Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

5.1Combining Functions 5.2Inverse Functions and Their Representations 5.3Exponential Functions and Models 5.4Logarithmic Functions and Models 5.5 Properties of Logarithms 5.6Exponential and Logarithmic Equations 5.7Constructing Nonlinear Models Exponential and Logarithmic Functions 5

4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Five Ways of Combining Two Functions f and g AdditionAddition SubtractionSubtraction MultiplicationMultiplication DivisionDivision CompositionComposition

5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Definitions Addition Subtraction Multiplication Division Composition If f(x) and g(x) both exist, the sum, difference, product, quotient and composition of two functions f and g are defined by

7 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example of Addition of Functions: Let f(x) = x 2 + 2x and g(x) = 3x - 1 Find the symbolic representation for the function f + g and use this to evaluate (f + g)(2)Find the symbolic representation for the function f + g and use this to evaluate (f + g)(2) x 2 + 2x)3x  1) (f + g)(x) = (x 2 + 2x) + (3x  1) x 2 + 5x  1 (f + g)(x) = x 2 + 5x  1 2 2 + 5(  1 = 13 (f + g)(2) = 2 2 + 5(2)  1 = 13 or (f + g)(2) = f(2) + g(2) = 2 2 + 2(2) + 3(2)  1 = 2 2 + 2(2) + 3(2)  1 = 13 = 13

8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example of Subtraction of Functions: Let f(x) = x 2 + 2x and g(x) = 3x  1  Find the symbolic representation for the function f  g and use this to evaluate (f  g)(2)  x 2 + 2x)  3x  1)(f  g)(x) = (x 2 + 2x)  (3x  1) (f  g)(x) = x 2  x + 1(f  g)(x) = x 2  x + 1 So  2 2  + 1 = 3So (f  g)(2) = 2 2  2 + 1 = 3

9 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example of Multiplication of Functions: Let f(x) = x 2 + 2x and g(x) = 3x  1 Find the symbolic representation for the function fg and use this to evaluate (fg)(2) x 2 + 2x)3x  1)(fg)(x) = (x 2 + 2x)(3x  1) x 3 + 6x 2  x 2  2x(fg)(x) = 3x 3 + 6x 2  x 2  2x x 3 + 5x 2  2x(fg)(x) = 3x 3 + 5x 2  2x So 3(2) 3 +5(2) 2  2(2) = 40So (fg)(2) = 3(2) 3 +5(2) 2  2(2) = 40

10 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example of Division of Functions: Let f(x) = x 2 + 2x and g(x) = 3x  1 Find the symbolic representation for the function and use this to evaluateFind the symbolic representation for the function and use this to evaluate SoSo

11 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example of Composition of Functions: Let f(x) = x 2 + 2x and g(x) = 3x - 1 Find the symbolic representation for the function f  g and use this to evaluate (f  g)(2)Find the symbolic representation for the function f  g and use this to evaluate (f  g)(2) (f  g)(x) = f(g(x)) = f(3x – 1) = (3x – 1) 2(f  g)(x) = f(g(x)) = f(3x – 1) = (3x – 1) 2 + 2(3x – 1) (f  g)(x) = (3x – 1) ( 3x – 1) + 6x – 2(f  g)(x) = (3x – 1) ( 3x – 1) + 6x – 2 (f  g)(x) = 9x 2 – 3x – 3x + 1 + 6x – 2(f  g)(x) = 9x 2 – 3x – 3x + 1 + 6x – 2 (f  g)(x) = 9x 2 – 1(f  g)(x) = 9x 2 – 1 So (f  g)(2) = 9(2) 2 – 1 = 35So (f  g)(2) = 9(2) 2 – 1 = 35

12 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Evaluating Combinations of Functions Numerically Given numerical representations for f and g in the tableGiven numerical representations for f and g in the table Evaluate combinations of f and g as specified.Evaluate combinations of f and g as specified.

15 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Evaluating Combinations of Functions Graphically Use graph of f and g below to evaluateUse graph of f and g below to evaluate (f + g) (1)(f + g) (1) (f –g) (1)(f – g) (1) (f  g) (1)(f  g) (1) (f/g) (1)(f/g) (1) (f  g) (1)(f  g) (1) y = g(x) y = f(x)

16 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley y = g(x) y = f(x)Answers: (f + g) (1) = f(1) + g(1) = 3 + 0 = 3 –––(f – g) (1) = f(1) – g(1) = 3 – 0 = 3 (fg) (1) = f(1)   g(1) = 3   0 = 0 (f/g) (1) is undefined, because division by 0 is undefined. (f  g) (1) = f(g(1)) = f(0) = 2

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inverse Functions and Their Representations ♦Calculate inverse operations ♦Identify one-to-one functions ♦Find inverse functions symbolically ♦Use other representations to find inverse functions 5.2

18 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inverse Operations Actions:Actions: Put on socks and put on shoesPut on socks and put on shoes Put a gift inside a box and wrap the boxPut a gift inside a box and wrap the box Multiply x by 3 and add 2Multiply x by 3 and add 2 Take cube root of x and subtract 1Take cube root of x and subtract 1 Inverse ActionsInverse Actions Take off shoes and take off socksTake off shoes and take off socks Unwrap the box and take the gift out of the boxUnwrap the box and take the gift out of the box Subtract 2 from x and divide by 3Subtract 2 from x and divide by 3 Add 1 to x and cube the resultAdd 1 to x and cube the result

19 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Reminder of the definition of a function y = f(x) means that given an input x, there is just one corresponding output y.y = f(x) means that given an input x, there is just one corresponding output y. Graphically, this means that the graph passes the vertical line test.Graphically, this means that the graph passes the vertical line test. Numerically, this means that in a table of values for y = f(x) there are no x-values repeated.Numerically, this means that in a table of values for y = f(x) there are no x-values repeated.

