MATHS NOTES The Institute of Eduction 06 SUBJECT: Mths LEVEL: Higher TEACHER: Aidn Rontree Topics Covered: Powers nd Logs About Aidn: Aidn is our senior Mths techer t the Institute, where he hs been teching Mths nd Applied Mths for over 6 yers. As the uthor of books, including the most recent Effective Mths Books &, he is cquinted with every detil of the new Mths course. Aidn hs lso ppered on rdio progrmmes to give dvice to students prior to ems. He is regulrly clled upon by brnches of the Irish Mthemtics Techers Assocition to give tlks to techers on spects of the Mths courses.
Unit 8 Algebr Powers nd Logs 8. Powers nd the Lws of Indices. Definition of Powers nd Indices In the sttement 8 * is clled power, i.e. to the power of, * is the bse of the power, * is the inde, or eponent, of the power, * 8 is the vlue of the power.. Lws of Indices The rules for deling with powers re clled the Lws of Indices. They re given on pge of the Formule nd Tbles. Lws of Indices p q p q.. +, e.g.. p p q e.g. q p. ( ) q p q n+ e.g. ( ).. 6. 0 p q e.g. p e.g. q e.g.. + 7 8 8 ( n+ ) n+ 0 8 p p q q p q 7. ( ) e.g. 7 7 ( 7 ) 8. ( b) p p b p e.g. 9. p b b p p e.g. ( ) 8 ( ) 6 y y y Pge
Emple Epress ech of the following in the form k, where k R. (i) 6 (ii) Solution (i) 6 (ii) ( ) (iii).8 ( ) ( ). 6 8 8 ( ) (iv) ( ) 7 ( ) 7 7 7. 6 (iii).8 (iv) 8. Eercises 8. Simplify ech of the following, without using clcultor:.... + +. 7. ( b ) 8. ( ) 8. + ( ) 6. y ( ) y 9. b b + p q q 0. Evlute n+ n ( ) 0(8 ). n k. If f ( n ) ( ), show tht f ( n + k) f ( n).. Epress + + + in the form k, without using clcultor. Pge
8. Equtions with the Unknown in the Inde. Single Power Equl to Single Power To solve n eqution such s + (i) write ech side with the sme bse, (ii) put the indices equl to ech other, nd so solve for. Emple Solve for R, 8. Solution (Write ech side s single power with bse of.) ( ) ( ) (( ) ) 6 6 ( ) 6... now ech side is power with the sme bse 6... equting indices 6 6 7 6 7. Equtions which become Qudrtic * An eqution such s + ( ) + 0 cn be converted to qudrtic eqution by letting y be the bsic power present, i.e. y. * To write + in terms of y :. + ( ) (8) 8y * We cn now write the given eqution s 8y y + 0. Pge
* We now solve this eqution for y. Don t forget to finish by clculting the corresponding vlues of. Emple Solve the eqution + + ( ) + 0 for R. Solution Let y. Then (i) (ii) +. 9( ) 9y +. ( ) y Then the eqution cn be written y (9 y) + 0 y 6y + 0 (7 y )(9 y ) 0 7 y 0 or 9 y 0 y or y 7 9 Then or or or. Eercises 8. Solve the following equtions for R.... + 7. 7. + 9 + 7 6. 6 + 8 7. By letting y nd converting to qudrtic eqution, solve 0( ) 6 0 +, for R. 8. By letting y nd converting to qudrtic eqution, solve 8( ) 9 0 + +, for R. Pge
8. Logrithms. Definition of Logrithm (Log) Logrithm, or log, is nother term for inde. For emple, in the inde sttement 8, the log is.. Inde Sttement nd Log Sttement * The inde sttement 8 cn be written s log sttement in the form log 8, i.e. the bse,, is written s subscript nd the vlue of the power, 8, is written directly fter the log. * The mening of the symbol log is the inde tht must be plced bove the bse to give the vlue. Definition of Logs y log y. Evluting Log To evlute log, without using clcultor: (i) let the log be, (ii) write the log sttement in inde form, (iii) solve this eqution for. Emple Evlute, without using clcultor, log 7. Solution Let log 7. Then 7 ( ). Pge
. Lws of Logs Lws of Logs. log ( y) log + log y. log log log y q. log q log. log nd log 0 logb. log (Chnge of Bse Lw) log b y The Lws of Logs re relted to the Lws of Indices nd cn be derived from these lws. These lws re often used in conjunction with ech other. Emple If log0 nd y log0, epress in terms of nd y: 000 (i) log 0, (ii) log 0. 7 9 Solution (i) log0 log0 7 7 log0 7 Lw ( log 0 log 0 7 ) Lw ( log 0 log 0 ) ( log 0 log 0 ) Lw ( y) log 000 log 9 000 log 9 Lw log (000 ) log (ii) 0 0 0 0 0 0 0 0 log 0 + log log Lw log0 0 + log0 log0 Lw + y. Lw Pge 6
Emple (i) Show tht log b. logb (ii) Show tht + + log ( bc). log logb logc Solution (i) By Lw, the Chnge of Bse Lw, logb b log b logb... Lw log b (ii) By prt (i), log log, log b, logb log c logc Then + + log + log b + log c log log log b c log ( b c ) log ( bc ). Eercises 8. Without using clcultor, evlute ech of the following.. log6. log 7. log 8. log6 8 Write ech of the following s single log. log. + log y 6. 7. log log log log log b b + c 8. log + log y log z 9. (i) If log log y, show tht y. (ii) Hence show tht if log log y then y. 0. If p log7 q, epress in terms of p (i) q (ii) log9 q (iii) log q 9. Pge 7
