Section 0.4 Inverse functions and logarithms (5/3/07) Overview: Some applications require not onl a function that converts a numer into a numer, ut also its inverse, which converts ack into. In this section we analze inverse functions and discuss logarithms, which are the inverses of eponential functions. Topics: Inverse functions Changing variales Restricting domains Logarithms The common and natural logarithms Laws of logarithms Inverse functions The function of Figure converts the ear t into the numer N = f(t) (millions) of cars that were registered in California at that time. () For eample, starting with t = 980 on the horizontal t-ais, moving verticall to the graph, and then horizontall to the vertical N-ais, we otain N = f(980) = 7.4. At the eginning of 980 there were 7.4 million cars registered in California. 30 N (millions) N = f(t) 000 t (ear) t = f (N) 0 7.4 980 0 960 940 940 000 980 960 t 0 7.4 30 N (millions) FIGURE FIGURE The inverse of f, denoted f and read f inverse, has the opposite effect. It converts the numer N into the time t = f (N) when there were N cars registered. We could use Figure to stud f having its variale (the independent variale) e on the vertical N-ais and its values on the horizontal t-ais. For eample, we could start with N = 7.4 on the vertical ais, move horizontall to the graph, and then verticall to the horizontal t ais to otain t = 980. This gives f (7.4) = 980, which means that numer of cars registered was 7.4 million at the eginning of 980. it is used. () Data adapted from CALPERG Citizen Agenda, Vol. 8, No., Los Angeles: CALPERG, 000, p. 5. Because the smol f also denotes the reciprocal /f of f, its meaning must e determined from the contet in which
p. (5/3/07) Section 0.4, Inverse functions and logarithms This procedure for studing the inverse is, however, not convenient ecause we prefer to have the independent variale on the horizontal ais. To achieve this, we flip the drawing aout the diagonal line that makes a 45 -degree angle with the positive - and -aes. This ields the graph t = f (N) of the inverse function in Figure, where the variale N is on the horizontal ais and the values t of the function are on the vertical ais. The graph of f in Figure is the mirror image of the graph of f in Figure with respect to the diagonal line. Here is a general definition: Definition Suppose that = f() is a function such that for each in its range, the equation = f() has one and onl one solution. Then is the value of the inverse function = f () at. Thus, = f () = f(). () Question Suppose that f has an inverse f, that f() = 0, and that f (30) =. What are the values of f (0) and f()? The domain of f is the range of f and the range of f is the domain of f. Figures 3 and 4 illustrate how the domains and ranges of a function and its inverse are related. The domain of = f() in Figure 3 is the interval a on the horizontal -ais and its range is the interval c d on the vertical -ais. The domain of = f () in Figure 4 is the interval c d on the horizontal -ais and its range is the interval a on the vertical -ais. = f() = f () d c a a c d FIGURE 3 FIGURE 4 The function f and its inverse f undo each other in the sense that f (f()) = for in the domain of f () f ( f () ) = for in the domain of f. (3) Eample Figure 5 shows the graph = h() of the function h() = 3. Find a formula for the inverse function = h () and draw its graph in a -plane. To find a formula for = h (), we solve = 3 for. We otain first 3 = and then = 6 3. Thus, the inverse function = h () is given h () = 6 3. Its graph in the -plane of Figure 6 is the line of slope 3 with -intercept 6 and is the mirror image aout the diagonal line of the graph of = h() in Figure 5. (The diagonal line that makes equal angles with the positive aes in Figures 5 and 6 is = ecause the scales on the aes are equal.) The smol = means implies, = means is implied, and means if and onl if.
Section 0.4, Inverse functions and logarithms p. 3 (5/3/07) = = 6 6 4 = h() 4 = h () 4 6 4 6 h() = 3 h () = 6 3 FIGURE 5 FIGURE 6 Changing variales In Definition and Eample, the variale of the inverse function = f () was the letter used for values of the original function = f().in some cases it is convenient to have e the variale of oth the function and its inverse. A formula for the inverse function can then often e found the following rule. Rule To find a formula with variale for the inverse of = f(), first solve the equation = f() to otain the inverse = f () with variale. Then interchange the letters and to otain a formula for = f (). The procedure descried in Rule is illustrated in Figures 7 through 9. We egin with the function = f() in the -plane of Figure 7. We solve for to otain the inverse function = f () with variale in the -plane of Figure 8. Then we interchange the letters and to otain the function = f () with variale in the -plane of Figure 9. = f() = f () = f () d c a a a c d c d FIGURE 7 FIGURE 8 FIGURE 9 Question How can Figure 9 e otained directl from Figure 7? Notice that interchanging and to otain = f () from = f () is not the same as solving the latter equation for. That would take us ack to the original equation = f(). When we interchange and, we are not solving an equation, we are just interchanging the letters used for the variale and for values of the function.