20 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Given y 2 = x, is y = f(x)? That is, is y a function of x?Given y 2 = x, is y = f(x)? That is, is y a function of x? No, because if x = 4, y could be 2 or – 2.No, because if x = 4, y could be 2 or – 2. Note that the graph fails the vertical line test.Note that the graph fails the vertical line test. Note that there is a value of x in the table for which there are two different values of y (that is, x-values are repeated.)Note that there is a value of x in the table for which there are two different values of y (that is, x-values are repeated.)xy4 – 2 1 – 1 00 11 42

21 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Idea Behind a One-to-One Function Given a function y = f(x), f is 1-1 (pronounced “one-to-one”) means thatGiven a function y = f(x), f is 1-1 (pronounced “one-to-one”) means that given an output y there was just one input x which produced that output.given an output y there was just one input x which produced that output. Graphically, this means that the graph passes the horizontal line test. (Every horizontal line intersects the graph at most once.)Graphically, this means that the graph passes the horizontal line test. (Every horizontal line intersects the graph at most once.) Numerically, this means the there are no y-values repeated in a table of values.Numerically, this means the there are no y-values repeated in a table of values.

22 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Given y = f(x) = |x|, is f 1-1?Given y = f(x) = |x|, is f 1-1? No, because if y = 2, x could be 2 or – 2.No, because if y = 2, x could be 2 or – 2. Note that the graph fails the horizontal line test.Note that the graph fails the horizontal line test. Note that there is a value of y in the table for which there are two different values of x (that is, y-values are repeated.)Note that there is a value of y in the table for which there are two different values of x (that is, y-values are repeated.) (2,2) (-2,2) xy – 2 2 – 1 1 00 11 22

23 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Formal Definition of One-to-One Function A function f is a one-to-one function if, for elements c and d in the domain of f,A function f is a one-to-one function if, for elements c and d in the domain of f, c ≠ d implies f(c) ≠ f(d) Example: Given y = f(x) = |x|, f is not 1-1 because –2 ≠ 2 yet | –2 | = | 2 |Example: Given y = f(x) = |x|, f is not 1-1 because –2 ≠ 2 yet | –2 | = | 2 |

24 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Given a 1-1 function f f -1 is a symbol for the inverse of the function f, not to be confused with the reciprocal.f -1 is a symbol for the inverse of the function f, not to be confused with the reciprocal. If f -1 (x) does NOT mean 1/ f(x), what does it mean?If f -1 (x) does NOT mean 1/ f(x), what does it mean? y = f -1 (x) means that x = f(y)y = f -1 (x) means that x = f(y) Note that y = f -1 (x) is pronounced “y equals f inverse of x.”Note that y = f -1 (x) is pronounced “y equals f inverse of x.”

25 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example of An Inverse Function Let F be Fahrenheit temperature and let C be Centigrade temperature.Let F be Fahrenheit temperature and let C be Centigrade temperature. F = f(C) = (9/5)C + 32F = f(C) = (9/5)C + 32 C = f -1 (F) = ?????C = f -1 (F) = ????? The function f multiplies an input C by 9/5 and adds 32.The function f multiplies an input C by 9/5 and adds 32. To undo multiplying by 9/5 and adding 32, one shouldTo undo multiplying by 9/5 and adding 32, one should subtract 32 and divide by 9/5subtract 32 and divide by 9/5 So C = f -1 (F) = (F – 32)/(9/5)So C = f -1 (F) = (F – 32)/(9/5) So C = f -1 (F) = (5/9)(F – 32)So C = f -1 (F) = (5/9)(F – 32)

26 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley F = f(C) = (9/5)C + 32F = f(C) = (9/5)C + 32 C = f -1 (F) = (5/9)(F – 32)C = f -1 (F) = (5/9)(F – 32) Evaluate f(0) and interpret.Evaluate f(0) and interpret. f(0) = (9/5)(0) + 32 = 32f(0) = (9/5)(0) + 32 = 32 When the Centigrade temperature is 0, the Fahrenheit temperature is 32.When the Centigrade temperature is 0, the Fahrenheit temperature is 32. Evaluate f -1 (32) and interpret.Evaluate f -1 (32) and interpret. f -1 (32) = (5/9)(32 - 32) = 0f -1 (32) = (5/9)(32 - 32) = 0 When the Fahrenheit temperature is 32, the Centigrade temperature is 0.When the Fahrenheit temperature is 32, the Centigrade temperature is 0. Note that f(0) = 32 and f -1 (32) = 0Note that f(0) = 32 and f -1 (32) = 0 Example of An Inverse Function (con’t)

27 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 2 To undo multiplying by 3 and adding 2, one mustTo undo multiplying by 3 and adding 2, one must subtract 2 andsubtract 2 and divide by 3divide by 3 So if f(x) = 3x + 2, then f -1 (x) = (x – 2)/3So if f(x) = 3x + 2, then f -1 (x) = (x – 2)/3 Let’s checkLet’s check f(1) = 3(1) + 2 = 5 and f -1 (5) = (5 – 2)/3 = 1f(1) = 3(1) + 2 = 5 and f -1 (5) = (5 – 2)/3 = 1 So f(1) = 5 and f -1 (5) = 1So f(1) = 5 and f -1 (5) = 1

28 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 3 Let f(x) compute the distance traveled in miles after x hours by a car with a velocity of 60 miles per hour. Explain what f -1 computes. We are given that distance = f(time) so time = f -1 (distance). f -1 computes the time it takes a car with a velocity of 60 mph to travel x miles.

