Inverse, Exponential, and Logarithmic Functions PDF

5 Inverse, Inverse, Exponential, Exponential, and and Logarithmic Logarithmic Functions Functions Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 5.1 Inverse Functions Objectives: One-to-One Functions Inverse Functions Equations...

5 Inverse, Inverse, Exponential, Exponential, and and Logarithmic Logarithmic Functions Functions Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 5.1 Inverse Functions Objectives: One-to-One Functions Inverse Functions Equations of Inverses Copyright © 2013, 2009, 2005 Pearson Education, Inc. 24.1 - 2  Pre-Questions 1) What the different between the two functions F={(1,2),(3,2),(4,5)}, G={(1,2),(3,4),(5,6)}? 2) Is every function has an inverse function? 3) What the type of function that has inverse? Copyright © 2013, 2009, 2005 Pearson Education, Inc. 3  One-to-One Functions Suppose we define the function F  {(2, 2), ( 1,1), (0, 0), (1, 3), (2, 5)}. We can form another set of ordered pairs from F by interchanging the x- and y-values of each pair in F. We call this set G, so G  {(2, 2), (1, 1), (0, 0), (3,1), (5, 2)}. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 4  One-to-One Functions To show that these two sets are related, G is called the inverse of F. For a function  to have an inverse,  must be a one-to-one function. In a one-to-one function, each x-value corresponds to only one y-value, and each y-value corresponds to only one x-value. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 5 One-to-One Functions This function is not one-to-one because the y-value 7 corresponds to two x-values, 2 and 3. That is, the ordered pairs (2, 7) and (3, 7) both belong to the function. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 6 One-to-One Functions This function is one-to-one. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 7 Quiz 1 Which function is not one-to-one ? A.  0,1 , 1,2 ,  2,3 , 3,4 B.  0,1 , 1,0 ,  2,0 , 3,2 C.  0,0 , 1,1 ,  2,2 , 3,3 D.  0,1 , 1,0 ,  2,3 , 3,2 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 8 One-to-One Function A function  is a one-to-one function if, for elements a and b in the domain of , a  b implies f (a)  f (b). Copyright © 2013, 2009, 2005 Pearson Education, Inc. 9  One-to-One Functions Using the concept of the contrapositive from the study of logic, the last line in the preceding box is equivalent to f (a )  f (b ) implies a  b. We use this statement to decide whether a function  is one-to-one in the next example. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 10 Not One-to-One Function A function  is not one-to-one function if, for elements a and b in the domain of , a  b implies f (a)  f (b). Copyright © 2013, 2009, 2005 Pearson Education, Inc. 11  Example 1 DECIDING WHETHER FUNCTIONS ARE ONE-TO-ONE Decide whether each function is one-to-one. (a) f ( x )   4 x  12 Solution We must show that (a) = (b) leads to the result a = b. f (a )  f ( b )  4a  12   4b  12 f ( x )   4 x  12  4a   4b Subtract 12. ab Divide by –4. By the definition, f ( x )   4 x  12 is one-to-one. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 12  Example 1 DECIDING WHETHER FUNCTIONS ARE ONE-TO-ONE Decide whether each function is one-to-one. (b) f ( x )  25  x 2 Solution If we choose a = 3 and b = – 3, then 3 ≠ – 3, but f (3)  25  32  25  9  16  4 and f ( 3)  25  ( 3)  25  9  4. 2 Here, even though 3 ≠ –3, (3) = (–3 ) = 4. By the definition,  is not a one-to-one function. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 13  Horizontal Line Test As illustrated in Example 1(b), a way to show that a function is not one-to-one is to produce a pair of different domain elements that lead to the same function value. There is also a useful graphical test, the horizontal line test, that tells whether or not a function is one-to-one. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 14 Horizontal Line Test A function is one-to-one if every horizontal line intersects the graph of the function at most once. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 15  Note In Example 1(b), the graph of the function is a semicircle, as shown in the figure. Because there is at least one horizontal line that intersects the graph in more than one point, this function is not one-to-one. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 16  One-to-One Functions Notice that the function graphed in Example 2(b) decreases on its entire domain. In general, a function that is either increasing or decreasing on its entire domain, such as (x) = –x, g(x) = x3, and h(x) = x , must be one- to-one. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 17 Tests to Determine Whether a Function is One-to-One 1. Show that (a) = (b) implies a = b. This means that  is one-to-one. (Example 1(a)) 2. In a one-to-one function every y-value corresponds to no more than one x-value. To show that a function is not one-to-one, find at least two x-values that produce the same y-value. (Example 1(b)) Copyright © 2013, 2009, 2005 Pearson Education, Inc. 18 Tests to Determine Whether a Function is One-to-One 3. Sketch the graph and use the horizontal line test. (Example 2) 4. If the function either increases or decreases on its entire domain, then it is one-to-one. A sketch is helpful here, too. (Example 2(b)) Copyright © 2013, 2009, 2005 Pearson Education, Inc. 19  Inverse Functions Consider the functions 1 5 f ( x )  8 x  5 and g ( x )  x . 8 8 Let us choose an arbitrary element from the domain of f, say 10. Evaluate f(10). f (10)  8  10  5  85 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 20  Inverse Functions 1 5 f ( x )  8 x  5 and g ( x )  x . 8 8 Now, we evaluate g(85). 1 5 g(85)  (85)  Let x = 85. 8 8 85 5   Multiply. 8 8 80  Subtract. 8 g(85)  10 Divide. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 21  Inverse Functions Starting with 10, we “applied” function  and then “applied” function g to the result, which returned the number 10. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 22  Inverse Functions As further examples, check that f (3)  29 and g ( 29)  3, f ( 5)  35 and g ( 35)  5, 3  3 g ( 2)   and f   2. 8  8 In particular, for this pair of functions, f ( g (2))  2 and g ( f ( 2))  2. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 23  Inverse Functions In fact, for any value of x, f ( g ( x ))  x and g ( f ( x ))  x. Using the notation for composition introduced in Section 2.4, these two equations can be written as follows. ( f g )( x )  x and (g f )( x )  x. Because the compositions of f and g yield the identity function, they are inverses of each other. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 24 Inverse Function Let  be a one-to-one function. Then g is the inverse function of  if (f g )( x )  x for every x in the domain of g, and (g f )( x )  x for every x in the domain of . Copyright © 2013, 2009, 2005 Pearson Education, Inc. 25 Inverse Function The condition that  is one-to-one in the definition of inverse function is essential. Otherwise, g will not define a function. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 26  Example 2 DECIDING WHETHER TWO FUNCTIONS ARE INVERSES Let functions  and g be defined by f ( x )  x 3  1 and g ( x )  x  1 , respectively. 3 Is g the inverse function of ? Solution The horizontal line test applied to the graph indicates that  is one-to-one, so the function does have an inverse. Since it is one-to- one, we now find ( ◦ g)(x) and (g ◦ )(x). Copyright © 2013, 2009, 2005 Pearson Education, Inc. 27  Example 2 DECIDING WHETHER TWO FUNCTIONS ARE INVERSES Let functions  and g be defined by f ( x )  x 3  1 and g ( x )  x  1 , respectively. 3 Is g the inverse function of ? Solution   3 (f g )( x )  f ( g ( x ))  3 x 1 1  x  1 1 x Copyright © 2013, 2009, 2005 Pearson Education, Inc. 28  Example 2 DECIDING WHETHER TWO FUNCTIONS ARE INVERSES Let functions  and g be defined by f ( x )  x 3  1 and g ( x )  x  1 , respectively. 3 Is g the inverse function of ? Solution (g f )( x )  g ( f ( x ))  ( x  1)  1 3 3  3 x3 x Since ( ◦ g)(x) = x and (g ◦ )(x) = x, function g is the inverse of function . Copyright © 2013, 2009, 2005 Pearson Education, Inc. 29  Special Notation A special notation is used for inverse functions: If g is the inverse of a function , then g is written as –1 (read “-inverse”). For (x) = x3 – 1, f 1( x )  3 x  1. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 30  Caution Do not confuse the –1 in –1 with a negative exponent. The symbol 1  (x) does not represent f ( x ) ;it represents –1 the inverse function of . Copyright © 2013, 2009, 2005 Pearson Education, Inc. 31  Inverse Function By the definition of inverse function, the domain of  is the range of –1, and the range of  is the domain of –1. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 32  Equations of Inverses The inverse of a one-to-one function is found by interchanging the x- and y- values of each of its ordered pairs. The equation of the inverse of a function defined by y = (x) is found in the same way. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 33 Finding the Equation of the Inverse of y = (x) For a one-to-one function  defined by an equation y = (x), find the defining equation of the inverse as follows. (If necessary, replace (x) with y first. Any restrictions on x and y should be considered.) Step 1 Interchange x and y. Step 2 Solve for y. Step 3 Replace y with –1(x). Copyright © 2013, 2009, 2005 Pearson Education, Inc. 34  Example 3 FINDING EQUATIONS OF INVERSES Decide whether each equation defines a one-to-one function. If so, find the equation of the inverse. (a) f ( x )  2 x  5 Solution The graph of y = 2x + 5 is a non- horizontal line, so by the horizontal line test,  is a one-to-one function. To find the equation of the inverse, follow the steps in the preceding box, first replacing (x) with y. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 35  Example 3 FINDING EQUATIONS OF INVERSES Decide whether each equation defines a one-to-one function. If so, find the equation of the inverse. Solution y  2x  5 Let y = (x). x  2y  5 Interchange x and y. x  5  2y Solve for y. x 5 y 2 1 1 5 f (x)  x  Replace y with -1(x). 2 2 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 36  Example 3 FINDING EQUATIONS OF INVERSES Decide whether each equation defines a one-to-one function. If so, find the equation of the inverse. Solution In the function defined by y = 2x + 5, the value of y is found by starting with a value of x, multiplying by 2, and adding 5. The form 1 x 5 f (x)  for the equation of the inverse 2 has us subtract 5 and then divide by 2. This shows how an inverse is used to “undo” what a function does to the variable x. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 37  Example 3 FINDING EQUATIONS OF INVERSES Decide whether each equation defines a one-to-one function. If so, find the equation of the inverse. (b) y  x 2  2 Solution The equation has a parabola opening up as its graph, so some horizontal lines will intersect the graph at two points. For example, both x = 3 and x = –3 correspond to y = 11. Because of the presence of the x2-term, there are many pairs of x- values that correspond to the same y-value. This means that the function defined by y = x2 + 2 is not one-to-one and does not have an inverse. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 38  Example 3 FINDING EQUATIONS OF INVERSES Decide whether each equation defines a one-to-one function. If so, find the equation of the inverse. (b) y  x 2  2 Solution The steps for finding the equation of an inverse lead to the following. y  x2  2 Remember x  y2  2 Interchange x and y. both roots. x 2 y 2 Solve for y.  x 2 y Square root property Copyright © 2013, 2009, 2005 Pearson Education, Inc. 39  Example 3 FINDING EQUATIONS OF INVERSES Decide whether each equation defines a one-to-one function. If so, find the equation of the inverse. (b) y  x 2  2 Solution  x 2 y The last step shows that there are two y-values for each choice of x greater than 2, so the given function is not one-to-one and cannot have an inverse. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 40  Example 3 FINDING EQUATIONS OF INVERSES Decide whether each equation defines a one-to-one function. If so, find the equation of the inverse. (c) f ( x )  ( x  2)3 Solution The figure shows that the horizontal line test assures us that this horizontal translation of the graph of the cubing function is one-to-one. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 41  Example 3 FINDING EQUATIONS OF INVERSES Decide whether each equation defines a one-to-one function. If so, find the equation of the inverse. Solution f ( x )  ( x  2) 3 y  ( x  2) 3 Replace (x) with y. x  ( y  2)3 Interchange x and y. 3 x  ( y  2) 3 3 Take the cube root on each side. 3 x  y 2 3 x 2 y Solve for y by adding 2. f 1( x )  3 x  2 Replace y with -1(x). Rewrite. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 42  Inverse Function One way to graph the inverse of a function  whose equation is known follows. Step 1 Find some ordered pairs that are on the graph of . Step 2 Interchange x and y to get ordered pairs that are on the graph of –1. Step 3 Plot those points, and sketch the graph of –1 through them. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 43  Inverse Function Another way is to select points on the graph of  and use symmetry to find corresponding points on the graph of –1. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 44  Inverse Function For example, suppose the point (a, b) shown here is on the graph of a one-to-one function . Copyright © 2013, 2009, 2005 Pearson Education, Inc. 45  Inverse Function Then the point (b, a) is on the graph of –1. The line segment connecting (a, b) and (b, a) is perpendicular to, and cut in half by, the line y = x. The points (a, b) and (b, a) are “mirror images” of each other with respect to y = x. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 46  Inverse Function Thus, we can find the graph of –1 from the graph of  by locating the mirror image of each point in  with respect to the line y = x. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 47  Example 4 GRAPHING f–1 GIVEN THE GRAPH OF f In each set of axes, the graph of a one-to- one function  is shown in blue. Graph –1 in red. Solution On the next slide, the graphs of two functions  shown in blue are given with their inverses shown in red. In each case, the graph of –1 is a reflection of the graph of  with respect to the line y = x. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 48  Example 4 GRAPHING f–1 GIVEN THE GRAPH OF f Solution Copyright © 2013, 2009, 2005 Pearson Education, Inc. 49 Important Facts About Inverses 1. If  is one-to-one, then -1 exists. 2. The domain of  is the range of -1,and the range of  is the domain of -1. 3. If the point (a, b) lies on the graph of , then (b, a) lies on the graph of -1. The graphs of  and -1 are reflections of each other across the line y = x. 4. To find the equation for -1, replace (x) with y, interchange x and y, and solve for y. This gives -1 (x). Copyright © 2013, 2009, 2005 Pearson Education, Inc. 50  Exponential Functions 5.2 Exponents and Properties Exponential Functions Exponential Equations Copyright © 2013, 2009, 2005 Pearson Education, Inc. 51 4.1 - 51  Exponents and Properties Recall the definition of am/n: if a is a real number, m is an integer, n is a positive integer, n and a is a real number, then  a m For example, a mn  n.   3 16 34  16 4  23  8, 1 3 1 1 1 27  13  3  , 27 27 3 1 1 1 and 641 2  12  . 64 64 8 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 52  Exponents and Properties In this section we extend the definition of ar to include all real (not just rational) values of the 3 exponent r. For example, 2 might be evaluated by approximating the exponent 3 with the rational numbers 1.7, 1.73, 1.732, and so on. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 53  Exponents and Properties Since these decimals approach the value of 3 3 more and more closely, it seems reasonable that 2 should be approximated more and more closely by the numbers 21.7, 21.73, 21.732, and so on. 17 (Recall, for example, that 21.7 = 217/10 = 10 2.   Copyright © 2013, 2009, 2005 Pearson Education, Inc. 54  Exponents and Properties To show that this assumption is reasonable, see the graphs of the function (x) = 2x with three different domains. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 55  Exponents and Properties Using this interpretation of real exponents, all rules and theorems for exponents are valid for all real number exponents, not just rational ones. In addition to the rules for exponents presented earlier, we use several new properties in this chapter. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 56 Additional Properties of Exponents For any real number a > 0, a ≠ 1, the following statements are true. (a) ax is a unique real number for each real number x. (b) ab = ac if and only if b = c. (c) If a > 1 and m < n, then am < an. (d) If 0 < a < 1 and m < n, then am > an. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 57  Properties of Exponents Properties (a) and (b) require a > 0 so that ax is always defined. For example, (–6)x is not a real number if x = ½. This means that ax will always be positive, since a must be positive. In property (a), a cannot equal 1 because 1x = 1 for every real number value of x, so each value of x leads to the same real number, 1. For property (b) to hold, a must not equal 1 since, for example, 14 = 15, even though 4 ≠ 5. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 58  Properties of Exponents Properties (c) and (d) say that when a > 1, increasing the exponent on “a” leads to a greater number, but when 0 < a < 1, increasing the exponent on “a” leads to a lesser number. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 59  Example 1 EVALUATING AN EXPONENTIAL EXPRESSION If (x) = 2x, find each of the following. (a) f ( 1) Solution 1 1 f ( 1)  2  Replace x with –1. 2 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 60  Example 1 EVALUATING AN EXPONENTIAL EXPRESSION If (x) = 2x, find each of the following. (b) f (3) Solution f (3)  2  8 3 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 61  Example 1 EVALUATING AN EXPONENTIAL EXPRESSION If (x) = 2x, find each of the following. 5 (c) f   2 Solution 5 f    25 / 2  (25 )1/ 2 2  32 1/ 2  32  16  2  4 2 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 62  Example 1 EVALUATING AN EXPONENTIAL EXPRESSION If (x) = 2x, find each of the following. (d) f (4.92) Solution f (4.92)  2 4.92  30.2738447 Use a calculator. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 63 Exponential Function If a > 0 and a ≠ 1, then f (x)  a x defines the exponential function with base a. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 64  Motion Problems Note We do not allow 1 as the base for an exponential function. If a = 1, the function becomes the constant function defined by (x) = 1, which is not an exponential function. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 65  Exponential Functions Slide 7 showed the graph of (x) = 2x with three different domains. We repeat the final graph (with real numbers as domain) here. The y-intercept is y = 20 = 1. Since 2x > 0 for all x and 2x 0 as x –, the x-axis is a horizontal asymptote. As the graph suggests, the domain of the function is (–, ) and the range is (0, ). The function is increasing on its entire domain, and is one-to-one. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 66 EXPONENTIAL FUNCTION f ( x )  a x Domain: (–, ) ‫ثابت ال يتغير‬ Range: (0, ) ‫قابل للتغير‬ For (x) = 2x: x (x) –2 ¼ –1 ½ 0 1 1 2 2 4 (x) = ax , for a > 1, is increasing 3 8 and continuous on its entire domain, (–, ). Copyright © 2013, 2009, 2005 Pearson Education, Inc. 67 EXPONENTIAL FUNCTION f ( x )  a x Domain: (–, ) Range: (0, ) For (x) = 2x: x (x) –2 ¼ –1 ½ 0 1 1 2 2 4 The x-axis is a horizontal 3 8 asymptote as x – . Copyright © 2013, 2009, 2005 Pearson Education, Inc. 68 EXPONENTIAL FUNCTION f ( x )  a x Domain: (–, ) Range: (0, ) For (x) = 2x: x (x) –2 ¼ –1 ½ 0 1 1 2 2 4 The graph passes through the points 3 8  1  1,  , (0,1), and (1, a).  a Copyright © 2013, 2009, 2005 Pearson Education, Inc. 69 EXPONENTIAL FUNCTION f ( x )  a x Domain: (–, ) Range: (0, ) For (x) = (½)x: x (x) –3 8 –2 4 –1 2 0 1 1 ½ (x) = ax , for 0 < a < 1, is 2 ¼ decreasing and continuous on its entire domain, (–, ). Copyright © 2013, 2009, 2005 Pearson Education, Inc. 70 EXPONENTIAL FUNCTION f ( x )  a x Domain: (–, ) Range: (0, ) For (x) = (½)x: x (x) –3 8 –2 4 –1 2 0 1 1 ½ The x-axis is a horizontal 2 ¼ asymptote as x . Copyright © 2013, 2009, 2005 Pearson Education, Inc. 71 EXPONENTIAL FUNCTION f ( x )  a x Domain: (–, ) Range: (0, ) For (x) = (½)x: x (x) –3 8 –2 4 –1 2 0 1 1 ½ The graph passes through the points 2 ¼  1  1,  , (0,1), and (1, a).  a Copyright © 2013, 2009, 2005 Pearson Education, Inc. 72  Exponential Function From Section 2.7, the graph of y = f(–x) is the graph of y = f(x) reflected across the y-axis. Thus, we have the following. If (x) = 2x, then (–x) = 2–x = (2–1)x = 2– 1·x = (½)x. This is supported by the graphs shown. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 73  Exponential Function The graph of (x) = 2x is typical of graphs of (x) = ax where a > 1. For larger values of a, the graphs rise more steeply, but the general shape is similar. When 0 < a < 1, the graph decreases in a manner similar to the graph of f(x) = (1/2)x. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 74  Exponential Function The graphs of several typical exponential functions illustrate these facts. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 75 Characteristics of the Graph of (x) = ax  1 1. The points  1,  , (0,1), and (1, a) are on the  a graph. 2. If a > 1, then  is an increasing function. If 0 < a < 1, then  is a decreasing function. 3. The x-axis is a horizontal asymptote. 