p. 4 (5/3/07) Section 0.4, Inverse functions and logarithms Eample Figure 0 shows the graph of the function = 3 with the diagonal line that makes equal angles with the positive aes. Find a formula for the inverse of = 3 with as variale and draw its graph. = 3 = 3 = 3 FIGURE 0 FIGURE FIGURE We solve = 3 taking cue roots of oth sides to find the inverse = 3 with as variale. The graph of this function in the -plane of Figure is otained flipping all of Figure 0 aout the dashed line in that drawing. We interchange the letters and to otain = 3. The graph of this function in the -plane of Figure can e otained interchanging and in Figure or reflecting the curve in Figure 0 aout the diagonal line without moving the aes. The net two eamples illustrate the fact that a function and its inverse undo each other if one is applied immediatel after the other. Eample 3 Epress the solutions of 3h( ) = in terms of the function h. We want to use the fact that h (h()) =, so we first divide oth sides of 3h( ) = 3 to otain h( ) = 4. Then, when we appl the function = h () to oth sides, we otain h ( h( ) ) = h (4), which simplifies to = h (4). Finall, taking square roots of oth sides gives the solutions = ± h (4). Eample 4 Epress the solution of S ( + ) = 5 in terms of the function S. It would do us no good to appl the function S to the given equation ecause of the square root. Instead, we first square oth sides to otain S ( + ) = 5. Appling S to oth sides of this equation then gives S ( S ( + ) ) = S(5), which simplifies to + = S(5). We sutract from oth sides to otain = S(5). Restricting domains If a horizontal line = c intersects the graph of a function = f() at more than one point, then the function does not have an inverse, ecause the inverse function cannot have more than one value at = c. The squaring function = in Figure 3, for eample, does not have an inverse ecause the horizontal line = c for an positive c intersects the graph at two points.
Section 0.4, Inverse functions and logarithms p. 5 (5/3/07) = = F() = c F() = for 0 FIGURE 3 FIGURE 4 To otain an inverse, we define F to e the function with values = ut defined onl for 0 (Figure 4). Then the horizontal line = c for an c 0 intersects the graph at onl one point. The inverse of F with variale is the square-root function = in Figure 5 and its inverse with variale is = in Figure 6. = = FIGURE 5 FIGURE 6 Eample 5 (a) Eplain wh the function = 3 4 of Figure 7 does not have an inverse. () Draw the graph of the function = G() defined G() = 3 4 for 0. (c) Find a formula for the inverse = G () and draw its graph. = 3 4 3 = G() = G () 3 G() = 3 4, 0 G () = 4 3 FIGURE 7 FIGURE 8 FIGURE 9
p. 6 (5/3/07) Section 0.4, Inverse functions and logarithms Question 3 (a) The function = 3 4 of Figure 7 does not have an inverse ecause the horizontal line at an < 3 intersects the graph at two points. () The graph of = G() in Figure 8 is the portion of = 3 4 for 0. (c) The equation = G() is equivalent to = 3 4 with 0. To solve for, we sutract 3 from oth sides of the equation to have 3 = 4 with 0. Multipling oth sides and taking fourth roots gives = ± 4 3 and then = 4 3 since 0. Therefore, G () = 4 3, and interchanging and ields G () = 4 3. The graph of = G () in Figure 9 can e otained reflecting the graph of G in Figure 8 aout the diagonal line at 45 to the positive -and -aes. Figure 0 shows the graph of = H() defined H() = 3 4 for 0. Find a formula for = H () and draw its graph. = H() 3 FIGURE 0 H() = 3 4, 0 Logarithms For an constant >, the eponential function = of Figure has an inverse, which is called the logarithm to the ase and is denoted = log (Figure ). Thus, = if and onl if = log. (4) Here is the variale of the eponential function and and is the variale of the logarithm. = = = log = log FIGURE FIGURE FIGURE 3 Interchanging and in Figure gives the graph = log in Figure 3 of the logarithm with variale. It is the mirror image of = aout the diagonal line in Figure. Because = is defined for all and its range is the interval > 0, = log is defined for > 0 and its range is the set of all numers. The fact that = and = log are inverse functions is also epressed in the two equations, log ( ) = for all (5) log = for all > 0. (6)
