Modelling Exercises 6.5 Inroducion This Secion provides examples and asks employing exponenial funcions and logarihmic funcions, such as growh and decay models which are imporan hroughou science and engineering. Prerequisies Before saring his Secion you should... Learning Oucomes On compleion you should be able o... be familiar wih he laws of logarihms have knowledge of logarihms o base 1 be able o solve equaions involving logarihms and exponenials develop exponenial growh and decay models 38 HELM (8): Workbook 6: Exponenial and Logarihmic Funcions
1. Exponenial increase Task (a) Look back a Secion 6. o review he definiions of an exponenial funcion and he exponenial funcion. (b) Lis examples in his Workbook of conexs in which exponenial funcions are appropriae. Your soluion Answer (a) (b) An exponenial funcion has he form y = a x where a >. T he exponenial funcion has he form y = e x where e =.7188... I is saed ha exponenial funcions are useful when modelling he shape of a hanging chain or rope under he effec of graviy or for modelling exponenial growh or decay. We will look a a specific example of he exponenial funcion used o model a populaion increase. Task Given ha P = 1e.1 ( 5) where P is he number in he populaion of a ciy in millions a ime in years answer hese quesions. (a) Wha does his model imply abou P when =? (b) (c) (d) (d) Wha is he saed upper limi of validiy of he model? Wha does he model imply abou values of P over ime? Wha does he model predic for P when = 1? Commen on his. Wha does he model predic for P when = 5? Commen on his. HELM (8): Secion 6.5: Modelling Exercises 39
Your soluion (a) (b) (c) (d) (e) Answer (a) A =, P = 1 which represens he iniial populaion of 1 million. (Recall ha e = 1.) (b) (c) (d) (e) The ime inerval during which he model is valid is saed as ( 5) so he model is inended o apply for 5 years. This is exponenial growh so P will increase from 1 million a an acceleraing rae. P (1) = 1e 1 33 million. This is geing very large for a ciy bu migh be aainable in 1 years and jus abou susainable. P (5) = 1e.5 146 million. This is unrealisic for a ciy. Noe ha exponenial populaion growh of he form P = P e k means ha as becomes large and posiive, P becomes very large. Normally such a populaion model would be used o predic values of P for >, where = represens he presen or some fixed ime when he populaion is known. In Figure 6, values of P are shown for <. These correspond o exrapolaion of he model ino he pas. Noe ha as becomes increasingly negaive, P becomes very small bu is never zero or negaive because e k is posiive for all values of. The parameer k is called he insananeous fracional growh rae. 3 5 15 1 5 P P = 1e.1 1 5 5 1 Figure 6: The funcion P = 1e.1 4 HELM (8): Workbook 6: Exponenial and Logarihmic Funcions
For he model P = 1e k we see ha k =.1 is unrealisic, and more realisic values would be k =.1 or k =.. These would be similar bu k=. implies a faser growh for > han k =.1. This is clear in he graphs for k =.1 and k =. in Figure 7. The funcions are ploed up o years o emphasize he increasing difference as increases. P 5 P = 1e. 5 P = 1e.1 5 1 15 Figure 7: Comparison of he funcions P = 1e.1 and P = 1e. The exponenial funcion may be used in models for oher ypes of growh as well as populaion growh. A general form may be wrien y = ae bx a >, b >, c x d where a represens he value of y a x =. The value a is he inercep on he y-axis of a graphical represenaion of he funcion. The value b conrols he rae of growh and c and d represen limis on x. In he general form, a and b represen he parameers of he exponenial funcion which can be seleced o fi any given modelling siuaion where an exponenial funcion is appropriae.. Linearisaion of exponenial funcions This subsecion relaes o he descripion of log-linear plos covered in