8. Log Equtions. Definition of Log Eqution A log eqution is n eqution contining logs. To solve such equtions, we usully need to eliminte the logs from the eqution.. Solving Log Equtions To eliminte logs from n eqution, we cn write the eqution in one or other of the following two formts. * Single log equl to number: log X Y Thus Y X * Single log equl to single log to the sme bse: log X log Y Thus X Y.. Checking Answers As long s is positive number, log is only defined if > 0. Thus when we solve log eqution, ny nswer must be checked to mke sure tht we re not tking the log of negtive number or zero. Emple Solve the eqution log ( ) + log ( + ), 6 6 for R. Solution (We will write the eqution s single log equl to number.) log ( ) ( ) 6 + log6 + log ( )( ) 6 + ( log 6) 6 6 6 0 ( )( + ) 0 0 or + 0 or Check: OK Not OK ( log ( ) 6 6 is not defined.) Answer:. Pge 8
Emple Solve the eqution log ( + ) log ( ), for R. Solution (We need to write both sides with the sme bse, sy.) log ( + ) log ( + )... Chnge of Bse Lw log log ( + ) Then the eqution is log ( + ) log ( ) log ( + ) log ( ) log ( + ) log ( )... Lw + + 0 ( ) 0 0 or 0 0 or Check: 0 Not OK (The given eqution contins log ( ).) OK Ans:.. Using Logs to Solve Inde Equtions Logs cn be used to solve equtions with the unknown in the inde, if the bse nd the vlue cnnot be esily epressed s powers with the sme bse. Emple Find the vlue of t R if t 06. Give your nswer correct to two deciml plces. Solution Writing in log form, nd using clcultor, t 06 t log 06 t 0,... using the log function on clcultor correct to two deciml plces. Pge 9
Eercises 8. Solve the following equtions for R.. log + log log 8 log ( ) log ( )... log ( ) log ( + ) log ( + ) log ( + ). log ( + ) log ( ) + log ( 7) 6. log ( + ) log ( ) log ( + ) 7. (i) Show tht log ( + ) log ( + ). (ii) Hence, or otherwise, solve the eqution log ( + ) log ( + ) for R. 8. (i) Show tht log 9( ) log ( ). (ii) Hence, or otherwise, solve the eqution log ( + ) log ( ) 9 for R. 9. Find the vlue of R for which 6 9 7. Give your nswer correct to two deciml plces. 0. Find the vlue of R for which 8 67. Give your nswer correct to two deciml plces. 8. Eponentil Functions nd Reltionships A Eponentil Functions nd Reltionships. Definition of n Eponentil Function * An eponentil function is function of the form f : ( b), where, b R, b > 0. * For emple, f ( ) () nd f ( ) 86( 0) re both eponentil functions. * There re mny rel-life emples of eponentil functions, e.g. the growth of n investment under compound interest, the decy of rdioctive substnce. Pge 0
. Eponentil Reltionship from Dt Tble The following tble gives corresponding vlues for two vribles nd y. 0 y 0 6 8 99 6 9 6 * As the chnges in the independent vrible,, re fied, we cn check if there is n eponentil reltionship between nd y by clculting the rtios of successive terms. If these re constnt, then there is n eponentil reltionship between nd y, i.e. y is n eponentil function of. * Checking, (i) 6 0, (ii) 8, (iii) 6 99 6, (iv) 8 9 6. 99 6 As these rtios re constnt, y is n eponentil function of. Definition of n Eponentil Reltionship The dt in dt tble represents n eponentil reltionship between the vribles if (i) for fied chnges in the independent vrible, (ii) the rtios of successive terms is constnt.. Finding the Eqution of n Eponentil Reltionship from Dt Tble * If there is n eponentil reltionship between nd y, we cn write y ( b). * We cn then use two dt pirs from the tble to evlute nd b. Emple The tble below gives corresponding vlues for the vribles nd y. 0 y 0 6 8 99 6 9 6 If there is n eponentil reltionship between nd y, find the vlues of the constnts nd b if y ( b). Solution [] 0 when y 0. 0 ( b) 0 0... s 0 b Pge
[] when y 6. 6 0( b) 6 0b b Thus y 0( ). (Note tht b, the bse of the eponentil function, is lwys the sme s the constnt rtio, s long s the differences in the vlues is.) B Eponentil Problems. Eponentil Chnge in Rel Life A quntity y chnges eponentilly with respect to nother quntity if the rte of chnge of y is proportionl to the current vlue of. For emple, * ech hour, biologicl smple increses by % of its size t the beginning of tht hour, * ech yer, quntity of rdioctive substnce decrese by % of the quntity present t the beginning of tht yer.. Eponentil Growth * If quntity y increses eponentilly s nother quntity increses, then we cn write y ( b), where b >. * The bse of the power being greter thn is the key to eponentil growth. In prcticl problems, it is tken tht > 0.. Eponentil Decy * If quntity y decreses eponentilly s nother quntity increses, then we cn write y ( b), where b <. * The bse of the power being less thn is the key to eponentil decy. In prcticl problems, it is tken tht > 0.. Finding n Inde Logs cn be used to evlute n inde. For emple, if 67 88 ( ) then 087 nd log 087 089, correct to four deciml plces. Pge