Section 0.4, Inverse functions and logarithms p. 7 (5/3/07) Notice that, while these formulas are visuall quite comple, all the state is that = and = log undo each other. The asic properties of log follow from properties of discussed in the last section. For an >, an n, and an positive and, log () = 0 (7) log () = (8) log () = log + log (9) ) log ( = log log (0) log ( n ) = nlog. () Also, logarithms with different ases, > and c >, are related the formula log = [log c][log c ] = log c log c. () To rememer formulas (), notice that the is higher the in all three epressions and that one c is higher than the other in each of the last two epresssions. Eample 6 Solve = 5 for. = 5 for = log (5). Question 4 Solve log 0 = for. The common logarithm The logarithm to the ase 0, = log 0, is called the common logarithm and is often denoted = log with no suscript. It was emploed to simplif calculations involving products and powers efore there were calculators and computers, and is still used in some applications. Eample 7 In chemistr the ph of a solution is defined ph = log 0 [H + ], where [H + ] is the concentration of hdrogen ions in the solution, measured in moles per liter. ( mole = 6 0 3 molecules.) The hdrogen ion concentration of water is 0 7 moles per liter. What is the ph of water? The ph of water is log 0 (0 7 ) = ( 7) = 7. The natural logarithm The logarithm that is most convenient for calculus is = log e, with ase equal to the ase e of the natural eponential function = e that was introduced in the last section. This logarithm is called the natural logarithm and is usuall denoted = ln (read ell n ): Since = e and = ln are inverse functions, ln = log e for all > 0. (3) e ln = for all > 0 (4) ln(e ) = for all. (5) The graphs = e and = ln in Figure 4 are mirror images of each other aout the line = ecause there are equal scales on the aes in that drawing. Eample 8 Solve the equation e 3 = 00 for. C Question 5 We take the natural logarithm of oth sides of the equation to otain ln(e 3 ) = ln(00). Since the natural logarithm and eponential function are inverses, this gives 3 = ln(00). Finall, we divide 3 to otain the solution = 3 ln(00). Check the result of Eample 8 using a calculator or computer to find the approimate decimal value of the solution and then to evaluate e 3 with this numer.
p. 8 (5/3/07) Section 0.4, Inverse functions and logarithms = e = FIGURE 4 = ln Eample 9 Find all solutions of (ln ) =. We cannot use the fact that = ln and = e are inverses appling the eponential function to oth sides of the given equation ecause the logarithm is squared. Instead, we first take the square roots of oth sides to have ln = ±. Then appling the eponential function gives the solutions = e = e and = e = /e. C Question 6 Generate the cuve = (ln ) with the line = in the window 5, 0.5.5 to illustrate the results of Eample 9. The natural logarithm can e used to find values of logarithms to an ase > using () with c = e: log = ln for all > 0. (6) ln To rememer this formula, notice that the is higher than the on oth sides of the equation. C Eample 0 Give the approimate decimal value of log (3). Formula (6) with = and = 3 gives log (3) = ln(3) ln(). =.5849650. Responses 0.4 Response f (0) = ecause f() = 0 f() = 30 ecause f (30) = Response To otain the graph of = f () in Figure 9, reflect the graph of = f() in Figure 7 aout the diagonal line =, without moving the - and -aes. Response 3 Solve = 3 4 for with 0. 4 = 3 = 4 3 H () = 4 3 H () = 4 3 The graph of = H () in Figure R3 is the mirror image aout the dashed line of the graph of = H() in Figure 0. = H ().5 = (ln ) = 3 0.5 e e 4 Figure R3 Figure R6
Section 0.4, Inverse functions and logarithms p. 9 (5/3/07) Response 4 log 0 = for = 0 = 0 Response 5 = 3 ln(00). =.53505679 e 3(.53505679). = 00.000000 Response 6 Figure R6 shows the curve = (ln ), the line =, and their intersections, which are at = e. =.7 and = e. = 0.37. Interactive Eamples 0.4 Interactive solutions are on the we page http//www.math.ucsd.edu/ ashenk/.. Find approimate values of (a) G(4), () G (0, 000), and (c) G (G(3) + 3000)), where G is the function whose graph is in Figure 5. 5000 = G() 0000 5000 FIGURE 5 3 4. Solve e 4 = 000 for. Then generate = e 4 and = 000 in the window.5.5, 50 500 and use an intersect or trace operation to check our answer. 3. Solve + = + for. Then generate = + and = + in the window, 7, and use an intersect or trace operation to check our answer. 4. Solve ln( 5 ) = for. 5. Solve 4 = 5(8 ) for. 6. Solve [ln(5)] = 9 for 7. Solve F (F(F ())) = under the assumption that = F() and its inverse = F () are defined for all. Eercises 0.4 A Answer provided. CONCEPTS: O Outline of solution provided. C Graphing calculator or computer required.. Simplif the epresions (a) f ( f (f()) ) and () f ( f ( f () )), where is a numer in the domains of f and f.. (a) What are the values of log 0 (00), log 0 (000), and log 0 (00,000)? () How are the numers from part (a) related and what propert of logarithms is illustrated this relationship? 3. (a) What are the values of ln(e 3 ), ln(e 5 ), and ln(e 3 e 5 )? () How are the numers from part (a) related and what propert of logarithms is illustrated this relationship? 4. (a) What are the values of 3 ln(e 5 ), and ln ( (e 5 ) 3)? () How are the numers from part (a) related and what propert of logarithms is illustrated this relationship? In the pulished tet the interactive solutions of these eamples will e on an accompaning CD disk which can e run an computer rowser without using an internet connection.