29 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 4 Describe verbally the inverse of the statement. Then express both the statement and its inverse symbolically.Describe verbally the inverse of the statement. Then express both the statement and its inverse symbolically. Take the cube root of x and add 1.Take the cube root of x and add 1. To undo taking a cube root and adding 1, one must subtract 1 and cube the result.To undo taking a cube root and adding 1, one must subtract 1 and cube the result. Expressing the given statement symbolicallyExpressing the given statement symbolically Expressing the inverse symbolicallyExpressing the inverse symbolically

31 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Note that: Since f(8) = 3 and f -1 (3) = 8 (f -1 o f)(8) = f -1 (f(8)) = f -1 (3) = 8(f -1 o f)(8) = f -1 (f(8)) = f -1 (3) = 8 (f o f -1 )(3) = f(f -1 (3)) = f (8) = 3(f o f -1 )(3) = f(f -1 (3)) = f (8) = 3

32 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Let f be a 1-1 function. Then f -1 is the inverse function of f, if (f -1 o f)(x) = f -1 (f(x)) = x for every x in the domain of f(f -1 o f)(x) = f -1 (f(x)) = x for every x in the domain of f (f o f -1 )(x) = f(f -1 (x)) = x for every x in the domain of f -1(f o f -1 )(x) = f(f -1 (x)) = x for every x in the domain of f -1 Formal Definition of an Inverse Function

35 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley f -1 Procedure for finding Symbolic Representation of f -1 Make sure that f is a 1-1 function. If not, f -1 does not exist.Make sure that f is a 1-1 function. If not, f -1 does not exist. Solve the equation y = f(x) for x, resulting in the equation x = f -1 (y)Solve the equation y = f(x) for x, resulting in the equation x = f -1 (y) Interchange x and y to obtain y = f -1 (x)Interchange x and y to obtain y = f -1 (x) Example.Example. f(x) = 3x + 2f(x) = 3x + 2 y = 3x + 2 y = 3x + 2 Solving for x gives: 3x = y – 2Solving for x gives: 3x = y – 2 x = (y – 2)/3 x = (y – 2)/3 Interchanging x and y gives: y = (x – 2)/3Interchanging x and y gives: y = (x – 2)/3 So f -1 (x) = (x – 2)/3So f -1 (x) = (x – 2)/3

36 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Recall To undo multiplying by 3 and adding 2, one mustTo undo multiplying by 3 and adding 2, one must subtract 2 andsubtract 2 and divide by 3divide by 3 So if f(x) = 3x + 2, then f -1 (x) = (x – 2)/3So if f(x) = 3x + 2, then f -1 (x) = (x – 2)/3

37 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Why does f need to be 1-1 to have an inverse function? Suppose f(x) = x 2 f(3) = 9 and f(  3) = 9f(3) = 9 and f(  3) = 9 If we defined f -1 (9), it could have two different values, namely 3 and –3 and thus f -1 would not be a function.If we defined f -1 (9), it could have two different values, namely 3 and –3 and thus f -1 would not be a function.

38 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Evaluating an inverse function numerically The function is 1-1 so f -1 exists.The function is 1-1 so f -1 exists. f -1 (–5) = 1f -1 (–5) = 1 f -1 (–3) = 2f -1 (–3) = 2 f -1 (0) = 3f -1 (0) = 3 f -1 (3) = 4f -1 (3) = 4 f -1 (5) = 5f -1 (5) = 5 xf(x) 1–5 2–3 30 43 55

39 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley f(2)=4 Evaluating an inverse function graphically The graph of f below passes the horizontal line test so f is 1-1. Evaluate f -1 (4).The graph of f below passes the horizontal line test so f is 1-1. Evaluate f -1 (4). Since the point (2,4) is on the graph of f, the point (4,2) will be on the graph of f -1 and thus f -1 (4) = 2Since the point (2,4) is on the graph of f, the point (4,2) will be on the graph of f -1 and thus f -1 (4) = 2

40 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphs of Functions and Their Inverses The graph of f -1 is a reflection of the graph of f across the line y = xThe graph of f -1 is a reflection of the graph of f across the line y = x Note that the domain of f equals the range of f -1 and the range of f equals the domain of f -1

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exponential Functions and Models ♦Distinguish between linear and exponential growth ♦Model data with exponential functions ♦Calculate compound interest ♦Use the natural exponential functions in applications 5.3

42 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Population Growth by a Constant Number vs by a Constant Percentage 500 people per year500 people per year What is the population in Jan 2005?What is the population in Jan 2005? 10,000 + 500 = 10,50010,000 + 500 = 10,500 What is the population in Jan 2006?What is the population in Jan 2006? 10,500 + 500 = 11,00010,500 + 500 = 11,000 5% per year5% per year What is the population in Jan 2005?What is the population in Jan 2005? 10,000 +.05(10,000) = 10,000 + 500 = 10,50010,000 +.05(10,000) = 10,000 + 500 = 10,500 What is the population in Jan 2006?What is the population in Jan 2006? 10,500 +.05(10,500) = 10,500 + 525 = 11,02510,500 +.05(10,500) = 10,500 + 525 = 11,025 Suppose a population is 10,000 in January 2004. Suppose the population increases by…

43 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Suppose a population is 10,000 in Jan 2004. Suppose the population increases by 500 per year. What is the population in …. Jan 2005?Jan 2005? 10,000 + 500 = 10,50010,000 + 500 = 10,500 Jan 2006?Jan 2006? 10,000 + 2(500) = 11,00010,000 + 2(500) = 11,000 Jan 2007?Jan 2007? 10,000 + 3(500) = 11,50010,000 + 3(500) = 11,500 Jan 2008?Jan 2008? 10,000 + 4(500) = 12,00010,000 + 4(500) = 12,000