Emple A biologicl smple strts with size of P nd grows t the rte of % ech dy fter tht. Let f ( ) represent the size of the smple dys lter. (i) Determine wht fctor P is multiplied by fter period of one dy. (ii) Epress f ( ) in terms of. (iii) Find, correct to the nerest hour, how long it tkes for the smple to double in size. (iv) Find n epression for in terms of n if f ( ) np. Solution (i) At the end of the first dy, the size of the smple is f () P + % of P f () P + 0 0P f () P( + 0 0) f () P( 0) Thus the size of the smple is multiplied by the fctor 0 over period of one dy. (ii) Hence f () f () ( 0) P( 0) ( 0) P( 0) nd f ( ) P( 0) (iii) (We wnt to find the vlue of for which f ( ) P, i.e. when the smple size is twice its originl size.) f ( ) P P( 0) P ( 0) Re-writing this eqution in log form, log 0 dys... using the log function on clcultor dys nd hours, correct to the nerest hour. (iv) Given f ( ) np P( 0) np ( 0) n n. log 0 Pge
Eercises 8.. The tble below gives corresponding vlues for the vribles nd y. 0 y 00 0 0 76 06 (i) Use the dt in the tble to show tht there is n eponentil reltionship between nd y. (ii) If y ( b), find the vlues of the constnts nd b.. The tble below gives corresponding vlues for the vribles nd y. 0 6 8 y 00 80 60 8 68 678 (i) Use the dt in the tble to show tht there is n eponentil reltionship between nd y. (ii) If y ( b), find the vlues of the constnts nd b.. A lbortory obtins 0 kg smple of rdioctive substnce, which decys over time. The following mesurements of weight, W, in grms, were obtined on yerly bsis. Let represent the number of yers from when the substnce ws obtined. / yers 0 W / grms 0000 9000 800 790 66 (i) Verify tht W is n eponentil function of. (ii) Epress W in terms of. (iii) Use your epression for W to clculte the weight of rdioctive substnce present fter 7 yers. (iv) Use logs to find the number of yers it will tke for the mount of rdioctive substnce to reduce to hlf the originl mount. (This is clled the hlf life of the substnce.) Give your nswer in yers to three deciml plces. (v) Determine the number of yers, correct to three deciml plces, tht it will tke for the weight of rdioctive substnce to reduce to kg.. Ech yer % of the quntity of rdioctive substnce present decys. The mount of rdioctive substnce present t the strt ws 00 grms. Let y f ( ) be the mount of rdioctive substnce present fter yers. (i) Write down n epression for f (), the mount present fter one yer. (ii) By writing down epressions for f () nd f (), find n epression for f ( ). (iii) Find the number of yers it tkes for the mount of rdioctive substnce to reduce to 00 grms. Give your nswer correct to the nerest yer. Pge
. The number of bcteri present in culture increses by % ech hour. The number of bcteri present initilly is N. (i) If f ( t ) represents the number of bcteri present fter t hours, show tht f ( t) N( 0) t. (ii) Find the number of hours it tkes for the number of bcteri present to increse to N. Give your nswer correct to the nerest hour. (iii) If the number present fter 6 hours is clculted to be 86, find the vlue of N correct to the nerest unit. 6. A sum of money P, invested in finncil institution, grows by % ech yer fter tht. Let A( t ) be the mount to which the sum hs grown fter t yers. (i) Find n epression for A( t ). (ii) Find t, the vlue of t for which the sum of money hs grown by 0%. (iii) Investigte if A( t) P. (iv) If A( t + t) P, find the vlue of t. 7. A new mchine costs 00,000. Its vlue deprecites by 8% per yer fter tht. Let f ( t ) be the vlue of the mchine t yers fter being purchsed. (i) Epress f ( t ) in the form ( b ) t. (ii) Find the number of yers, correct to two deciml plces, t which the vlue of the mchine is hlf its originl vlue. (iii) The compny tht buys the mchine plns on replcing it when it reches 0% of its originl vlue. Clculte the number of yers the mchine will hve been in use before being replced. 8.6 Eponentil nd Log Grphs A Eponentil Grphs. Eponentil Grphs from Tble of Vlues nd its Properties * Like other functions, we cn construct n eponentil grph by forming tble of vlues. * For emple, consider the function f ( ) nd the corresponding grph y. * We cn construct tble of vlues for this function. 0 y 8 6 6 8 Pge
* From this tble, we cn construct the grph of the curve y. * The key fetures of this curve re s follows. (i) The curve lies completely bove the -is. (ii) On the left, the curve gets closer nd closer to the -is. The -is is clled n symptote of the curve. An symptote of curve is line which pproimtes the curve s the curve tends to infinity. (iii) On the right, the curve increses more nd more rpidly s increses.. Eponentil Growth Grphs * The eqution of n eponentil growth grph is of the form y ( b) where > 0 nd b >. * Some of these grphs re shown opposite. Notice tht: 0 8 6 y y y ( ) y ( ) (i) The lrger the bse, b, the more steeply the grph rises on the right. (ii) The grph intersects the -is t the point (,0).. Eponentil Decy Grphs * The eqution of n eponentil decy grph is of the form y ( b) where > 0 nd 0 < b <. * Some of these grphs re shown opposite. Notice tht: (i) The smller the bse, b, the more rpidly the grph reduces on the right. (ii) The grph intercts the -is t the point (,0). y (0 ) y (0 7) y 0 () y (0 ) Pge 6