Secion 6.6. Frequenly in engineering, he quesion arises of how he parameers of an exponenial funcion migh be found from given daa. The mehod follows from he fac ha i is possible o undo he exponenial funcion and obain a linear funcion by means of he logarihmic funcion. Before showing he implicaions of his mehod, i may be necessary o remind you of some rules for manipulaing logarihms and exponenials. These are summarised in Table 1 on he nex page, which exacly maches he general lis provided in Key Poin 8 in Secion 6.3 (page.) HELM (8): Secion 6.5: Modelling Exercises 41
Table 1: Rules for manipulaing base e logarihms and exponenials Number Rule Number Rule 1a ln(xy) = ln(x) + ln(y) 1b e x e y = e x+y a ln(x/y) = ln(x) ln(y) b e x /e y = e x y 3a ln(x y ) = y ln(x) 3b (e x ) y = e xy 4a ln(e x ) = x 4b e ln(x) = x 5a ln(e) = 1 5b e 1 = e 6a ln(1) = 6b e = 1 We will ry undoing he exponenial in he paricular example P = 1e.1 We ake he naural logarihm (ln) of boh sides, which means logarihm o he base e. So ln(p ) = ln(1e.1 ) The resul of using Rule 1a in Table 1 is ln(p ) = ln(1) + ln(e.1 ). The naural logarihmic funcions undoes he exponenial funcion, so by Rule 4a, ln(e.1 ) =.1 and he original equaion for P becomes ln(p ) = ln(1) +.1. Compare his wih he general form of a linear funcion y = ax + b. y = ax + b ln(p ) =.1 + ln(1) If we regard ln(p ) as equivalen o y,.1 as equivalen o he consan a, as equivalen o x, and ln(1) as equivalen o he consan b, hen we can idenify a linear relaionship beween ln(p ) and. A plo of ln(p ) agains should resul in a sraigh line, of slope.1, which crosses he ln(p ) axis a ln(1). (Such a plo is called a log-linear or log-lin plo.) This is no paricularly ineresing here because we know he values 1 and.1 already. Suppose, hough, we wan o ry using he general form of he exponenial funcion P = ae b (c d) o creae a coninuous model for a populaion for which we have some discree daa. The firs hing o do is o ake logarihms of boh sides ln(p ) = ln(ae b ) Rule 1 from Table 1 hen gives ln(p ) = ln(a) + ln(e b ) (c d). (c d). Bu, by Rule 4a, ln(e b ) = b, so his means ha ln(p ) = ln(a) + b (c d). 4 HELM (8): Workbook 6: Exponenial and Logarihmic Funcions
So, given some populaion versus ime daa, for which you believe can be modelled by some version of he exponenial funcion, plo he naural logarihm of populaion agains ime. If he exponenial funcion is appropriae, he resuling daa poins should lie on or near a sraigh line. The slope of he sraigh line will give an esimae for b and he inercep wih he ln(p ) axis will give an esimae for ln(a). You will have carried ou a logarihmic ransformaion of he original daa for P. We say he original variaion has been linearised. A similar procedure will work also if any exponenial funcion raher han he base e exponenial funcion is used. For example, suppose ha we ry o use he funcion P = A B (C D), where A and B are consan parameers o be derived from he given daa. We can ake naural logarihms again o give ln(p ) = ln(a B ) Rule 1a from Table 1 hen gives ln(p ) = ln(a) + ln( B ) Rule 3a hen gives and so ln( B ) = B ln() = B ln() ln(p ) = ln(a) + B ln() (C D). (C D). (C D). Again we have a sraigh line graph wih he same inercep as before, ln A, bu his ime wih slope B ln(). Task The amoun of money M o which 1 grows afer earning ineres of 5% p.a. for N years is worked ou as M = 1.5 N Find a linearised form of his equaion. Your soluion Answer Take naural logarihms of boh sides. ln(m) = ln(1.5 N ). Rule 3b gives ln(m) = N ln(1.5). So a plo of ln(m) agains N would be a sraigh line passing hrough (, ) wih slope ln(1.5). HELM (8): Secion 6.5: Modelling Exercises 43