44 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Suppose a population is 10,000 in Jan 2004 and increases by 500 per year. Let t be the number of years after 2004. Let P(t) be the population in year t. What is the symbolic representation for P(t)? We know…Let t be the number of years after 2004. Let P(t) be the population in year t. What is the symbolic representation for P(t)? We know… Population in 2004 = P(0) = 10,000 + 0(500)Population in 2004 = P(0) = 10,000 + 0(500) Population in 2005 = P(1) = 10,000 + 1(500)Population in 2005 = P(1) = 10,000 + 1(500) Population in 2006 = P(2) = 10,000 + 2(500)Population in 2006 = P(2) = 10,000 + 2(500) Population in 2007 = P(3) = 10,000 + 3(500)Population in 2007 = P(3) = 10,000 + 3(500) Population t years after 2004 = P(t) = 10,000 + t(500)Population t years after 2004 = P(t) = 10,000 + t(500)

45 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Population is 10,000 in 2004; increases by 500 per yr P(t) = 10,000 + t(500) P is a linear function of t.P is a linear function of t. What is the slope?What is the slope? 500 people/year500 people/year What is the y-intercept?What is the y-intercept? number of people at time 0 (the year 2004) = 10,000number of people at time 0 (the year 2004) = 10,000 When P increases by a constant number of people per year, P is a linear function of t.

46 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Suppose a population is 10,000 in Jan 2004. More realistically, suppose the population increases by 5% per year. What is the population in …. Jan 2005?Jan 2005? 10,000 +.05(10,000) = 10,000 + 500 = 10,50010,000 +.05(10,000) = 10,000 + 500 = 10,500 Jan 2006?Jan 2006? 10,500 +.05(10,500) = 10,500 + 525 = 11,02510,500 +.05(10,500) = 10,500 + 525 = 11,025 Jan 2007?Jan 2007? 11,025 +.05(11,025) = 11,025 + 551.25 = 11,576.2511,025 +.05(11,025) = 11,025 + 551.25 = 11,576.25

47 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Suppose a population is 10,000 in Jan 2004 and increases by 5% per year. Let t be the number of years after 2004. Let P(t) be the population in year t. What is the symbolic representation for P(t)? We know…Let t be the number of years after 2004. Let P(t) be the population in year t. What is the symbolic representation for P(t)? We know… Population in 2004 = P(0) = 10,000Population in 2004 = P(0) = 10,000 Population in 2005 = P(1) = 10,000 +.05 (10,000) = 1.05(10,000) = 1.05 1 (10,000) =10,500Population in 2005 = P(1) = 10,000 +.05 (10,000) = 1.05(10,000) = 1.05 1 (10,000) =10,500 Population in 2006 = P(2) = 10,500 +.05 (10,500) = 1.05 (10,500) = 1.05 (1.05)(10,000) = 1.05 2 (10,000) = 11,025Population in 2006 = P(2) = 10,500 +.05 (10,500) = 1.05 (10,500) = 1.05 (1.05)(10,000) = 1.05 2 (10,000) = 11,025 Population t years after 2004 = P(t) = 10,000(1.05) tPopulation t years after 2004 = P(t) = 10,000(1.05) t

48 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Population is 10,000 in 2004; increases by 5% per yr P(t) = 10,000 (1.05) t P is an EXPONENTIAL function of t. More specifically, an exponential growth function.P is an EXPONENTIAL function of t. More specifically, an exponential growth function. What is the base of the exponential function?What is the base of the exponential function? 1.051.05 What is the y-intercept?What is the y-intercept? number of people at time 0 (the year 2004) = 10,000number of people at time 0 (the year 2004) = 10,000 When P increases by a constant percentage per year, P is an exponential function of t.

49 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Linear vs. Exponential Growth A Linear Function adds a fixed amount to the previous value of y for each unit increase in xA Linear Function adds a fixed amount to the previous value of y for each unit increase in x For example, in f(x) = 10,000 + 500x 500 is added to y for each increase of 1 in x.For example, in f(x) = 10,000 + 500x 500 is added to y for each increase of 1 in x. An Exponential Function multiplies a fixed amount to the previous value of y for each unit increase in x.An Exponential Function multiplies a fixed amount to the previous value of y for each unit increase in x. For example, in f(x) = 10,000 (1.05) x y is multiplied by 1.05 for each increase of 1 in x.For example, in f(x) = 10,000 (1.05) x y is multiplied by 1.05 for each increase of 1 in x.

50 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Definition of Exponential Function A function represented by f(x) = Ca x, a > 0, a not 1, and C > 0 is an exponential function with base a and coefficient C.A function represented by f(x) = Ca x, a > 0, a not 1, and C > 0 is an exponential function with base a and coefficient C. If a > 1, then f is an exponential growth functionIf a > 1, then f is an exponential growth function If 0 < a < 1, then f is an exponential decay functionIf 0 < a < 1, then f is an exponential decay function

51 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Caution Don’t confuse f(x) = 2 x with f(x) = x 2Don’t confuse f(x) = 2 x with f(x) = x 2 f(x) = 2 x is an exponential function.f(x) = 2 x is an exponential function. f(x) = x 2 is a polynomial function, specifically a quadratic function.f(x) = x 2 is a polynomial function, specifically a quadratic function. The functions and consequently their graphs are very different.The functions and consequently their graphs are very different. f(x) = 2 x f(x) = x 2

53 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Linear Function Linear Function - Slope is constant. y = 10000 + 500x x y x  y  x y   0 10000 1 10500 1 500 500/1= 500 2 11000 1 500 500/1= 500 3 11500 1 500 500/1= 500 4 12000 1 500 500/1= 500 5 12500 1 500 500/1= 500 6 13000 1 500 500/1= 500

54 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exponential Function Note that this constant is the base of the exponential function. Exponential Function - Ratios of consecutive y-values (corresponding to unit increases in x) are constant, in this case 1.05. Y = 10000 (1.05) x

55 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Which function is linear and which is exponential? x y -3 3/8 -2 3/4 -1 3/2 0 3 1 6 2 12 3 24 x y -3 9 -2 7 -1 5 0 3 1 1 2 -1 3 -3 For the linear function, tell the slope and y-intercept. For the exponential function, tell the base and the y-intercept. Write the equation of each.