B Logrithmic Grphs. Eponentil nd Log Functions re Inverses * The functions y nd y log re inverse functions of ech other. * To demonstrte this, we show tht performing one fter the other returns us to. (i) log log () log (ii) To find, let log y Then log y log y Thus log.. Grph of Log s Inverse of Eponentil * As discussed bove, the functions f ( ) nd g( ) log re inverses of ech other. * Becuse of this, the grph of ech is the reflection of the other in the line y. y y y log * The two grphs, y nd y log, re shown opposite.. Other Log Grphs * The digrm opposite shows the three log grphs: y log, y log nd y log0. * Ech of these grphs (i) is only defined for > 0, (ii) crosses the -is t (,0), (iii) tends to s tends to 0 from the plus side (the y-is is n symptote) (iv) hs y co-ordinte of when the co-ordinte is equl to the bse. 0 y y log y log y log 0 6 8 0 6 8 Pge 7
. Constructing Log Grphs We cn construct log grph by (i) mking tble of vlues, nd (ii) using our knowledge of the shpe of log grph, s outlined bove. C Solving Equtions Using Intersecting Grphs We cn obtin pproimte solutions of n eqution contining eponentil functions or log functions by finding the intersection of the types of grphs we hve discussed previously. Emple (i) Sketch rough grph of the function f : ( ). (ii) Using the sme es nd the sme scles, sketch rough grph of the function g :. (iii) Use your grph to estimte the solutions of the eqution ( ). (iv) Wht other methods, if ny, could hve been used to solve this eqution? Solution (i) The function is f ( ) ( ) nd the corresponding curve is y ( ). Constructing tble of vlues: y 0 6 y ( ) The grph is shown below. y (ii) The function is g( ) nd the corresponding curve is y. This curve is lso shown on the grph opposite. (iii) The eqution ( ) my be written f ( ) g( ). Pge 8
The solutions of this eqution re the co-ordintes of the points of intersection of the curves y f ( ) nd y g( ). From the grph, the solutions re nd, correct to one deciml plce. (iv) We hve no lgebric method on our course to solve the eqution ( ). One other possible method is to estimte the solutions from tbles of vlues, but the grphicl method is undoubtedly the best vilble. Eercises 8.6 +. f : R R : (). (i) By completing the following tble of vlues, construct grph of the curve y f ( ), for. y f ( ) 0. (ii) Using the sme es nd the sme scles, construct grph of the function g : +, for. (iii) Use your grphs to find the solutions of the eqution () +. (iv) Wht other methods tht we hve seen cn be used to solve this eqution? Discuss. + f : R R :. (i) By completing the following tble of vlues, construct grph of the curve y f ( ), for. y f ( ) 0 (ii) Using the sme es nd the sme scles, construct grph of the function g : +, for. (iii) Use your grphs to find the solutions of the eqution +. (iv) Wht other methods tht we hve seen cn be used to solve this eqution? Pge 9
. Two functions re + f : R R : ( ) + nd g : R R : (0 ). (i) By completing the following tble of vlues, construct grph of the curve y f ( ), for. y f ( ) 0 (ii) Using the sme es nd the sme scles, sketch grph of y g( ), for. (iii) Use your grphs to find the solution of the eqution f ( ) g( ). (iv) Check your nswer to prt (iii) by using lgebr.. f : R + R : log. (i) Copy nd complete the tble: y f ( ) Hence construct the grph y f ( ). (ii) Using the sme es nd the sme scles, sketch the grph of g :. (iii) Use your grph to find the solutions of the eqution log. (iv) Wht other methods tht we hve seen cn be used to solve this eqution? Discuss. 8.7 Nturl Eponentil nd Nturl Log. The Number e * The number e is fmous nd importnt mthemticl constnt. Its vlue is pproimtely e 78. * The origin of this number will be seen in lter chpter.. Nturl Eponentil Function * The nturl eponentil function, or sometimes simply the eponentil function, is f ( ) e, where e is pproimtely 78. Pge 0
* The nturl eponentil function obeys ll the usul Lws of Indices, e.g. (i) e. e e + (ii) ( e ) e.. Nturl Log Function * The nturl log function is f ( ) log e where e is pproimtely 78. * log e cn lso be written ln. Thus we cn lso write f ( ) ln. * The nturl log function obeys ll the usul Lws of Logs, e.g. (i) ln(. y) ln + ln y (ii) ln ln.. Nturl Eponentil nd Nturl Log re Inverses Like other eponentil nd log functions with the sme bse, the nturl eponentil nd nturl log functions re inverses. Thus (i) ln e (ii) ln e. y y e. Grphs * As inverse functions, the grphs y e nd y ln re reflections of ech other in the line y. 0 y log ln e * These grphs re shown opposite. 6. Other Eponentil nd Log Grphs * Like ny eponentil or log grphs, we cn plot nturl eponentil grph, e.g. y e, or nturl log grph, e.g. y ln( + ), by mking tble of vlues. * Note tht y e rises more rpidly thn decrese more rpidly thn y y e, for > 0. e, for > 0, nd tht y e will Pge