The linearisaion procedure also works if logarihms oher han naural logarihms are used. We sar again wih P = A B (C D). and will ake logarihms o base 1 insead of naural logarihms. Table presens he laws of logarihms and indices (based on Key Poin 8 page ) inerpreed for log 1. Table : Rules for manipulaing base 1 logarihms and exponenials Number Rule Number Rule 1a log 1 (AB) = log 1 A + log 1 B 1b 1 A 1 B = 1 A+B a log 1 (A/B) = log 1 A log 1 B b 1 A /1 B = 1 A B 3a log 1 (A k ) = k log 1 A 3b (1 A ) k = 1 ka 4a log 1 (1 A ) = A 4b 1 log 1 A = A 5a log 1 1 = 1 5b 1 1 = 1 6a log 1 1 = 6b 1 = 1 Taking logs of P = A B gives: log 1 (P ) = log 1 (A B ) Rule 1a from Table hen gives log 1 (P ) = log 1 (A) + log 1 ( B ) Use of Rule 3a gives he resul log 1 (P ) = log 1 (A) + B log 1 () (C D). (C D). (C D). Task (a) (b) Wrie down he sraigh line funcion corresponding o aking logarihms of he general exponenial funcion P = ae b (c d) by aking logarihms o base 1. Wrie down he slope of his line. Your soluion Answer (a) log 1 (P ) = log 1 (a) + (b log 1 (e)) (c d) (b) b log 1 (e) I is no usually necessary o declare he subscrip 1 when indicaing logarihms o base 1. If you mee he erm log i will probably imply o he base 1. In he remainder of his Secion, he subscrip 1 is dropped where log 1 is implied. 44 HELM (8): Workbook 6: Exponenial and Logarihmic Funcions
3. Exponenial decrease Consider he value, D, of a car subjec o depreciaion, in erms of he age A years of he car. The car was bough for 15. The funcion D = 15e.5A ( A 6) could be considered appropriae on he ground ha (a) D had a fixed value of 15 when A =, (b) D decreases as A increases and (c) D decreases faser when A is small han when A is large. A plo of his funcion is shown in Figure 8. 1 1 D pounds 8 6 4 1 3 4 5 6 A years Figure 8: Plo of car depreciaion over 6 years Task Produce he linearised model of D = 15e.5A. Your soluion Answer ln D = ln 15 + ln(e.5a ) so ln D = ln 15.5A HELM (8): Secion 6.5: Modelling Exercises 45
Engineering Example Exponenial decay of sound inensiy Inroducion The rae a which a quaniy decays is imporan in many branches of engineering and science. A paricular example of his is exponenial decay. Ideally he sound level in a room where here are subsanial conribuions from reflecions a he walls, floor and ceiling will decay exponenially once he source of sound is sopped. The decay in he sound inensiy is due o absorbion of sound a he room surfaces and air absorpion alhough he laer is significan only when he room is very large. The conribuions from reflecion are known as reverberaion. A measuremen of reverberaion in a room of known volume and surface area can be used o indicae he amoun of absorpion. Problem in words As par of an emergency es of he acousics of a concer hall during an orchesral rehearsal, consulans asked he principal rombone o play a single noe a maximum volume. Once he sound had reached is maximum inensiy he player sopped and he sound inensiy was measured for he nex. seconds a regular inervals of. seconds. The iniial maximum inensiy a ime was 1. The readings were as follows: ime..4.6.8.1.1.14.16.18. inensiy 1.63.35..13.8.5.3..1.5 Draw a graph of inensiy agains ime and, assuming ha he relaionship is exponenial, find a funcion which expresses he relaionship beween inensiy and ime. Mahemaical saemen of problem If he relaionship is exponenial hen i will be a funcion of he form I = I 1 k and a log-linear graph of he values should lie on a sraigh line. Therefore we can plo he values and find he gradien and he inercep of he resuling sraigh-line graph in order o find he values for I and k. k is he gradien of he log-linear graph i.e. k = change in log 1 (inensiy) change in ime and I is found from where he graph crosses he verical axis log 1 (I ) = c Mahemaical analysis Figure 9(a) shows he graph of inensiy agains ime. 46 HELM (8): Workbook 6: Exponenial and Logarihmic Funcions
