 20: Transformations of the Graphs of Logarithmic and Exponential Functions Exit Ticket 1. Express () = log (2) in the general form of a logarithmic function, () = + log (). Identify,,, and. 2. Use the structure of when written in general form to describe the graph of as a transformation of the graph of () = log (). 3. Graph and on the same coordinate axes. Lesson 20: Transformations of the Graphs of Logarithmic and Exponential Functions 329
28 Lesson 20 Exit Ticket Sample Solutions 1. Express () = () in the general form of a logarithmic function, () = + (). Identify,,, and. Since () = () + () = (), the function is () = (), and =, =, =, and =. 2. Use the structure of when written in general form to describe the graph of as a transformation of the graph of () = (). The graph of is the graph of reflected about the horizontal axis and translated down unit. 3. Graph and on the same coordinate axes. The graph of is shown in blue, and the graph of is shown in green. Lesson 20: Transformations of the Graphs of Logarithmic and Exponential Functions 330
29 Lesson 20 Problem Set Sample Solutions Students should complete these problems without the use of a calculator. 1. Describe each function as a transformation of the graph of a function in the form () = (). Sketch the graph of and the graph of by hand. Label key features such as intercepts, intervals where is increasing or decreasing, and the equation of the vertical asymptote. a. () = () The graph of is the graph of () = () translated horizontally units to the right. The function is increasing on (, ). The intercept is, and the vertical asymptote is =. b. () = () The graph of is the graph of () = () translated vertically up units. The function is increasing on (, ). The intercept is. The vertical asymptote is =. Lesson 20: Transformations of the Graphs of Logarithmic and Exponential Functions 331
30 Lesson 20 c. () = The graph of is the graph of () = () reflected about the horizontal axis and translated vertically up units. The function is decreasing on (, ). The intercept is. The vertical asymptote is =. The reflected graph and the final graph are both shown in the solution. d. () = (() ) for > The graph of is the graph of () = () stretched vertically by a factor of and translated horizontally units to the right. The function is increasing on (, ). The intercept is, and the vertical asymptote is =. Lesson 20: Transformations of the Graphs of Logarithmic and Exponential Functions 332
31 Lesson Each function graphed below can be expressed as a transformation of the graph of () = (). Write an algebraic function for and, and state the domain and range. In Figure, () = () for >. The domain of is >, and the range of is all real numbers. In Figure, () = + () for >. The domain of is >, and the range of is all real numbers. Figure 1: Graphs of () = () and the function Figure 2: Graphs of () =() and the function Lesson 20: Transformations of the Graphs of Logarithmic and Exponential Functions 333
32 Lesson Describe each function as a transformation of the graph of a function in the form () =. Sketch the graph of and the graph of by hand. Label key features such as intercepts, intervals where is increasing or decreasing, and the horizontal asymptote. a. () = The graph of is the graph of () = scaled vertically by a factor of and translated vertically down unit. The equation of the horizontal asymptote is =. The intercept is, and the intercept is approximately.. The function is increasing for all real numbers. b. () = + The graph of is the graph of () = translated vertically up units. The equation of the horizontal asymptote is =. The intercept is. There is no intercept. The function is increasing for all real numbers. Lesson 20: Transformations of the Graphs of Logarithmic and Exponential Functions 334
33 Lesson 20 c. () = The graph of is the graph of () = translated horizontally units to the right OR the graph of scaled vertically by a factor of. The equation of the horizontal asymptote is =. The intercept is. There is no intercept, and the function is increasing for all real numbers. d. () = + The graph of is the graph of () = reflected about the horizontal axis and then translated vertically up unit. The equation of the horizontal asymptote is =. The intercept is, and the intercept is also. The function is decreasing for all real numbers. Lesson 20: Transformations of the Graphs of Logarithmic and Exponential Functions 335
34 Lesson Using the function () =, create a new function whose graph is a series of transformations of the graph of with the following characteristics: The function is decreasing for all real numbers. The equation for the horizontal asymptote is =. The intercept is. One possible solution is () = Using the function () =, create a new function whose graph is a series of transformations of the graph of with the following characteristics: The function is increasing for all real numbers. The equation for the horizontal asymptote is =. The intercept is. One possible solution is () = ( )+. 6. Consider the function () = : a. Write the function as an exponential function with base. Describe the transformations that would take the graph of () = to the graph of. = ( ) = = Thus, () =. The graph of is the graph of reflected about the vertical axis and scaled vertically by a factor of. b. Write the function as an exponential function with base. Describe two different series of transformations that would take the graph of () = to the graph of. = ( ) = () = = Thus, () =, or () = (). To obtain the graph of from the graph of, you can scale the graph horizontally by a factor of, reflect the graph about the vertical axis, and scale it vertically by a factor of. Or, you can scale the graph horizontally by a factor of, reflect the graph about the vertical axis, and translate the resulting graph horizontally units to the right. MP.3 7. Explore the graphs of functions in the form () = ( ) for >. Explain how the graphs of these functions change as the values of increase. Use a property of logarithms to support your reasoning. The graphs appear to be a vertical scaling of the common logarithm function by a factor of. This is true because of the property of logarithms that states ( )= (). 8. Use a graphical approach to solve each equation. If the equation has no solution, explain why. a. () = () This equation has no solution because the graphs of = () and = () are horizontal translations of each other. Thus, their graphs do not intersect, and the corresponding equation has no solution. Lesson 20: Transformations of the Graphs of Logarithmic and Exponential Functions 336