Emple Show tht Solution ( ) ( ) ln e + ln. ln e ln + ln e ln + ln e + ln. 7. Eponentil Growth In scientific work, it is very common to see quntity, Q( t ) which grows eponentilly with time, t, being epressed in the form bt Q( t) Ae, where b is positive constnt. 8. Eponentil Decy Also in scientific work, it is very common to see quntity, Q( t ), which decys eponentilly with time, t, being epressed in the form Q( t) Ae bt, where b is positive constnt. Emple In lbortory eperiment, quntity, Q( t ), of chemicl ws observed t vrious points in time, t. Time is mesured in minutes from the initil observtion. The tble below gives the results. t 0 Q( t) 9000 80 99 80 60 Q follows rule of the form Q( t) Ae b t, where A nd b re constnts. (i) Use two of the observtions from your tble to evlute A nd b. Verify your vlues by tking nother observtion from the tble. (ii) Hence find the vlue of Q fter 0 minutes, i.e. Q (0). (iii) Estimte the time tht it tkes for the quntity of the chemicl to reduce to 0% of the originl mount. Give your nswer in minutes, correct to three deciml plces. t (iv) It is suggested tht nother model for Q is Q( t) Ap. Epress p in terms of b, correct to five deciml plces. Pge
Solution (i) Q (0) 9 : (ii) b Ae (0) 9 0 Ae 9 A 9 () Q () 80 : 9e b 80 b 80 e 9 b e 0 88788 b ln 0 88788 b 0 b 0 0 t Thus Q( t) 9e To check: 0 () 0 8 Q() 9e 9e 60 Correct. 0 (0) Q(0) 9e 9e 0 789 (iii) 0% of the originl mount is 0 9. Thus Q( t ) 0 9 (iv) 0 9e t 0 9 0 t e 0 0 t ln 0 0 t 08 t minutes. 0 t Q( t) 9e Q( t) 9 0 ( e ) t Q( t ) 9(0 887) t Thus p 0 887, correct to five deciml plces. Eercises 8.7 Simplify ech of the following epressions.. log ( ). e e. ln( + ). log sin e e + cos. loge ( e ) sin ln( e ) 6. ln e Pge
7. A chemicl rection strts with 000 grms of rectnt X, which then reduces eponentilly s the rection proceeds. The mount of X, in grms, present fter t minutes is given by A( t) Ce b t, where C nd b re constnts. The following tble gives some vlues of A. t 0 A( t) 000 67 88 (i) Find the vlue of C nd the vlue of b. Hence complete the tble bove. (ii) Find the vlue of t for which A( t ) 0. (iii) Another rectnt, Y, lso strts with 000 grms nd reduces eponentilly with time, so tht the mount of Y remining, B( t ), is given by B( t) Ce dt. If fter minutes, there is less of Y remining thn of X, suggest possible vlue for the constnt d. 8. The mount compny owes on lon grows eponentilly. The mount owed fter t yers, A( t ), is given by the following tble. t 0 A( t) 7 6 98 08 bt A is given by rule of the form A( t) Pe, where P nd b re constnts. (i) Find the vlue of P nd the vlue of b, nd hence complete the tble. (ii) Find the number of yers for the mount owed to double from its initil mount. (iii) If A( t ) cn lso be written A( t) P( + i) t, find the vlue of the constnt i. 9. A bcteri colony grows by % per dy. (i) If the size t time t 0 is 00, epress Q( t ), the size fter t dys, in the form bt Q( t) Ae, where A nd b re constnts. (ii) Find, correct to two deciml plces, the time it tkes for the colony to double in size. Q( t + k) (iii) Show tht is independent of t. Q( t) 0. In 000, 00 grms of rdium, rdioctive substnce, were stored. This quntity decys eponentilly over time. Let Q( t) Ae b t represent the mount of rdium remining t yers lter. (i) The hlf-life of rdium is 60 yers, i.e. it tkes 60 yers for hlf of the originl mount to decy. Find the vlues of the constnts A nd b. (ii) Find the mss of rdium tht will remin in the yer 000, correct to two deciml plces. (iii) Find the mss of rdium tht will decy between the yers 000 nd 000. Pge
8.8 Scientific Nottion. Definition of Scientific Nottion * A number is sid to be written in scientific nottion, or in stndrd form, when it is in the form 0 n, where < 0 nd n Z, the set of integers. * This form is generlly used for very lrge or very smll numbers. * For emple, (i) 0 nd (ii) 7 0 re both numbers written in scientific nottion. Scientific Nottion 0 n, where < 0 nd n Z.. Converting from Deciml Form to Scientific Nottion To write number in scientific nottion, we shift the deciml point to the right or the left, to get number between nd 0, nd multiply by the pproprite power of 0. For emple, (i) 800 8 0 move plces to the left (ii) 0 00076 76 0 move plces to the right. Converting from Scientific Nottion to Deciml Form To convert number from scientific nottion, 0 n, to ordinry deciml form, move the deciml point n spces, to the right if n is positive nd to the left if n is negtive. Add zeros before or zeros behind s necessry. For emple, (i) 8 7 8 0 780000000 move 8 plces to the right (ii) 6 7 0 6 0 0000067 move 6 plces to the left Pge