We calculae he log 1 (inensiy) o creae he able below: ime..4.6.8.1.1.14.16.18. log 1 (inensiy) -. -.46 -.66 -.89-1.1-1.3-1.5-1.7 -. -. Figure 9(b) shows he graph of log (inensiy) agains ime. Inensiy Log(Inensiy) (, ) 1 (.,.) (a).1. Time (b).1. Time Figure 9: (a) Graph of sound inensiy agains ime (b) Graph of log 1 (inensiy) agains ime and a line fied by eye o he daa. The line goes hrough he poins (, ) and (.,.). We can see ha he second graph is approximaely a sraigh line and herefore we can assume ha he relaionship beween he inensiy and ime is exponenial and can be expressed as I = I 1 k. The log 1 of his gives log 1 (I) = log 1 (I ) + k. From he graph (b) we can measure he gradien, k using k = change in log 1 (inensiy) change in ime giving k =.. = 11 The poin a which i crosses he verical axis gives log 1 (I ) = I = 1 = 1 Therefore he expression I = I 1 k becomes I = 1 11 Inerpreaion The daa recorded for he sound inensiy fi exponenial decaying wih ime. log-linear plo o obain he approximae funcion: I = 1 11 We have used a HELM (8): Secion 6.5: Modelling Exercises 47
4. Growh and decay o a limi Consider a funcion inended o represen he speed of a parachuis afer he opening of he parachue where v m s 1 is he insananous speed a ime s. An appropriae funcion is v = 1 8e 1.5 ( ), We will look a some of he properies and modelling implicaions of his funcion. Consider firs he value of v when = : v = 1 8e = 1 8 = 4 This means he funcion predics ha he parachuis is moving a 4 m s 1 when he parachue opens. Consider nex he value of v when is arbirarily large. For such a value of, 8e 1.5 would be arbirarily small, so v would be very close o he value 1. The modelling inerpreaion of his is ha evenually he speed becomes very close o a consan value, 1 m s 1 which will be mainained unil he parachuis lands. The seady speed which is approached by he parachuis (or anyhing else falling agains air resisance) is called he erminal velociy. The parachue, of course, is designed o ensure ha he erminal velociy is sufficienly low (1 m s 1 in he specific case we have looked a here) o give a reasonably genle landing and avoid injury. Now consider wha happens as increases from near zero. When is near zero, he speed will be near 4 m s 1. The amoun being subraced from 1, hrough he erm 8e 1.5, is close o 8 because e = 1. As increases he value of 8e 1.5 decreases fairly rapidly a firs and hen more gradually unil v is very nearly 1. This is skeched in Figure 1. In fac v is never equal o 1 bu ges impercepibly close as anyone would like as increases. The value shown as a horizonal broken line in Figure 1 is called an asympoic limi for v. 15 v (m s 1 ) 1 5 1 3 4 5 (s) Figure 1: Graph of a parachuis s speed agains ime The model concerned he approach of a parachuis s velociy o erminal velociy bu he kind of behaviour porrayed by he resuling funcion is useful generally in modelling any growh o a limi. A general form of his ype of growh-o-a-limi funcion is y = a be kx (C x D) where a, b and k are posiive consans (parameers) and C and D represen values of he independen variable beween which he funcion is valid. We will now check on he properies of his general funcion. When x =, y = a be = a b. As x increases he exponenial facor e kx ges smaller, so y will increase from he value a b bu a an ever-decreasing rae. As be kx becomes very small, 48 HELM (8): Workbook 6: Exponenial and Logarihmic Funcions