4. The domain is (–, ), and the range is (0, ). Copyright © 2013, 2009, 2005 Pearson Education, Inc. 76 Characteristics of the Graph of (x) =- ax Copyright © 2013, 2009, 2005 Pearson Education, Inc. 77  Example 2 GRAPHING REFLECTIONS AND TRANSLATIONS Graph each function. Show the graph of y = 2x for comparison. Give the domain and range. (a) f ( x )  2 x Solution The graph of (x) = –2x is that of (x) = 2x reflected across the x-axis. The domain is (–, ), and the range is (–, 0). Copyright © 2013, 2009, 2005 Pearson Education, Inc. 78 Characteristics of the Graph of (x) =ax-h+k, a>1 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 79 Characteristics of the Graph of (x) =-ax-h+k, Copyright © 2013, 2009, 2005 Pearson Education, Inc. 80  Logarithmic Functions 5.3 Logarithms Logarithmic Equations Logarithmic Functions Properties of Logarithms Copyright © 2013, 2009, 2005 Pearson Education, Inc. 81 4.1 - 81  Logarithms The previous section dealt with exponential functions of the form y = ax for all positive values of a, where a ≠ 1. The horizontal line test shows that exponential functions are one-to-one, and thus have inverse functions. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 82  Logarithms The equation defining the inverse of a function is found by interchanging x and y in the equation that defines the function. Starting with y = ax and interchanging x and y yields x  ay. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 83  Logarithms x a y Here y is the exponent to which a must be raised in order to obtain x. We call this exponent a logarithm, symbolized by the abbreviation “log.” The expression logax represents the logarithm in this discussion. The number a is called the base of the logarithm, and x is called the argument of the expression. It is read “logarithm with base a of x,” or “logarithm of x with base a.” Copyright © 2013, 2009, 2005 Pearson Education, Inc. 84 Logarithm For all real numbers y and all positive numbers a and x, where a ≠ 1, y  loga x if and only if x  ay. The expression logax represents the exponent to which the base a must be raised in order to obtain x. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 85  WRITING EQUIVALENT LOGARITHMIC Example 1 AND EXPONENTIAL FORMS The table shows several pairs of equivalent statements, written in both logarithmic and exponential forms. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 86  WRITING EQUIVALENT LOGARITHMIC Example 1 AND EXPONENTIAL FORMS To remember the relationships among a, x, and y in the two equivalent forms y = logax and x = ay, refer to these diagrams. Exponent Logarithmic form: y = loga x Base Exponent Exponential form: ay = x Base Copyright © 2013, 2009, 2005 Pearson Education, Inc. 87  Homework 1 SOLVING LOGARITHMIC EQUATIONS Solve each equation. 8 (a) logx 3 27 8 Solution logx 3 27 8 x  3 Write in exponential 27 form. 3 2 8 2 3 x   3   3 27  3  2 x Take cube roots 3 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 88 Logarithmic Function If a > 0, a ≠ 1, and x > 0, then f ( x )  loga x defines the logarithmic function with base a. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 89  Logarithmic Functions Exponential and logarithmic functions are inverses of each other. The graph of y = 2x is shown in red. The graph of its inverse is found by reflecting the graph of y = 2x across the line y = x. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 90  Logarithmic Functions The graph of the inverse function, defined by y = log2 x, shown in blue, has the y-axis as a vertical asymptote. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 91  Logarithmic Functions Since the domain of an exponential function is the set of all real numbers, the range of a logarithmic function also will be the set of all real numbers. In the same way, both the range of an exponential function and the domain of a logarithmic function are the set of all positive real numbers. Thus, logarithms can be found for positive numbers only. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 92 LOGARITHMIC FUNCTION f ( x )  loga x Domain: (0, ) ‫يتغير‬ Range: (– , ) ‫ثابت ال يتغير‬ For (x) = log2 x: ‫لجميع الدوال اللوغاريتمية‬ x (x) ¼ –2 ½ –1 1 0 2 1 4 2  (x) = loga x, for a > 1, is 8 3 increasing and continuous on its entire domain, (0, ). Copyright © 2013, 2009, 2005 Pearson Education, Inc. 93 LOGARITHMIC FUNCTION f ( x )  loga x Domain: (0, ) Range: (– , ) For (x) = log2 x: x (x) ¼ –2 ½ –1 1 0 2 1 4 2  The y-axis is a vertical asymptote 8 3 as x 0 from the right. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 94 LOGARITHMIC FUNCTION f ( x )  loga x Domain: (0, ) Range: (– , ) For (x) = log2 x: x (x) ¼ –2 ½ –1 1 0 2 1 4 2 8 3  The graph passes through the points 1   , 1 , 1,0  , and  a,1. a  Copyright © 2013, 2009, 2005 Pearson Education, Inc. 95 LOGARITHMIC FUNCTION f ( x )  loga x Domain: (0, ) Range: (– , ) For (x) = log1/2 x: x (x) ¼ 2 ½ 1 1 0 2 –1 4 –2  (x) = loga x, for 0 < a < 1, is 8 –3 decreasing and continuous on its entire domain, (0, ). Copyright © 2013, 2009, 2005 Pearson Education, Inc. 96 LOGARITHMIC FUNCTION f ( x )  loga x Domain: (0, ) Range: (– , ) For (x) = log1/2 x: x (x) ¼ 2 ½ 1 1 0 2 –1 4 –2  The y-axis is a vertical 8 –3 asymptote as x 0 from the right. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 97 LOGARITHMIC FUNCTION f ( x )  loga x Domain: (0, ) Range: (– , ) For (x) = log1/2 x: x (x) ¼ 2 ½ 1 1 0 2 –1 4 –2  The graph passes through 8 –3 the points  1 , 1 , 1,0  , and  a,1.   a  Copyright © 2013, 2009, 2005 Pearson Education, Inc. 98 Characteristics of the Graph of f ( x )  loga x 1  1. The points  , 1 , 1,0  , and  a,1 are on the a  graph. 2. If a > 1, then  is an increasing function. If 0 < a < 1, then  is a decreasing function. 3. The y-axis is a vertical asymptote. 4. The domain is (0,), and the range is (–, ). Copyright © 2013, 2009, 2005 Pearson Education, Inc. 99 Characteristics of the Graph ‫الحالة العامة‬ of f ( x )  loga ( x  h)  K 1. The 1 points   h, 1  k  , 1  h ,0  k , and  a  h,1  a  are on the graph. 2. If a > 1, then  is an increasing function. If 0 < a < 1, then  is a decreasing function. 3. The x=h is a vertical asymptote. 4. The domain is (h,), and the range is (–, ). Copyright © 2013, 2009, 2005 Pearson Education, Inc. 100  Example 2 GRAPHING LOGARITHMIC FUNCTIONS Graph each function. (a) f ( x )  log1 2 x Solution First graph y = (½)x which defines the inverse function of , by plotting points. The graph of (x) = log1/2x is the reflection of the graph of y = (½)x across the line y = x. The ordered pairs for y = log1/2x are found by interchanging the x- and y-values in the ordered pairs for y = (½)x. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 101  Example 2 GRAPHING LOGARITHMIC FUNCTIONS Graph each function. (a) f ( x )  log1 2 x Solution Copyright © 2013, 2009, 2005 Pearson Education, Inc. 102  Properties of Logarithms The properties of logarithms enable us to change the form of logarithmic statements so that products can be converted to sums, quotients can be converted to differences, and powers can be converted to products. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 103 Properties of Logarithms For x > 0, y > 0, a > 0, a ≠ 1, and any real number r, the following properties hold. Property Description The logarithm of the Product Property product of two numbers is equal to the sum of loga xy  loga x  loga y the logarithms of the numbers. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 104 Properties of Logarithms For x > 0, y > 0, a > 0, a ≠ 1, and any real number r, the following properties hold. Property Description The logarithm of the Quotient Property quotient of two x numbers is equal to the loga  loga x  loga y difference between the y logarithms of the numbers. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 105 Properties of Logarithms For x > 0, y > 0, a > 0, a ≠ 1, and any real number r, the following properties hold. Property Description The logarithm of a Power Property number raised to a power loga x  r loga x r is equal to the exponent multiplied by the logarithm of the number. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 106 Properties of Logarithms For x > 0, y > 0, a > 0, a ≠ 1, and any real number r, the following properties hold. Property Logarithm of 1 loga1  0 Base a Logarithm of a logaa  1 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 107  Example 3 USING THE PROPERTIES OF LOGARITHMS Rewrite each expression. Assume all variables represent positive real numbers, with a ≠ 1 and b ≠ 1. (a) log6 (7  9) Solution log6 (7  9)  log6 7  log6 9 Product property Copyright © 2013, 2009, 2005 Pearson Education, Inc. 108  Example 3 USING THE PROPERTIES OF LOGARITHMS Rewrite each expression. Assume all variables represent positive real numbers, with a ≠ 1 and b ≠ 1. 