. Entering Numbers in Scientific Nottion on Clcultor () Csio Nturl Disply Clcultors To enter 7 0 7, press the following keys: 7 0 ( ) 7 (b) Shrp Write View Clcultors To enter 7 0 7, press the following keys:. 7 Ep ( ) 7. Adding nd Subtrcting Numbers in Scientific Nottion * Numbers given in scientific nottion cn be dded or subtrcted on clcultor. The nswer my be given in deciml form, nd if the nswer is required in scientific nottion, we will hve to convert it mnully. * To dd or subtrct the numbers without clcultor, we cn either convert them into deciml form, or write ech with the sme power of 0. This is so tht we cn tke out the power of 0 s common fctor. For emple, ( 6 0 ) + ( 9 0 ) (0 6 0 ) + ( 9 0 )... one plce to the left, ( ) 0 6 + 9 0 06 0 thus we chnge 0 to 0 6. Multiplying nd Dividing Numbers in Scientific Nottion To multiply or divide two numbers in scientific nottion, without using clcultor, del with the s nd the 0 n s seprtely. It my be necessry to djust the nswer to leve it in scientific nottion. For emple, 6 6 ( 0 ) (6 8 0 ) ( 6 8) (0 0 ) 0 76 0 76 0... moving the deciml point one plce 0 to the left chnges 0 to 0 Emple Without using clcultor, epress ( 0 ) + ( 6 0 ) 7 0 in the form 0 n, where < 0 nd n Z. Pge 6
Solution ( 0 ) + ( 6 0 ) ( 0 ) + (0 6 0 ) 7 0 7 0 ( + 0 6) 0 7 0 6 0 7 0 6 0 ( ) 7 8 0 0 7 0 Emple A quntity of rdioctive substnce strts with 0 toms. Ech month fter tht % of the remining substnce decys. Write in the form 0 n, where < 0 nd n Z, the number of rdioctive toms remining fter one yer. Solution Let P( t ) be the number of toms remining fter t months. Then P( t ) ( 0 )( 0 ) t ( 0 )(0 ) t After one yer, t, nd P () ( 0 ) (0 ) ( 0 ) (7 66 0 ) ( 7 66) 0 8 888 0 8888 0 0 Eercises 8.8 Epress ech of the following in the form your clcultor. 0 n, where < 0 nd n Z. You my use. 6000000. 80000. 68000. 00000. 000 6. 89000000 7. 0 0000 8. 0 008 9. 0 00000067 0. 0 00096. 0 008. 0 067 Pge 7
Without using clcultor, write ech of the following in scientific nottion nd in deciml form. You should check your nswer by using your clcultor... 7. 9.. ( 0 ) + (8 7 0 ). 6 (9 0 ) (8 0 ) 6. 6 ( 8 0 ) ( 0 ) 8. ( 0 ) + ( 6 0 ) 7 0 ( 0 ) + ( 0 ) 8 8 0 0.. 7 8 (8 0 ) + ( 7 0 ) ( 8 0 ) (8 7 0 ) 7 ( 0 ) ( 0 ) ( 0 ) + ( 0 ) 77 0 ( 7 0 ) + (7 0 0 ) 0. The distnce from erth to the Andromed gly is pproimtely million light yers, nd light yer is pproimtely 9 7 0 km. (i) Epress the distnce from erth to Andromed in kilometres in scientific nottion. 8 (ii) In the future, new spceship is cpble of trvelling t 0 km/h. How long would it tke this spceship to rech Andromed from erth?. The country of Bnnistn hs ntionl debt of 7 billion euro ( billion is equl to 000 million). The popultion of Bnnistn is million. (i) Clculte the verge mount owed by every citizen of Bnnistn. (ii) If the working popultion of Bnnistn is 7 million, clculte the mount owed by every working person in Bnnistn. 9. The number of cells in bcteri colony is initilly estimted to be 0. Ech hour fter tht the number of cells grows by 6%. Estimte the number of cells in the colony fter (i) hour, (ii) dy. Give your nswers in scientific nottion. 6. A mnufcturing compny hs number of mchines which produce widgets. One 6 mchine produces 0 widgets in dy, two other mchines ech produce 6 7 7 0 widgets in dy, nd one super mchine produces 0 widgets in dy. (i) Clculte the number of widgets produced by the compny in dy. (ii) If % of the widgets fil qulity test, clculte the number of good widgets produced by the compny in five dy week. 7. A lrge compny invests billion euro in reserch nd development in 0. It intends to increse this mount by 8% ech yer fter tht. Write in scientific nottion the mount it intends to invest in reserch nd development between 0 nd 0 inclusive. Pge 8
Revision Eercises 8. Solve the eqution. + 0 giving your nswer (i) in log form (ii) correct to three deciml plces.. Show tht n n 8. n n. (i) Solve the eqution 0 +. (ii) Hence solve the eqution y y 0 e + e.. Solve the eqution +, giving your nswer correct to three deciml plces.. Siobhn is given dose of rdioctive medicine, which decys t the rte of % per hour fter tht. If the originl dose is of 80 mg, nd f ( ) mg represents the mount left in her body hours fter receiving the dose, (i) clculte the mount left fter hours, (ii) find the vlue of when f ( ) 0, (iii) the mount by which f ( ) decreses in the third hour. 6. f ( ) Pe k gives the mount of rdioctive substnce present yers fter it strts to decy. (i) Wht is the initil mount present, i.e. f (0)? (ii) If it tkes 0000 yers for hlf the mount of rdioctive substnce to decy, epress the