56 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley y is an exponential function of x because the ratio of consecutive values of y is constant, namely 2. Thus the base is 2. The y-intercept is 3. Thus the equation is y = 3·2 x 3/(3/2) = 2 (3/2)/(3/4) = 2 x y -3 3/8 -2 3/4 -1 3/2 0 3 1 6 2 12 3 24 24/12 = 2 12/6 = 2 6/3 = 2 (3/4)/(3/8) = 2 Which function is linear and which is exponential? continued

57 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley y is a linear function of x because the slope is constant, namely –2/1 = –2. The y-intercept is 3. Thus the equation is y = –2x + 3 x y -3 9 -2 7 -1 5 0 3 1 1 2 -1 3 -3 Which function is linear and which is exponential? continued

58 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exponential Growth vs Decay Example of exponential growth functionExample of exponential growth function f(x) = 3 2 x Example of exponential decay functionExample of exponential decay function

59 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Recall In the exponential function f(x) = Ca xIn the exponential function f(x) = Ca x If a > 1, then f is an exponential growth functionIf a > 1, then f is an exponential growth function If 0 < a < 1, then f is an exponential decay functionIf 0 < a < 1, then f is an exponential decay function

60 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exponential Growth Function f(x) = Ca x where a > 1 ExampleExample f(x) = 3 2 xf(x) = 3 2 x Properties of an exponential growth function Domain: (-∞, ∞)Domain: (-∞, ∞) Range: (0, ∞)Range: (0, ∞) f increases on (-∞, ∞)f increases on (-∞, ∞) The negative x-axis is a horizontal asymptote.The negative x-axis is a horizontal asymptote. y-intercept is (0,3).y-intercept is (0,3).

61 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exponential Decay Function f(x) = Ca x where 0 < a < 1 ExampleExampleProperties of an exponential decay function Domain: (-∞, ∞)Domain: (-∞, ∞) Range: (0, ∞)Range: (0, ∞) f decreases on (-∞, ∞)f decreases on (-∞, ∞) The positive x-axis is a horizontal asymptote.The positive x-axis is a horizontal asymptote. y-intercept is (0,3).y-intercept is (0,3).

62 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example of exponential decay - Carbon-14 dating The time it takes for half of the atoms to decay into a different element is called the half-life of an element undergoing radioactive decay.The time it takes for half of the atoms to decay into a different element is called the half-life of an element undergoing radioactive decay. The half-life of carbon-14 is 5700 years.The half-life of carbon-14 is 5700 years. Suppose C grams of carbon-14 are present at t = 0. Then after 5700 years there will be C/2 grams present.Suppose C grams of carbon-14 are present at t = 0. Then after 5700 years there will be C/2 grams present.

63 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Recall the half-life of carbon-14 is 5700 years. Let t be the number of years.Let t be the number of years. Let A =f(t) be the amount of carbon-14 present at time t.Let A =f(t) be the amount of carbon-14 present at time t. Let C be the amount of carbon-14 present at t = 0.Let C be the amount of carbon-14 present at t = 0. Then f(0) = C and f(5700) = C/2.Then f(0) = C and f(5700) = C/2. Thus two points of f are (0,C) and (5700, C/2)Thus two points of f are (0,C) and (5700, C/2) Using the point (5700, C/2) and substituting 5700 for t and C/2 for A in A = f(t) = Ca t yields: C/2 = C a 5700Using the point (5700, C/2) and substituting 5700 for t and C/2 for A in A = f(t) = Ca t yields: C/2 = C a 5700 Dividing both sides by C yields: 1/2 = a 5700Dividing both sides by C yields: 1/2 = a 5700

65 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Generalizing this If a radioactive sample containing C units has a half-life of k years, then the amount A remaining after x years is given byIf a radioactive sample containing C units has a half-life of k years, then the amount A remaining after x years is given by

66 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example of Radioactive Decay Radioactive strontium-90 has a half-life of about 28 years and sometimes contaminates the soil. After 50 years, what percentage of a sample of radioactive strontium would remain?Radioactive strontium-90 has a half-life of about 28 years and sometimes contaminates the soil. After 50 years, what percentage of a sample of radioactive strontium would remain? Note calculator keystrokes : Since C is present initially and after 50 years.29C remains, then 29% remains.

67 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example of Exponential Growth - Compound Interest Suppose $10,000 is deposited into an account which pays 5% interest compounded annually. Then the amount A in the account after t years is: A(t) = 10,000 (1.05) tSuppose $10,000 is deposited into an account which pays 5% interest compounded annually. Then the amount A in the account after t years is: A(t) = 10,000 (1.05) t Note the similarity with: Suppose a population is 10,000 in 2004 and increases by 5% per year. Then the population P, t years after 2004 is: P(t) = 10,000 (1.05) tNote the similarity with: Suppose a population is 10,000 in 2004 and increases by 5% per year. Then the population P, t years after 2004 is: P(t) = 10,000 (1.05) t

68 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Frequencies of Compounding (Adding Interest) annually (1 time per year)annually (1 time per year) semiannually (2 times per year)semiannually (2 times per year) quarterly (4 times per year)quarterly (4 times per year) monthly (12 times per year)monthly (12 times per year) daily (365 times per year)daily (365 times per year)