y, approaches he value a. This value represens he limi, owards which y grows. If a funcion of his general form was being used o creae a model of populaion growh o a limi, hen a would represen he limiing populaion, and a b would represen he saring populaion. There are hree parameers, a, b, and k in he general form. Knowledge of he iniial and limiing populaion only gives wo pieces of informaion. A value for he populaion a some non-zero ime is needed also o evaluae he hird parameer k. As an example we will obain a funcion o describe a food-limied bacerial culure ha has 3 cells when firs couned, has 6 cells afer 3 minues bu seems o have approached a limi of 4 cells afer 18 hours. We sar by assuming he general form of growh-o-a-limi funcion for he baceria populaion, wih ime measured in hours P = a be k ( 18). When = (he sar of couning), P = 3. Since he general form gives P = a b when =, his means ha a b = 3. The limi of P as ges large, according o he general form P = a b k, is a, so a = 4. From his and he value of a b, we deduce ha b = 37. Finally, we use he informaion ha P = 6 when (measuring ime in hours) =.5. Subsiuion in he general form gives 6 = 4 37e.5k 34 = 37e.5k 34 37 = e.5k Taking naural logs of boh sides: ln ( 34 37 ) =.5k so k = ln( 34 37 ) =.1691 Noe, as a check, ha k urns ou o be posiive as required for a growh-o-a-limi behaviour. Finally he required funcion may be wrien P = 4 37e.1691 ( 18). As a check we should subsiue = 18 in his equaion. The resul is P = 384 which is close o he required value of 4. HELM (8): Secion 6.5: Modelling Exercises 49
Task Find a funcion ha could be used o model he growh of a populaion ha has a value of 3 when couns sar, reaches a value of 6 afer 1 year bu approaches a limi of 1 afer a period of 1 years. (a) Firs find he modelling equaion: Your soluion Answer Sar wih P = a be k ( 1). where P is he number of members of he populaion a ime years. The given daa requires ha a is 1 and ha a b = 3, so b = 9. The corresponding curve mus pass hrough ( = 1, P = 6) so 6 = 1 9e k e k = 1 6 9 = 3 so e k = (e k ) = So he populaion funcion is ( ) P = 1 9 ( 1). 3 ( ) (using Rule 3b, Table 1, page 4) 3 Noe ha P (1) according o his formula is approximaely 1184, which is reasonably close o he required value of 1. (b) Now skech his funcion: 5 HELM (8): Workbook 6: Exponenial and Logarihmic Funcions
Your soluion Answer P 1 1 5 4 6 8 1 (s) 5. Inverse square law decay Engineering Example 3 Inverse square law decay of elecromagneic power Inroducion Engineers are concerned wih using and inerceping many kinds of wave forms including elecromagneic, elasic and acousic waves. In many siuaions he inensiy of hese signals decreases wih he square of he disance. This is known as he inverse square law. The power received from a beacon anenna is expeced o conform o he inverse square law wih disance. Problem in words Check wheher he daa in he able below confirms ha he measured power obeys his behaviour wih disance. Power received, W.393.9.4.1.13.8 Disance from anenna, m 1 3 4 5 6 HELM (8): Secion 6.5: Modelling Exercises 51
Mahemaical saemen of problem Represen power by P and disance by r. To show ha he daa fi he funcion P = A r where A is a consan, plo log(p ) agains log(r) (or plo he raw daa on log-log axes) and check (a) how close he resuling graph is o ha of a sraigh line (b) how close he slope is o. Mahemaical analysis The values corresponding o log(p ) and log(r) are log(p ) -.48-1.41-1.399-1.653-1.851 -.1 log(r).31.499.6.694.778 These are ploed in Figure 11 and i is clear ha hey lie close o a sraigh line..5 1 log(p ) 1.5.5..4.6.8 log(r) Figure 11 The slope of a line hrough he firs and hird poins can be found from 1.399 (.48).499 =.35 The negaive value means ha he line slopes downwards for increasing r. I would have been possible o use any pair of poins o obain a suiable line bu noe ha he las poin is leas in line wih he ohers. Taking logarihms of he equaion P = A gives log(p ) = log(a) n log(r) rn The inverse square law corresponds o n =. In his case he daa yield n =.35. Where log(r) =, log(p ) = log(a). This means ha he inercep of he line wih he log(p ) axis gives he value of log(a) =.48. So A = 1.48 =.393. Inerpreaion If he power decreases wih disance according o he inverse square law, hen he slope of he line should be. The calculaed value of n =.35 is sufficienly close o confirm he inverse square law. The values of A and n calculaed from he daa imply ha P varies wih r according o P =.4 r The slope of he line on a log-log plo is a lile larger han. Moreover he poins a 5 m and 6 m range fall below he line so here may be addiional aenuaion of he power wih disance compared wih predicions of he inverse square law. 5 HELM (8): Workbook 6: Exponenial and Logarihmic Funcions
