Higher. Exponentials and Logarithms 160

Transcription

1 hsn uknt Highr Mthmtics UNIT UTCME Eponntils nd Logrithms Contnts Eponntils nd Logrithms 6 Eponntils 6 Logrithms 6 Lws of Logrithms 6 Eponntils nd Logrithms to th Bs 65 5 Eponntil nd Logrithmic Equtions 66 6 Grphing with Logrithmic As 69 7 Grph Trnsformtions 7 HSN This documnt ws producd spcill for th HSNuknt wbsit, nd w rquir tht n copis or drivtiv works ttribut th work to Highr Still Nots For mor dtils bout th copright on ths nots, pls s

2 Highr Mthmtics Unit Eponntils nd Logrithms UTCME Eponntils nd Logrithms Eponntils W hv lrd mt ponntil functions in Unit utcom A function of th form f ( ) = whr, R nd > is known s n ponntil function to th bs If > thn th grph looks lik this: =, > (, ) This is somtims clld growth function If < < thn th grph looks lik this: =, < < (, ) This is somtims clld dc function Rmmbr tht th grph of n ponntil function f ( ) = lws psss through (, ) nd (, ) sinc f = =, f ( ) = = hsnuknt Pg 6 HSN

3 Highr Mthmtics Unit Eponntils nd Logrithms EXAMPLES Th ottr popultion on n islnd incrss b 6% pr r How mn full rs will it tk for th popultion to doubl? Lt u b th initil popultion u = 6 u (6% s dciml) u = 6 u = 6 6 u = 6 u u = 6 u = 6 6 u = 6 u u n n = 6 u For th popultion to doubl ftr n rs, w rquir u u W wnt to know th smllst n which givs 6 n vlu of or mor, sinc this will mk u n t lst twic s big s u Tr vlus of n until this is stisfid If n =, 6 = 5 < If n =, 6 = 56 < If n =, 6 = 8 < 5 If n = 5, 6 = > Thrfor ftr 5 rs th popultion will doubl n clcultor: 6 = n 6 ANS = Th fficinc of mchin dcrss b 5% ch r Whn th fficinc drops blow 75%, th mchin nds to b srvicd Aftr how mn rs will th mchin nd srvicd? Lt u b th initil fficinc u = 95 u (95% s dciml) u u u u = 95 = = 95 = 95 = = 95 n 95 n u u u u u = u Whn th fficinc drops blow 75u (75% of th initil vlu) th mchin must b srvicd So th mchin nds srvicd ftr n rs if 95 n 75 = hsnuknt Pg 6 HSN

4 Highr Mthmtics Unit Eponntils nd Logrithms Tr vlus of n until this is stisfid If n =, 95 = 9 > 75 If n =, 95 = 857 > 75 If n =, 95 = 85 > 75 5 If n = 5, 95 = 77 > 75 6 If n = 6, 95 = 75 < 75 Thrfor ftr 6 rs, th mchin will hv to b srvicd Logrithms Hving prviousl dfind wht logrithm is (s Unit utcom ) w wnt to look in mor dtil t th proprtis of ths importnt functions Th rltionship btwn logrithms nd ponntils is prssd s: = log = whr, > Hr, is th powr of which givs EXAMPLES Writ 5 = 5 in logrithmic form 5 = 5 = log55 Evlut log 6 Th powr of which givs 6 is, so log 6 = Lws of Logrithms Thr r thr lws of logrithms which ou must know Rul log + log = log whr,, > If two logrithmic trms with th sm bs numbr ( bov) r bing ddd togthr, thn th trms cn b combind b multipling th rgumnts ( nd bov) EXAMPLE Simplif log5 + log5 log + log = log5 = log 8 hsnuknt Pg 6 HSN

5 Highr Mthmtics Unit Eponntils nd Logrithms Rul ( ) log log = log whr,, > If logrithmic trm is bing subtrctd from nothr logrithmic trm with th sm bs numbr ( bov), thn th trms cn b combind b dividing th rgumnts ( nd in this cs) Not tht th rgumnt which is bing tkn w ( bov) pprs on th bottom of th frction whn th two trms r combind EXAMPLE Evlut log 6 log Rul log 6 log 6 = log = log = (sinc = = ) log n = nlog whr, > Th powr of th rgumnt (n bov) cn com to th front of th trm s multiplir, nd vic-vrs EXAMPLE Eprss log7 in th form log7 log 7 = log7 = log 9 7 Sqush, Split nd Fl You m find th following nms r simplr w to rmmbr th lws of logrithms log log log ( ) + = th rgumnts r squshd togthr b multipling log log log ( ) log = th rgumnts r split into frction n = nlog th powr of n rgumnt cn fl to th front of th log trm nd vic-vrs hsnuknt Pg 6 HSN

6 Highr Mthmtics Unit Eponntils nd Logrithms Not Whn working with logrithms, ou should rmmbr: EXAMPLE log = sinc Evlut log7 7 + log log 7 + log = + = 7 Combining svrl log trms =, log = sinc = Whn dding nd subtrcting svrl log trms in th form log b, thr is simpl w to combin ll th trms in on stp Multipl th rgumnts of th positiv log trms in th numrtor Multipl th rgumnts of th ngtiv log trms in th dnomintor EXAMPLES 5 Evlut log + log 6 log 5 log + log 6 log = log = log = 6 Evlut log6 + log6 log + log 6 6 = log + log 6 6 = log + log = log 9 = log 6 log = = (sinc 6 6) rgumnts of positiv log trms rgumnts of ngtiv log trms log R log + log 6 6 = log + log 6 6 = log + log 6 6 ( 6 6 ) ( ) = log + log = log 6 6 = log 6 + log + log 6 log 5 = (sinc log 6 = ) 6 hsnuknt Pg 6 HSN