69 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Compound Interest Formula If P dollars is deposited in an account paying an annual rate of interest r, compounded (paid) n times per year, then after t years the account will contain A dollars, whereIf P dollars is deposited in an account paying an annual rate of interest r, compounded (paid) n times per year, then after t years the account will contain A dollars, where

70 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Suppose $1000 is deposited into an account yielding 5% interest compounded at the following frequencies. How much money after t years? AnnuallyAnnually SemiannuallySemiannually QuarterlyQuarterly MonthlyMonthly

71 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Natural Exponential Function The function f, represented byThe function f, represented by f(x) = e x is the natural exponential function where is the natural exponential function where e  2.718281828 e  2.718281828

72 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Continuously Compounded Interest If a principal of P dollars is deposited in an account paying an annual rate of interest r (expressed in decimal form), compounded continuously, then after t years the account will contain A dollars, where A = Pe rt

73 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Suppose $100 is invested in an account with an interest rate of 8% compounded continuously. How much money will there be in the account after 15 years? A = Pe rt A = $100 e.08(15) A = $332.01

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Logarithmic Functions and Models ♦Evaluate the common logarithm function ♦Solve basic exponential and logarithmic equations ♦Evaluate logarithms with other bases ♦Solve general exponential and logarithmic equations 5.4

75 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Common Logarithm The common logarithm of a positive number x, denoted log x, is defined byThe common logarithm of a positive number x, denoted log x, is defined by logx = k if and only if x = 10 k logx = k if and only if x = 10 k where k is a real number. The function given by f(x) = log x is called the common logarithm function.The function given by f(x) = log x is called the common logarithm function.

76 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Evaluate each of the following. log10log10 log 100log 100 log 1000log 1000 log 10000log 10000 log (1/10)log (1/10) log (1/100)log (1/100) log (1/1000)log (1/1000) log 1log 1 1 because 10 1 = 101 because 10 1 = 10 2 because 10 2 = 1002 because 10 2 = 100 3 because 10 3 = 10003 because 10 3 = 1000 4 because 10 4 = 100004 because 10 4 = 10000 –1 because 10 -1 = 1/10–1 because 10 -1 = 1/10 –2 because 10 -2 = 1/100–2 because 10 -2 = 1/100 –3 because 10 -3 = 1/1000–3 because 10 -3 = 1/1000 0 because 10 0 = 10 because 10 0 = 1

77 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph of f(x) = log x xf(x)f(x).01-2.1 10 101 1002 Note that the graph of y = log x is the graph of y = 10 x reflected through the line y = x. This suggests that these are inverse functions.

78 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Inverse of y = log x Note that the graph of f(x) = log x passes the horizontal line test so it is a 1-1 function and has an inverse function.Note that the graph of f(x) = log x passes the horizontal line test so it is a 1-1 function and has an inverse function. Find the inverse of y = log xFind the inverse of y = log x Using the definition of common logarithm to solve for x givesUsing the definition of common logarithm to solve for x gives x = 10 y x = 10 y Interchanging x and y givesInterchanging x and y gives y = 10 x y = 10 x So yes, the inverse of y = log x is y = 10 xSo yes, the inverse of y = log x is y = 10 x

79 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inverse Properties of the Common Logarithm Recall that f -1 (x) = 10 x given f(x) = log x (f  f -1 )(x) = x for every x in the domain of f -1Since (f  f -1 )(x) = x for every x in the domain of f -1 log(log(10 x ) = x for all real numbers x. Since (f -1  f)(x) = x for every x in the domain of fSince (f -1  f)(x) = x for every x in the domain of f 10 logx = x for any positive number x10 logx = x for any positive number x

80 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Exponential Equations Using The Inverse Property log( Solving Exponential Equations Using The Inverse Property log(10 x ) = x Solve the equation 10 x = 35Solve the equation 10 x = 35 Take the common log of both sidesTake the common log of both sides log 10 x = log 35log 10 x = log 35 Using the inverse property log(Using the inverse property log(10 x ) = x this simplifies to x = log 35 Using the calculator to estimate log 35 we have x  1.54

81 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Logarithmic Equations Using The Inverse Property 10 logx = x Solve the equation log x = 4.2Solve the equation log x = 4.2 Exponentiate each side using base 10Exponentiate each side using base 10 10 logx = 10 4.210 logx = 10 4.2 Using the inverse property 10 logx = xUsing the inverse property 10 logx = x this simplifies to 10 4.2x = 10 4.2 10 4.2Using the calculator to estimate 10 4.2 we have x  15848.93

82 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Definition of Logarithm With Base a The logarithm with base a of a positive number x, denoted by log a x is defined byThe logarithm with base a of a positive number x, denoted by log a x is defined by log a x = k if and only if x = a k where a > 0, a ≠1, and k is a real number. The function given by f(x) = log a x is called the logarithmic function with base a.The function given by f(x) = log a x is called the logarithmic function with base a.

83 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Practice with the Definition Log b c = d means____Log b c = d means____ p = log y m means ____p = log y m means ____ True or false:True or false: True or false: log 2 8 = 3True or false: log 2 8 = 3 True or false: log 5 25 = 2True or false: log 5 25 = 2 True or false: log 25 5 = 1/2True or false: log 25 5 = 1/2 True or false: log 4 8 = 2True or false: log 4 8 = 2 b d = cb d = c y p = my p = m True because 2 3 =8True because 2 3 =8 True because 5 2 =25True because 5 2 =25 True because 25 1/2 =5True because 25 1/2 =5 False because 4 2 =16 not 8False because 4 2 =16 not 8 Practice Questions: Answers: What is the value of log 4 8? It is 3/2 because 4 3/2 = 8