Exercises 1. Skech he graphs of (a) y = e (b) y = e + 3 (c) y = e (d) y = e 1. The figure below shows he graphs of y = e, y = e and y = e. 16 14 y e e 1 1 8 6 e 4 1 1 Sae in words how he graphs of y = e and y = e relae o he graph of y = e. 3. The figures below show graphs of y = e, y = 4 e and y = 4 3e. 1 1 y 1 y = e y 4 3 y = 4 e 4 y y = 4 3e 3 1 1 1 1 4 1 Use he above graphs o help you o skech graphs of (a) y = 5 e (b) y = 5 e 4. (a) The graph (a) in he figure below has an equaion of he form y = A + e k, where A and k are consans. Wha is he value of A? (b) The graph (b) below has an equaion of he form y = Ae k where A and k are consans. Wha is he value of A? (c) Wrie down a possible form of he equaion of he exponenial graph (c) giving numerical values o as many consans as possible. (d) Wrie down a possible form of he equaion of he exponenial graph (d) giving numerical values o as many consans as possible. HELM (8): Secion 6.5: Modelling Exercises 53
y y 3 -------------------------- 5 (a) (b) y y 6 -------------------------------- (c) 3 1 ------------------------------ (d) Answers 1. y e + 3 e 1 e 4 e 1 1 1. (a) y = e is he same shape as y = e bu wih all y values doubled. (b) y = e is much seeper han y = e for > and much flaer for <. Boh pass hrough (, 1). Noe ha y = e = (e ) so each value of y = e is he square of he corresponding value of y = e. 3. (a) 6 4 y 5 e (b) y 3 4 3 e 4. (a) (b) 5 (c) y = 6 4e k (d) y = 1 + e k 54 HELM (8): Workbook 6: Exponenial and Logarihmic Funcions
6. Logarihmic relaionships Experimenal psychology is concerned wih observing and measuring human response o various simuli. In paricular, sensaions of ligh, colour, sound, ase, ouch and muscular ension are produced when an exernal simulus acs on he associaed sense. A nineeenh cenury German, Erns Weber, conduced experimens involving sensaions of hea, ligh and sound and associaed simuli. Weber measured he response of subjecs, in a laboraory seing, o inpu simuli measured in erms of energy or some oher physical aribue and discovered ha: (1) No sensaion is fel unil he simulus reaches a cerain value, known as he hreshold value. () Afer his hreshold is reached an increase in simulus produces an increase in sensaion. (3) This increase in sensaion occurs a a diminishing rae as he simulus is increased. Task (a) Do Weber s resuls sugges a linear or non-linear relaionship beween sensaion and simulus? Skech a graph of sensaion agains simulus according o Weber s resuls. (b) Consider wheher an exponenial funcion or a growh-o-a-limi funcion migh be an appropriae model. Answer (a) Non-lineariy is required by observaion (3). 1 S 5 4 6 8 1 P (b) An exponenial-ype of growh is no appropriae for a model consisen wih hese experimenal resuls, since we need a diminishing rae of growh in sensaion as he simulus increases. A growh-o-a-limi ype of funcion is no appropriae since he daa, a leas over he range of Weber s experimens, do no sugges ha here is a limi o he sensaion wih coninuing increase in simulus; only ha he increase in sensaion occurs more and more slowly. A lae nineeenh cenury German scienis, Gusav Fechner, sudied Weber s resuls. Fechner suggesed ha an appropriae funcion modelling Weber s findings would be logarihmic. He suggesed ha he variaion in sensaion (S) wih he simulus inpu (P ) is modelled by HELM (8): Secion 6.5: Modelling Exercises 55