15 (b) log9 7 Solution 15 log9  log915  log9 7 Quotient property 7 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 109  Example 3 USING THE PROPERTIES OF LOGARITHMS Rewrite each expression. Assume all variables represent positive real numbers, with a ≠ 1 and b ≠ 1. (c) log5 8 Solution 1 log5 8  log5 (8 )  log5 8 12 Power property 2 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 110  Example 3 USING THE PROPERTIES OF LOGARITHMS Rewrite each expression. Assume all variables represent positive real numbers, with a ≠ 1 and b ≠ 1. mnq (d) loga 2 4 pt Use parentheses to avoid errors. Solution mnq loga 2 4  loga m  loga n  logaq  (loga p  loga t ) 2 4 pt  logam  logan  logaq  (2 loga p  4 loga t )  logam  logan  logaq  2 loga p  4 loga t Be careful with signs. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 111  USING THE PROPERTIES OF Example 4 LOGARITHMS WITH NUMERICAL VALUES Assume that log10 2 = 0.3010. Find each logarithm. (a) log10 4 Solution log10 4  log10 22  2 log10 2  2(0.3010)  0.6020 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 112  Exponential and Logarithmic 5.4 Equations Exponential Equations Logarithmic Equations Copyright © 2013, 2009, 2005 Pearson Education, Inc. 113 4.1 - 113 Property of Logarithms If x > 0, y > 0, a > 0, and a ≠ 1, then the following holds. x = y is equivalent to loga x = loga y. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 114 Rules : ln( x )  loge ( x )(natural log) e  2.718281828459045 log( x )  log10 ( x ) ln x r  r ln x ln e x  x ,e ln x x y  loga x  x  a y Copyright © 2013, 2009, 2005 Pearson Education, Inc. 115  Example 1 SOLVING AN EXPONENTIAL EQUATION Solve 7x = 12. Give the solution to the nearest thousandth. Solution The properties of exponents given in Section 4.2 cannot be used to solve this equation, so we apply the preceding property of logarithms. While any appropriate base b can be used, the best practical base is base 10 or base e. We choose base e (natural) logarithms here. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 116  Example 1 SOLVING AN EXPONENTIAL EQUATION Solve 7x = 12. Give the solution to the nearest thousandth. Solution x 7  12 In 7  In 12 x Property of logarithms xIn 7  In 12 Power property In 12 x Divide by In 7. In 7 x  1.277 Use a calculator. The solution set is {1.277}. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 117  Example 2 SOLVING BASE e EXPONENTIAL EQUATIONS Solve each equation. Give solutions to the nearest thousandth. x2 (a) e  200 Solution x2 e  200 x2 Take the natural In e  In 200 logarithm on each side. x2 x 2  In 200 In e = x2 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 118  Example 2 SOLVING BASE e EXPONENTIAL EQUATIONS Solve each equation. Give solutions to the nearest thousandth. x2 (a) e  200 Remember both roots. Solution x   In 200 Square root property x  2.302 Use a calculator. The solution set is { 2.302}. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 119  Example 3 SOLVING LOGARITHMIC EQUATIONS Solve each equation. Give exact values. (a) 7ln x = 28 Solution 7ln x  28 ln x  4 Divide by 7. Write the natural logarithm in xe 4 exponential form. The solution set is {e4}. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 120  Example 3 SOLVING LOGARITHMIC EQUATIONS Solve each equation. Give exact values. (b) log2(x3 – 19) = 3 Solution log2 ( x  19)  3 3 x 3  19  23 Write in exponential form. x 3  19  8 Apply the exponent. x 3  27 Add 19. x  3 27 Take cube roots. 3 27  3 The solution set is {3}. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 121 Solving Exponential or Logarithmic Equations To solve an exponential or logarithmic equation, change the given equation into one of the following forms, where a and b are real numbers, a > 0 and a ≠ 1, and follow the guidelines. 1. a(x) = b Solve by taking logarithms on both sides. 2. loga (x) = b Solve by changing to exponential form ab = (x). 3. loga (x) = loga g(x) The given equation is equivalent to the equation (x) = g(x). Solve algebraically. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 122 Solving Exponential or Logarithmic Equations 4. In a more complicated equation, such as 2 x 1 4 x e e  3e in Example 3(b), it may be necessary to first solve for a(x) or loga (x) and then solve the resulting equation using one of the methods given above. 5. Check that each proposed solution is in the domain. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 123