p. 0 (5/3/07) Section 0.4, Inverse functions and logarithms BASICS: 5. Epress the solutions of (a) f() = 4 and () f () = 7 in terms of values of f and f. 6. O What are (a) P(), () P (0.), and (c) P ( P (0.) + ) if P is given the following tale? 0 3 4 5 P() 4. 3.9. 0..4 3 7. O (a) Find a formula for = f () where f() = /3. C () Generate the graphs of = f() and = f () with the line = in an -plane and cop them on our paper. Use a window with equal scales on the aes that includes the square,. 8. O Solve (a) log 0 ( ) = 4, () 0 = 5, (c) ln =, and (d) e = 7 for. Give eact answers and approimate decimal values. C Check our answers with a calculator or computer. 9. O Use properties of the common logarithm to show that each of the numers (a) (c) in the first column elow equals one of the numers (I) (III) in the second column. (a) log 0 (300) (I) log 0 (3) () log 0 (30) log 0 (3) (II) + log 0 (3) (c) log0 (8) (III) 0. O Figure 6 shows the graph of a function = H(), defined for 0 6. Draw the graph of its inverse = H (). 6 4 = H() FIGURE 6 4 6. The graph of = k() is in Figure 7. Draw the graph of its inverse = k (). = k() 4 FIGURE 7 4
Section 0.4, Inverse functions and logarithms p. (5/3/07). O The function r = G(v) of Figure 8 gives the rate of gasoline consumption of a car, measured in miles per gallon, as a function of the car s velocit v, measured in miles per hour. (a) What is the action of the inverse function v = G (r)? () What is approimate value of G (5)? 0 r (miles per gallon) r = G(v) 5 0 FIGURE 8 30 50 70 v (miles per hour) 3. O Figure 9 shows the graph of the function P = F(Q) that converts the numer of quarts Q in a volume into the numer of points P. (a) What is the action of Q = F (P)? () Give formulas for P = F(Q) and Q = F (P). (c) Draw the graph of Q = F (P) in a PQ-plane P (pints) P = F(Q) 6 4 FIGURE 9 4 6 Q (quarts) 4. A An airplane that is fling west at the constant speed of 500 miles per hour flies s = f(t) miles in t (hours) for t 0, where f(t) = 500t. (a) What is the action of t = f (s)? () Find a formula for t = f (s). Then draw the graph s = f(t) in a ts-plane and the graph of t = f (s) in an st-plane. 5. A grain silo contains 00 tons of wheat at the eginning of an eight-hour work da. During the da wheat is removed at the constant rate of 0 tons per hour, so that the weight of wheat in the silo t hours after the eginning of the work da is w = F(t) tons, where F(t) = 00 0t of (a) What does F (00) represent in terms of weight and time? () Give a formula for t = F (w). 6. A The volume of a sphere of radius r is V = G(r) with G(r) = 4 3 πr3 for r 0. Give a formula for r = G (V ). What does this function do? 7. If V gallons of a salt solution contains 5 pounds of salt, then the concentration of the solution is = f(v ) pounds per gallon, where f(v ) = 5/V. (a) What does the inverse function V = f () do? () Find a formula for V = f (). (c) Draw the graph of = f(v ) in a V -plane and the graph of V = f () in a V -plane. Use V 8 and 8 and have equal scales on the aes in oth drawings. 8. O (a) Find a formula for the inverse of = 3 + as a function of. C () Generate the graphs of oth functions with = in a window that includes 3 3, 3 3 and has equal scales on the aes. Cop the drawing on our paper. If our calculator or computer does not generate the graph of 3 + for <, enter it as the two functions (+) (/3) for and (as( + )) (/3) for <. Use the setting ZSquare under Zoom to get equal scales on a TI calculator.