S = A log(p/t ) ( < T 1) where T represens he hreshold of simulus inpu below which here is no sensaion and A is a consan. Noe ha when P = T, log(p/t ) = log(1) =, so his funcion is consisen wih iem (1) of Weber s resuls. Recall also ha log means logarihm o base 1, so when P = 1T, S = A log(1) = A. When P = 1T, S = A log(1) = A. The logarihmic funcion predics ha a enfold increase in he simulus inpu from T o 1T will resul in he same change in sensaion as a furher enfold increase in simulus inpu o 1T. Each enfold change is simulus resuls in a doubling of sensaion. So, alhough sensaion is prediced o increase wih simulus, he simulus has o increase a a faser and faser rae (i.e. exponenially) o achieve a given change in sensaion. These poins are consisen wih iems () and (3) of Weber s findings. Fechner s suggesion, ha he logarihmic funcion is an appropriae one for a model of he relaionship beween sensaion and simulus, seems reasonable. Noe ha he logarihmic funcion suggesed by Weber is no defined for zero simulus bu we are only ineresed in he model a and above he hreshold simulus, i.e. for values of he logarihm equal o and above zero. Noe also ha he logarihmic funcion is useful for looking a changes in sensaion relaive o simulus values oher han he hreshold simulus. According o Rule a in Table on page 4, Fechner s sensaion funcion may be wrien S = A log(p/t ) = A[log(P ) log(t )] (P T > ). Suppose ha he sensaion has he value S 1 a P 1 and S a P, so ha and S 1 = A[log(P 1 ) log(t )] (P 1 T > ), S = A[log(P ) log(t )] (P T > ). If we subrac he firs of hese wo equaions from he second, we ge S S 1 = A[log(P ) log(p 1 )] = A log(p /P 1 ), where Rule a of Table has been used again for he las sep. According o his form of equaion, he change in sensaion beween wo simuli values depends on he raio of he simuli values. We sar wih S = A log(p/t ) (1 T > ). Divide boh sides by A: S A = log P (1 T > ). T Undo he logarihm on boh sides by raising 1 o he power of each side: 1 S/A = 1 log(p/t ) = P T (1 T > ), using Rule 4b of Table. So P = T 1 S/A (1 T > ) which is an exponenial relaionship beween simulus and sensaion. A logarihmic relaionship beween sensaion and simulus herefore implies an exponenial relaionship beween simulus and sensaion. The relaionship may be wrien in wo differen forms wih he variables playing opposie roles in he wo funcions. The logarihmic relaionship beween sensaion and simulus is known as he Weber-Fechner Law of Sensaion. The idea ha a mahemaical funcion could describe our sensaions was sarling when 56 HELM (8): Workbook 6: Exponenial and Logarihmic Funcions
firs propounded. Indeed i may seem quie amazing o you now. Moreover i doesn always work. Neverheless he idea has been quie fruiful. Ou of i has come much quaniaive experimenal psychology of ineres o sound engineers. For example, i relaes o he sensaion of he loudness of sound. Sound level is expressed on a logarihmic scale. A a frequency of 1 khz an increase of 1 db corresponds o a doubling of loudness. Task Given a relaionship beween y and x of he form y = 3 log( x 4 ) he relaionship beween x and y. (x 4), find Your soluion Answer One way of answering is o compare wih he example preceding his ask. We have y in place of S, x in place of P, 3 in place of A, 4 in place of T. So i is possible o wrie down immediaely x = 4 1 y/3 (y ) Alernaively we can manipulae he given expression algebraically. Saring wih y = 3 log(x/4), divide boh sides by 3 o give y/3 = log(x/4). Raise 1 o he power of each side o eliminae he log, so ha 1 y/3 = x/4. Muliply boh sides by 4 and rearrange, o obain x = 4 1 y/3, as before. The associaed range is he resul of he fac ha x 4, so 1 y/3 1, so y/3 > which means y >. HELM (8): Secion 6.5: Modelling Exercises 57