vlue of k in scientific nottion. (iii) Epress f (000) in terms of P. 7. A compny wnts to replce n eisting mchine in five yers time. It estimtes tht the replcement will cost 00000 in five yers time. It decides to set side n mount P now so tht it will grow t the rte of % per yer nd will provide the finnce to purchse the replcement. Clculte the vlue of P. 8. A bcteri colony grows t the rte of 8% per dy. (i) How mny dys will it tke the colony to double in size? (ii) How mny dys will it tke the colony to treble in size? 7 (iii) If initilly there re 8 0 bcteri present, write in scientific nottion the number of bcteri present fter 0 dys. 9. A bcteri colony strts with size of A nd grows eponentilly. It doubles in size in 0 minutes. Write the size, Q( t ), of the colony fter t minutes in the form bt Q( t) Ae giving the vlue of b, correct to four deciml plces. Pge 9
Pge 0
Solutions to Eercises Eercises 8...... 6. 7. 8. 9. ( ) 9 ( ) 8 6 6 + + (+ ) + (+ ) +. + + ( ) (+ ) ( ) + b b + b ( ) ( + ) 6 b ( ) b 6 b 6 9 ( b ) ( ) ( b ) b y y y y ( y) y ( ) ( ) p q p ( p ) p q q ( q ) q 6 ( ) 7 8 0. ( n + ) 0(8 n ) (( ) n + ) 0 (( ) n ) 6n+ 6n ( ) 0( ) 6n 6n (. ) 0( ) n. f ( n ) ( ). n+ k f ( n + k) ( ) n k (. ) k n ( ) k f ( n) 6n 6n 0( ) 0( ) 0 9 + + + +.. Pge
Eercises 8... 7... 6. 6 6 6 6 + 7 ( ) + + + 9 + 7 + ( ) + ( ) + 6+ + 6 + 6 8 9 9 8 ( ) + Pge
( ) + 9 (+ 9) 8 ( ) ( 8) ( 8) 8 8 7. Let y. Then ( ) y The eqution is y 0y + 6 0 ( y )( y 6) 0 y or y 6 or or 8. Let y. Then + ( ) y The eqution is y 8y + 9 0 (y )( y 9) 0 y 0 or y 9 0 y or y 9 or or. Eercises 8.. log6 6. log 7 ( ) Pge
. log 8 7 ( ) 7 7 7 log 8. 6 7 ( ) ( ) 7 7 7 8 log + log y log y. 6. 7. log log b log b log log b + log c log log b + log c log c b 8. log + log y log z log + log y log z log y z 9. (i) log log y log y y (ii) log log y log y log log log log y log log log y y 0. (i) p log7 q q 7 p p (ii) log9 q log9 7 p log9 7 Let log9 7 9 7 ( ) y Pge
Then log9 q p log9 9 (iii) log 9 q log 9 q log9 q p p. Eercises 8.. log + log log 8 log log 8 8 log ( ) log ( ). log 9 9 6 7 log ( ) log ( + ). log ( ) log ( + ) log ( + ) ( + ) ( ) + 6 + 9 96 + 6 + 9 0 6 + 0 ( )( ) 0 0 or 0 or log ( + ) log ( + ). log ( + ) log ( + ) + + + 0 ( ) 0 0 or 0 0 or (OK) (OK) Ans: 0 or Pge
. log ( + ) log ( ) + log ( 7) log ( + ) log ( )( 7) + + 0 + 0 6 + 0 ( )( ) 0 0 or 0 or (OK) (not OK) Ans: log ( + ) log ( ) log ( + ) 6. + log log ( + ) + + + ( )( + ) + 0 6 ( )( + ) 0 0 or + 0 or (OK) (not OK) Ans: log ( + ) 7. (i) log ( + ) log ( + ) log (ii) log ( + ) log ( + ) log ( + ) log ( ) + log ( + ) log ( + ) log ( + ) log ( + ) + + + 0 ( ) 0 0 or log ( ) 8. (i) log 9( ) log ( ) log 9 (ii) log ( + ) log 9( ) log ( + ) log ( ) log ( + ) log ( ) ( + ) log Pge 6
+ 6 + 9 9 + 6 + 9 6 0 + 0 ( 6)( ) 0 6 0 or 0 6 or 9. 6 9 7 log 6 9 7 0. 8 67 log 8 67. Eercises 8.. (i) (ii). (i) (ii) 0 0 76 06 0, 0, 0, 0 00 0 0 76 As these rtios re constnt, y is n eponentil function of. y ( b) 0 0, y 00 : 00 ( b) 00, y 0 : 0 00( b) b 0 Thus y 00( 0). 80 60 8 68 678 0 96, 0 96, 0 96, 0 96 00 80 60 8 68 As these rtios re constnt, y is n eponentil function of. y ( b) 0, y 00 :, y 80 : 0 00 ( b) 00 80 00( b) b 0 96 b 0 9798 Thus y 00(0 9798).. (i) Clculting rtios of successive terms: 9000 800 790 66 0 9, 0 9, 0 9, 0 9 0000 9000 800 790 As the rtio is constnt, W is n eponentil function of. (ii) Let W ( b). Then b 0 9, nd 0 0000 (0 9) 0000 Thus Pge 7
W 0000(0 9) (iii) When 7, 7 W 0000(0 9) W 78 7 g (iv) Hlf the originl mount is 000 g. 000 0000(0 9) (0 9) 0 log 0 0 9 6 79 (v) 000 W : 000 0000(0 9) (0 9) 0 log0 9 0 8.. (i) f (0) 00 f () f (0) % of f (0) f () 96% of f (0) f () f (0) 0 96 00 0 96 (ii) f () f () 0 96 (iii) 00 [ ] f () f (0) 0 96 0 96 f (0) (0 96) f () f (0) (0 96) f ( ) f (0) (0 96) 00(0 96) 00(0 96) 0 96 0 log0 96 0, correct to the nerest yer.. (i) f (0) N f () N + % of N f () N( 0) f () N( 0) f ( t) N( 0) t (ii) N N ( 0) t t 0 t log 0 t, correct to the nerest hour 6 (iii) 86 N( 0) 86 N 6 0 N 000. 6. (i) A( t) P( + 0 0) t A( t) P( 0) t (ii) A( t) P Pge 8
P t ( 0) t ( 0) t log 0 P t 6 (iii) A( t) A( 8) P( 0) P P (iv) A(6 + t) P 6 + t P( 0) P 6 +t 0 0 8 6 + t log t 6. 7. (i) f ( t ) 00000( 0 08) t f ( t ) 00000(0 9) t (ii) f ( t ) 0000 t 00000(0 9) 0000 t (0 9) 0 t log0 9 0 t 8, correct to two deciml plces (iii) f ( t ) 0000 t 00000(0 9) 0000 t (0 9) 0 t log0 9 0 t 9 yers. Eercises 8.6. (i) f ( ) () Tble: 0 y f ( ) 6 8 The grph is constructed below. (ii) g( ) + The liner grph y + contins the points (, 0) nd (0,), nd is shown opposite. Pge 9