84 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Practice Evaluating Logarithms log 6 36log 6 36 log 36 6log 36 6 log 2 32log 2 32 log 32 2log 32 2 log 6 (1/36)log 6 (1/36) log 2 (1/32)log 2 (1/32) log 100log 100 log (1/10)log (1/10) log 1log 1 2 because 6 2 = 362 because 6 2 = 36 1/2 because 36 (1/2) = 61/2 because 36 (1/2) = 6 5 because 2 5 = 325 because 2 5 = 32 1/5 because 32 (1/5) = 21/5 because 32 (1/5) = 2 –2 because 6 –2 = 1/36–2 because 6 –2 = 1/36 –5 because 2 –5 = 1/32–5 because 2 –5 = 1/32 2 because 10 2 = 1002 because 10 2 = 100 –1 because 10 –1 = 1/10–1 because 10 –1 = 1/10 0 because 10 0 = 10 because 10 0 = 1 Evaluate Answers:

85 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Calculators and logarithms The TI-83 evaluates base 10 logarithms and base e logarithms.The TI-83 evaluates base 10 logarithms and base e logarithms. Base 10 logs are called common logs.Base 10 logs are called common logs. log x means log 10 x.log x means log 10 x. Notice the log button on the calculator.Notice the log button on the calculator. Base e logs are called natural logs.Base e logs are called natural logs. ln x means log e x.ln x means log e x. Notice the ln button on the calculator.Notice the ln button on the calculator.

86 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Evaluate each of the following without calculator. Then check with calculator. lnelne ln(e 2 )ln(e 2 ) ln1ln1. lne = log e e = 1 since e 1 = elne = log e e = 1 since e 1 = e ln(e 2 ) = log e (e 2 ) = 2 since 2 is the exponent that goes on e to produce e 2.ln(e 2 ) = log e (e 2 ) = 2 since 2 is the exponent that goes on e to produce e 2. ln1 = log e 1 = 0 since e 0 = 1ln1 = log e 1 = 0 since e 0 = 1 1/2 since 1/2 is the exponent that goes on e to produce e 1/21/2 since 1/2 is the exponent that goes on e to produce e 1/2

87 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Inverse of y = log a x Note that the graph of f(x) = log a x passes the horizontal line test so it is a 1-1 function and has an inverse function.Note that the graph of f(x) = log a x passes the horizontal line test so it is a 1-1 function and has an inverse function. Find the inverse of y = log a xFind the inverse of y = log a x Using the definition of common logarithm to solve for x givesUsing the definition of common logarithm to solve for x gives x = a y x = a y Interchanging x and y givesInterchanging x and y gives y = a x y = a x So the inverse of y = log a x is y = a xSo the inverse of y = log a x is y = a x

88 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inverse Properties of Logarithms With Base a Recall that f -1 (x) = a x given f(x) = log a x (f  f -1 )(x) = x for every x in the domain of f -1Since (f  f -1 )(x) = x for every x in the domain of f -1 log a (log a (a x ) = x for all real numbers x. Since (f -1  f)(x) = x for every x in the domain of fSince (f -1  f)(x) = x for every x in the domain of f a log a x = x for any positive number xa log a x = x for any positive number x

89 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Exponential Equations Using The Inverse Property log a (a x ) = x Solve the equation 4 x = 1/64Solve the equation 4 x = 1/64 Take the log of both sides to the base 4Take the log of both sides to the base 4 log 4 (4 x ) = log 4 (1/64)log 4 (4 x ) = log 4 (1/64) Using the inverse property log a (a x ) =xUsing the inverse property log a (a x ) =x this simplifies to log 4 (1/64)x = log 4 (1/64) Since 1/64 can be rewritten as –Since 1/64 can be rewritten as 4 –3 log 4 ( – )x = log 4 (4 –3 ) Using the inverse property log a (a x ) = xUsing the inverse property log a (a x ) = x this simplifies to –x = –3

90 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Exponential Equations Using The Inverse Property log a (a x ) = x Solve the equation e x = 15Solve the equation e x = 15 Take the log of both sides to the base eTake the log of both sides to the base e ln(e x ) = ln(15)ln(e x ) = ln(15) Using the inverse property log a (a x ) = xUsing the inverse property log a (a x ) = x this simplifies to ln15x = ln15 Using the calculator to estimate ln 15Using the calculator to estimate ln 15 x  2.71

91 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Logarithmic Equations Using The Inverse Property a log a x = x Solve the equation lnx = 1.5Solve the equation lnx = 1.5 Exponentiate both sides using base eExponentiate both sides using base e e lnx = e 1.5e lnx = e 1.5 Using the inverse property a log a x = xUsing the inverse property a log a x = x this simplifies to e 1.5x = e 1.5 Using the calculator to estimate e 1.5Using the calculator to estimate e 1.5 x  4.48

92 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Recall from section 5.3 Graph of f(x) = a x where a >1 Graph of f(x) = a x where 0 < a <1 Using the fact that the graph of a function and its inverse are symmetric with respect to the line y = x, graph f -1 (x) = log a x for the two types of exponential functions listed above. Looking at the two resulting graphs, what is the domain of a logarithmic function? What is the range of a logarithmic function? Exponential Growth Function Exponential Decay Function

93 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph of f(x) = a x where a >1 Graph of f(x) = a x where 0 < a < 1 Exponential Growth Function Exponential Decay Function Superimpose graphs of the inverses of the functions above similar to Figure 5.58 on page 422

95 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Property 1 log a 1 = 0 and log a a = 1log a 1 = 0 and log a a = 1 True because a 0 = 1 and a 1 = aTrue because a 0 = 1 and a 1 = a