p. (5/3/07) Section 0.4, Inverse functions and logarithms 9. O (a) Find a formula for the inverse of = / as a function of. () Draw the graphs of oth functions in separate -planes with 4 6, 4 6 and equal scales on the aes. 0. (a) Find a formula for the inverse of = 3 as a function of. C () Generate the graphs of oth functions with = in a window that has equal scales on the aes and that includes the square.5.5,.5.5. Cop the drawing on our paper. In Eercises through 3, (a) solve the required equations for. C () Then generate the curves in the given windows and use an intersect or trace operation or the curves as a partial check of our answers.. Solve ln(00 + ) = 6. (Generate = ln(00 + ) and = 6 in the window 50 750, 8 with -scale = 00.). Solve = 0. (Generate = and = 0 in the window.5.5, 0.5 5.) 3. Solve 50 (0 ) = 50. (Generate = (0 ) and = 00 in the window, 50 00 with -scale = 50.) Solve the equations in Eercises 4 through 30 for. 4. A 3 = 7 8. O e = 7 5. 4 = 5 9. ln( ) + ln( ) = 5 6. = 9 30. ln(e ) = 3 7. A 4 8 = 0 The functions in Eercises 3 through 33 and their inverses are defined for all. Solve the equations for. 3. A 3A() = 5 3. 3f() = 4 f() EXPLORATION: 33. P(P ()) = 4 34. A The 00 California state ta T (dollars) for a single person on taale income of I dollars with 0 I 3,65 is given T = f(i) with () f(i) = { 0.0I for 0 I 5748 57.48 + 0.0(I 5748) for 5459 I,939. (a) Sketch the graph T = f(i) in an IT-plane. () Descrie in terms of income and taes the action of the function f. (c) Find a formula similar to (7) for I = f (T) and sketch its graph in a TI-plane. C 00 35 A o that weighs 00 pounds on the surface of the earth weighs w = ( + 4000 pounds when h) it is h miles aove the surface of the earth. (a) Generate the graph of this function on our calculator or computer, using 000 h 0, 000, h-scale = 000, 0 w 50, and w-scale = 50, and cop it on our paper. () Give a formula for the height of the o aove the surface of the earth as a function of its weight. Generate the graph of this function in a wh-plane with the same ranges as in part (a), and cop it on our paper. 36. Find a formula for the inverse = h () of = h() defined h() = / for > 0. Then draw the graphs of oth functions with equal scales on the aes. () Data from 00 Personal Income Ta Booklet, Sacramento, CA: Franchise Ta Board, 00, p. 66.
Section 0.4, Inverse functions and logarithms p. 3 (5/3/07) In Eercises 37 through 39 (a) solve the equations for. C () Generate the graphs of the functions on oth sides of the equations in sutale windows to check our answers. 37. A ln( + ) = 38. A = 39. = 6( ) 40. The magnitude of an earthquake on the Richter scale is M = 0.67 log 0 (0.37E) +.46, where E is the energ of the earthquake in kilowatt hours. What is the energ (a A ) of an earthquake of magnitude 5 and () of an earthquake of magnitude 6? Give approimate decimal values. 4. The loudness of sound can e measured in deciels, where one deciel if supposed to e the smallest change in loudness that the average human can detect. (One el, named in honor of Aleander Graham Bell equals ten deciels.) Sound reaches the eardrum as an oscillation in the air pressure. If the variation in air presure is P pounds per square inch, then the loudness of the sound is D = 0log 0 [(3.45 0 8 )P] deciels. What is the variation in sound pressure caused (a) a whisper at 0 deciels and () a rock and at 0 deciels? 4. The magnitude of a star is defined the formula M =.5log 0 (ki), where k is a positive constant and I is the intensit of the light from the star. (a) Do righter stars have less or greater magnitudes than dimmer stars? () What is the ratio of the intensities of light from the rightst star Sirius, which has a magnitude of.6 and from the star Betelgeuse, which has a magnitude of 0.9? In Eercises 43 through 46 solve the equations for under the assumption that the functions and their inverses are defined for all. 43. O 4 G() = G() 44. A H()H () = 0 45. [R()] R() + = 0 46. 3S () S () + = 6 (End of Section 0.4)