35 Lesson 20 b. () = () This equation has no solution because () = () + (), which means that the graphs of = () and = () are a vertical translation of each other. Thus, their graphs do not intersect, and the corresponding equation has no solution. c. () = The solution is the coordinate of the intersection point of the graphs of = () and = () (). Since the graph of the function defined by the right side of the equation is a reflection across the horizontal axis and a vertical translation of the graph of the function defined by the left side of the equation, the graphs of these functions intersect in exactly one point with an coordinate approximately.. d. Show algebraically that the exact solution to the equation in part (c) is. () = () () () = () () = () () = = Since =, the exact solution is. Lesson 20: Transformations of the Graphs of Logarithmic and Exponential Functions 337
36 Lesson Make a table of values for () = () for >. Graph the function for >. Use properties of logarithms to explain what you see in the graph and the table of values.,, () = = = = = =, =, = = = = = We see that () = for all > because = () () = (). Therefore, when we substitute () into the expression (), we get (), which is equal to according to the definition of a logarithm. Lesson 20: Transformations of the Graphs of Logarithmic and Exponential Functions 338
37 Lesson 21 Name Date Lesson 21: The Graph of the Natural Logarithm Function Exit Ticket 1. a. Describe the graph of () = 2 ln( + 3) as a transformation of the graph of () = ln(). b. Sketch the graphs of and by hand. 2. Explain where the graph of () = log (2) would sit in relation to the graph of () = ln(). Justify your answer using properties of logarithms and your knowledge of transformations of graph of functions. Lesson 21: The Graph of the Natural Logarithm Function 347
38 Lesson 21 Exit Ticket Sample Solutions 1. a. Describe the graph of () = ( + ) as a transformation of the graph of () = (). The graph of is the graph of translated units to the left, reflected about the horizontal axis, and translated up units. b. Sketch the graphs of and by hand. 2. Explain where the graph of () = () would sit in relation to the graph of () = (). Justify your answer using properties of logarithms and your knowledge of transformations of graph of functions. Since () = = + (), the graph of would be a vertical shift and a vertical scaling by a factor greater than of the graph of. The graph of will lie vertically above the graph of. Problem Set Sample Solutions 1. Rewrite each logarithmic function as a natural logarithm function. a. () = () () = () () b. () = ( ) ( ) () = () Lesson 21: The Graph of the Natural Logarithm Function 348
39 Lesson 21 c. () = () = () () () () d. () = () () = () () e. () = ( + ) () = ( + ) () f. () = () () = + () () 2. Describe each function as a transformation of the natural logarithm function () = (). a. () = ( + ) The graph of is the graph of translated units to the left and scaled vertically by a factor of. b. () = ( ) The graph of is the graph of translated unit to the right, reflected about =, and then reflected about the horizontal axis. c. () = + ( ) The graph of is the graph of translated up units. d. () = () The graph of is the graph of translated up units and scaled vertically by a factor of (). Lesson 21: The Graph of the Natural Logarithm Function 349
40 Lesson Sketch the graphs of each function in Problem 2, and identify the key features including intercepts, intervals where the function is increasing or decreasing, and the vertical asymptote. a. The equation of the vertical asymptote is =. The intercept is. The function is increasing for all >. The intercept is approximately.. b. The equation of the vertical asymptote is =. The intercept is. The function is increasing for all <. The intercept is. Lesson 21: The Graph of the Natural Logarithm Function 350
41 Lesson 21 c. The equation of the vertical asymptote is =. The intercept is approximately.. The function is increasing for all >. d. The equation of the vertical asymptote is =. The intercept is.. The function is increasing for all >. Lesson 21: The Graph of the Natural Logarithm Function 351
42 Lesson Solve the equation = () graphically, without using a calculator. It appears that the two graphs intersect at the point (, ). Checking, we see that = =, and we know that () =, so when =, we have = (). Thus is a solution to this equation. From the graph, it is the only solution. 5. Use a graphical approach to explain why the equation () = () has only one solution. The graphs intersect in only one point (, ), so the equation has only one solution. Lesson 21: The Graph of the Natural Logarithm Function 352
43 Lesson Juliet tried to solve this equation as shown below using the change of base property and concluded there is no solution because (). Construct an argument to support or refute her reasoning. MP.3 () = () () () = () () () () = (()) () () = Juliet s approach works as long as (), which occurs when =. The solution to this equation is. When you divide both sides of an equation by an algebraic expression, you need to impose restrictions so that you are not dividing by. In this case, Juliet divided by (), which is not valid if =. This division caused the equation in the third and final lines of her solution to have no solution; however, the original equation is true when is. 7. Consider the function given by () = () for > and. a. What are the values of (), (), and? () =, () =, = b. Why is the value excluded from the domain of this function? The value is excluded from the domain because is not a base of an exponential function since it would produce the graph of a constant function. Since logarithmic functions by definition are related to exponential functions, we cannot have a logarithm with base. c. Find a value so that () =.. () =.. = =, The value of that satisfies this equation is,. d. Find a value so that () =. The value of that satisfies this equation is. e. Sketch a graph of = () for > and. Lesson 21: The Graph of the Natural Logarithm Function 353
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