96 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Property 2 log a m + log a n = log a (mn)log a m + log a n = log a (mn) The sum of logs is the log of the product.The sum of logs is the log of the product. Example: Let a = 2, m = 4 and n = 8Example: Let a = 2, m = 4 and n = 8 log a m + log a n = log 2 4 + log 2 8 = 2 + 3log a m + log a n = log 2 4 + log 2 8 = 2 + 3 log a (mn) = log 2 (4 · 8) = log 2 (32) = 5log a (mn) = log 2 (4 · 8) = log 2 (32) = 5

97 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Property 3 The difference of logs is the log of the quotient.The difference of logs is the log of the quotient. Example: Let a = 2, m = 4 and n = 8Example: Let a = 2, m = 4 and n = 8

101 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Change of Base Formula Example: Use the change of base formula to evaluate log 3 8Example: Use the change of base formula to evaluate log 3 8

103 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Modeling Compound Interest How long does it take money to grow from $100 to $200 if invested into an account which compounds quarterly at an annual rate of 5%?How long does it take money to grow from $100 to $200 if invested into an account which compounds quarterly at an annual rate of 5%? Must solve for t in the following equationMust solve for t in the following equation

104 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve for t Divide each side by 100 Take common logarithm of each side Property 4: log(m r ) = rlogm Divide each side by 4log1.0125 Approximate using calculator

105 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Alternatively Divide each side by 100 Take natural logarithm of each side Property 4: ln(m r ) = r lnm Divide each side by 4ln1.0125 Approximate using calculator Note that the only difference in this and the previous solution is that we took natural logs of both sides of the equation instead of common logs.

107 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve 3(1.2) x + 2 = 15 for x symbolically Divide each side by 3 Take common logarithm of each side (Could use natural logarithm) Property 4: log(m r ) = r logm Divide each side by log1.2 Approximate using calculator

108 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve e x+2 = 5 2x for x symbolically Take natural logarithm of each side Property 4: ln(m r ) = r lnm lne = 1 Subtract 2xln5 and 2 from each side Factor x from left-hand side Divide each side by 1 – 2ln5 Approximate using calculator

109 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving a Logarithmic Equation Symbolically In developing countries there is a relationship between the amount of land a person owns and the average daily calories consumed. This relationship is modeled by the formula C(x) = 280 ln(x+1) + 1925 where x is the amount of land owned in acres andIn developing countries there is a relationship between the amount of land a person owns and the average daily calories consumed. This relationship is modeled by the formula C(x) = 280 ln(x+1) + 1925 where x is the amount of land owned in acres and Source: D. Gregg: The World Food Problem Determine the number of acres owned by someone whose average intake is 2400 calories per day.Determine the number of acres owned by someone whose average intake is 2400 calories per day. Must solve for x in the equationMust solve for x in the equation 280 ln(x+1) + 1925 = 2400

110 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve 280 ln(x+1) + 1925 = 2400 Symbolically Subtract 1925 from each side Divide each side by 280 Divide each side by 280 Exponentiate each side base e Inverse property e lnk = k Subtract 1 from each side Approximate using calculator

111 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Logarithmic Equations Symbolically Definition of logarithm log a x = k if and only if x = a k Add x to both sides of equation Subtract 2 from both sides of equation

114 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exponential Model Example The National Institute for Automotive Service Excellence (ASE) reported that the number of females working in automotive repair is increasing. a) What type of function might model these data? b) Use least-squares regression to find an exponential function given by f(x) = ab x that model the data. c) Use f to estimate the number of certified female technicians in 2005. Round the result to the nearest hundred. Year19881989199019911992199319941995 Total556614654737849108613291592

115 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution a) Let y = number of female technicians Let x = 0 correspond to 1988, x = 1 to 1989 and so on, until x = 7 corresponds to 1995. A scatterplot of the data is shown. The data is rapidly increasing, and an exponential function might model these data. Year19881989199019911992199319941995 Total556614654737849108613291592

116 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued b) The formula f(x) = 507.1(1.166) x is found using the least-squares regression, as shown in the following figures. c) Since x = 17 corresponds to year 2005, f(17) = 507.1(1.166) 17  6902. According to the model, the number of certified female automotive technicians in 2005 could be about 6900.

117 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Logarithmic Model Example The table below lists the interest rates for certificates of deposit during January 1997. Use the data to complete the following. a) Make a scatterplot of the data. What type of function might model these data? b) Use least-squares regression to obtain a formula, f(x) = a + b ln x, that models these data. c) Graph f and the data in the same viewing rectangle. Time6 mo1 yr2.5 yr5 yr Yield4.75%5.03%5.25%5.54%

118 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution a) Enter the data points into your calculator. The data increase but are gradually leveling off. A logarithmic modeling function may be appropriate. b) The least-square regression has been used to find logarithmic function. f(x) = 5 + 0.33 lnx. c) A graph and data are shown. Time6 mo1 yr2.5 yr5 yr Yield4.75%5.03%5.25%5.54%

119 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Logistic Model Example One of the earliest studies about population growth was done using yeast plants in 1913. A small amount of yeast was placed in a container with a fixed amount of nourishment. The units of yeast were recorded every 2 hours. a) Make a scatterplot of the data. Describe the growth. b) Use least-squares regression to find a logistic function that models the data. c) Graph f and the data in the same viewing rectangle. d) Approximate graphically the time when the amount of yeast was 200 units. Time024681012141618 Yeast9.629.071.1174.6350.7513.3594.8640.8655.9661.8

120 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution a) The yeast increase slowly at first, then they grow more rapidly, until the amount of yeast gradually levels off. b) Least-squares regression to find a logistic function. Time024681012141618 Yeast9.629.071.1174.6350.7513.3594.8640.8655.9661.8

121 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued c) Data and graph in the same viewing rectangle. d) The graphs of y 1 = f(x) and y(2) = 200 intersect near (6.29, 200). The amount of yeast reached 200 units after about 6.29 hours.

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