. (i) (iii) From the grph, the solutions re pproimtely nd 0. (iv) We could compre vlues in the tbles of vlues, but there is no lgebric method we hve seen tht cn be used to solve this eqution. f ( ) Tble: 0 y f ( ) 0 0 0 67 7 The grph is constructed below. (ii) g( ) + The liner grph y y + contins the points (0,) y + nd (,), nd is shown opposite. (iii) From the grph, the solutions of the eqution re y pproimtely nd. (iv) We could compre vlues in the tbles of vlues, but there is no lgebric method we hve seen tht cn be used to solve this eqution.. (i) f ( ) ( ) Tble: 0 y f ( ) 0 0 8 The grph is constructed below. (ii) g( ) (0 ) Tble: y () y y + Pge 0
0 y g( ) 6 0 7 The grph is constructed below. (iii) From the grph, the only solution of the eqution is 0. (iv) f ( ) g( ) ( ) (0 ) log 0 9. (i) f ( ) log Tble: y f ( ) 0 The grph is constructed below. (ii) g( ) y The liner grph y contins the points (,0) y log nd (0, ), nd is shown opposite. (iii) From the grph, y the solutions of the eqution re pproimtely 0 nd 7. (iv) We could compre vlues in the tbles of vlues, but there is no lgebric method we hve seen tht cn be used to solve this eqution. y y ( ) y (0 ) Pge
Eercises 8.7..... 6. e e + e log ( e ) log log e log + ( )log e log e sin log e e + (sin + ) loge e sin + log ( e ) log + log e cos cos e e e e log + (cos )log e e loge + cos ln( + ) ln + ln( + ) ln + ln( + ) sin sin ln( e ) ln + ln e ln + sin (ln e) ln + sin ln ln ln e e ln ( ) ln e e e ln 7. (i) A( t) Ce bt A (0) 000 : A () 67 88 : Thus nd (ii) A( t ) 0 0 000 Ce C 000 67 88 000e b() b 0 6788 e b ln 0 6788 b 0 9999 b 0 0 t A( t) 000e A A A 0 () 000e 778 8 0 () 000e 606 0 7 () 000e 7 7 0 t 000e 0 0 t e 0 0 0 t ln 0 0 0 t 997 t 98 (iii) d > 0, e.g. d 0. Pge
8. (i) A( t) Pe bt b A () 7 6 : 7 6 Pe A () 98 08 : b 98 08 Pe Dividing by, b Pe 98 08 b Pe 7 6 b e 00778 b ln 00778 b 0 0997 : 7 6 P( 00778) P 00 Thus 0 0997t A( t) 00 e nd A(0) P 00 A A (ii) A( t) P b () Pe 60 0 b () Pe 6 9 0 0997t P Pe 0 0997t e 0 0997t ln 0 0997t 0 697806 t 7 (iii) A( t) P( + i) t ( + i) t e bt b + i e + i 00778 i 0 00778 9. (i) Q( t ) 00( 0) t b Let e 0 b ln 0 0 0907 0 0907t Thus Q( t) 00e (ii) Q( t ) 00 t 00( 0) 00 t ( 0) t log 7 67 0 t+ k Q( t + k) 00( 0) (iii) 0 t Q( t) 00( 0) which is independent of t. 0. (i) Q( t) Ae b t Q(60) A Ae A 60b e 0 60 b k Pge
60b ln 0 60b 0 697806 b 0 000676 nd A 00. (ii) In 000, t 000. 0 000676(000) Q(000) 00e Q (000) 09 grms (iii) In 000, t 000. 0 000676(000) Q(000) 00e Q (000) 6 88 The mss of rdium tht decreses between 000 nd 000 is 6 88 09 79 grms. Eercises 8.8..... 6. 7. 8. 9. 0...... 6. 7. 8. 7 6000000 6 0 6 80000 8 0 6 68000 6 8 0 8 00000 0 000 0 9 89000000 8 9 0 0 0000 0 0 008 8 0 0 00000067 6 7 0 7 0 00096 9 6 0 0 008 8 0 0 067 67 0 ( 0 ) + (8 7 0 ) ( 0 ) + (0 87 0 ) 6 7 0 6700 7 8 8 8 (8 0 ) + ( 7 0 ) (0 8 0 ) + ( 7 0 ) 8 0 000000 6 (9 0 ) (8 0 ) (9 0 ) (0 8 0 ) 8 8 0 0 000088 ( 8 0 ) (8 7 0 ) ( 8 0 ) (0 87 0 ) 98 0 980 6 ( 8 0 ) ( 0 ) 0 00000000 7 ( 0 ) ( 0 ) 7 8 0 7800 Pge
( 0 ) + ( 6 0 ) 6 0 9. 7 0 7 0 8 0 0 7 0 0000000 ( 0 ) + ( 0 ) 77 0 0. 77 0 77 0 9 0 00000000 ( 0 ) + ( 0 ) 7 0. 8 8 8 0 8 0 0 0 00 ( 7 0 ) + (7 0 0 ) 8 0. 0 0 8 0 0000000. (i) distnce (000000) (9 7 0 ) km (ii) Time 9 89 0 km distnce speed 9 89 0 8 0 0 7 0 hours 0 yers... ssuming 6 dys per yer totl debt. (i) Averge mount owed popultion 7 000000000 000000 7 0 6 0 9 (ii) Averge mount owed per working person 7 0 6 7 0 97 9. (i) No. of cells ( 0 ) ( 06) 9 7 0 0 ( 06) (ii) No. of cells ( ) 9 0 86 0 6. (i) In one dy, number of widgets 6 6 7 ( 0 ) + ( 7 0 ) + ( 0 ) 7 9 0 900000 Pge
(ii) Number of good widgets in one dy 7 (0 96) 9 0 ( ) 0000 Number of good widgets in one five dy week 00000 8 0 0 9 7. 0: 0 9 9 0: ( 0 ) 08 78 0 0: 9 9 ( 0 ) ( 08) 08 0 0: ( 0 ) ( 08) 09 0 Totl investment 9 ( + 78 + 08 + 09) 0 9 77 0 0 777 0 euros 77 billion. 9 9 Revision Eercises 8. (i).. + 0 + 0 log 0 + log 0 log 0 (ii) 6 n n n n 8 ( ) n n ( ) ( ) n 6n n. (i) 0 + Multiply by. + 0 0 + 0 ( )( ) 0 or (ii) y y 0 e + e y 0 e y e From prt (i), Pge 6
. +. y y e or e y ln or y ln log 70. (i) f ( ) 80( 0 ) 80(0 8) f () 80(0 8) 76 mg (ii) f ( ) 0 80(0 8) 0 (0 8) 0 log0 8 0 8 (iii) The third hour is from to. f () 80(0 8) 7 8 f Reduction in third hour f () f () 7 8 9 8 67 mg. 6. (i) f ( ) Pe k () 80(0 8) 9 0 f (0) Pe P (ii) f (0000) 0 P (iii) 0000k Pe 0 P 0000k e 0 0000k ln 0 0000k 0 69 k 0 000069 k 6 9 0 f (000) Pe (6 9 0 ) 0 0 069 f (000) Pe f (000) 0 90P. 7. P invested now t % per nnum compound interest in yers time will be worth P( 0) 66P If this is to py for the replcement mchine, then 66P 00000 Pge 7
00000 P 66 P 096 8. (i) Let P be the initil size of the colony nd f ( ) be the size of the colony fter dys. Then f ( ) P( 08) To double in size: f ( ) P P( 08) P ( 08) log 08 9 0 Thus it will tke 9 0 dys to double in size. (ii) To treble in size: f ( ) P (iii) bt 9. Q( t) Ae Q(0) A P( 08) P ( 08) log 08 7 Thus it will tke 7 dys to treble in size. 7 P 8 0 nd 7 0 f (0) [8 0 ] ( 08) 0b Ae A 0b e 0b ln 0b 0 69 b 0 069. 7 f (0) [8 0 ] 0 06 8 f (0) 8 0 0. Pge 8