7 Highr Mthmtics Unit Eponntils nd Logrithms Eponntils nd Logrithms to th Bs Th constnt is n importnt numbr in Mthmtics, nd occurs frquntl in modls of rl-lif situtions Its vlu is roughl (to 9 dp), nd is dfind s: ( n ) n + s n If ou tr lrg vlus of n on our clcultor, ou will gt clos to th vlu of Lik π, is n irrtionl numbr Throughout this sction, w will us in prssions of th form:, which is clld n ponntil to th bs, log, which is clld logrithm to th bs This is lso known s th nturl logrithm of, nd is oftn writtn s ln (i ln log ) EXAMPLES Clcult th vlu of log 8 log 8 = 8 (to dp) Solv log = 9 log = 9 so = 9 = 8 8 (to dp) Simplif log ( ) log ( ) prssing our nswr in th form + log b log c whr, b nd c r whol numbrs log ( ) log ( ) = log + log log log = log + log = + log log = + log log = + log 6 log 7 n clcultor: ln 8 = n clcultor: 9 = R log ( ) log ( ) = log ( ) log ( ) ( ) = log ( ) 6 Rmmbr = log 7 n n n ( b) = b 6 = log 7 = log + log 6 log 7 = + log 6 log 7 hsnuknt Pg 65 HSN

8 Highr Mthmtics Unit Eponntils nd Logrithms 5 Eponntil nd Logrithmic Equtions Mn mthmticl modls of rl-lif situtions us ponntils nd logrithms It is importnt to bcom fmilir with using th lws of logrithms to hlp solv qutions EXAMPLES Solv log + log = log 7 for > log + log = log 7 log = log 7 = 7 (sinc log = log = ) = Solv ( ) ( ) log + log = for > ( ) ( ) log + log = log + = + = = (sinc log = = ) + = ( ) + = 8 = 6 = + + = for p > Solv log ( p ) log ( p ) log ( p) log ( p + ) + log ( p ) = log ( p) log (( p + )( p ) ) = log ( p) ( p + )( p ) = p p p + p p = 6 p + = p p = 6 8 ( p )( p ) p = Sinc w rquir + 5 = or p 5 = p = 5 p >, p = 5 is th solution hsnuknt Pg 66 HSN

9 Highr Mthmtics Unit Eponntils nd Logrithms Dling with Constnts Somtims it m b ncssr to writ constnts s logs, in ordr to solv qutions EXAMPLE Solv log 7 = log + for > Writ in logrithmic form: = = log (sinc log = ) = log = log 8 Us this in th qution: log 7 = log + log 8 log 7 = log 8 7 = 8 = 7 8 R log 7 = log + log 7 log = 7 log = Convrting from log to ponntil form: 7 = = = Solving Equtions with Unknown Eponnts If n unknown vlu (g ) is th powr of trm (g or ), nd its vlu is to b clcultd, thn w must tk logs on both sids of th qution to llow it to b solvd Th sm solution will b rchd using n bs, but clcultors cn b usd for vluting logs ithr in bs or bs EXAMPLES 5 Solv = 7 Tking log of both sids R Tking log of both sids log = log 7 log = log 7 ( log = ) = log 7 = 96 (to dp) = log log 7 log = log 7 log 7 = log = 96 (to dp) hsnuknt Pg 67 HSN

10 Highr Mthmtics Unit Eponntils nd Logrithms 6 Solv 5 + = + = log 5 log ( + ) log 5 = log log + = log 5 + = 9 = 9 = (to dp) Not log could hv bn usd instd of log Eponntil Growth nd Dc Rcll from Sction tht ponntil functions r somtims known s growth or dc functions Ths oftn occur in modls of rl-lif situtions For instnc, rdioctiv dc cn b modlld using n ponntil function An importnt msurmnt is th hlf-lif of rdioctiv substnc, which is th tim tkn for th mss of th rdioctiv substnc to hlv 7 Th mss G grms of rdioctiv smpl ftr tim t rs is givn b t th formul G = () Wht ws th initil mss of rdioctiv substnc in th smpl? (b) Find th hlf-lif of th rdioctiv substnc () Th initil mss ws prsnt whn t = : G = = = So th initil mss ws grms (b) Th hlf-lif is th tim t t which G = 5, so 5 = t t = 5 = t = log ( ) (convrting to log form) t = (to dp) So th hlf-lif is rs, roughl 56 = 8 5 ds hsnuknt Pg 68 HSN

11 Highr Mthmtics Unit Eponntils nd Logrithms 8 Th world popultion, in billions, t rs ftr 95 is givn b 78t P = 5 () Wht ws th world popultion in 95? (b) Find, to th nrst r, th tim tkn for th world popultion to doubl () For 95, t = : P = 5 78 = 5 = 5 So th world popultion in 95 ws 5 billion (b) For th popultion to doubl: 78 t 5 = 5 78t = 78t = log (convrting to log form) t = 8 9 (to dp) So th popultion doubld ftr 9 rs (to th nrst r) 6 Grphing with Logrithmic As It is common in pplictions to find n ponntil rltionship btwn vribls; for instnc, th rltionship btwn th world popultion nd tim in th prvious mpl Givn som dt (g from n primnt) w would lik to find n plicit qution for th rltionship Rltionships of th form = b Suppos w hv n ponntil grph = b, whr, b > = b hsnuknt Pg 69 HSN

12 Highr Mthmtics Unit Eponntils nd Logrithms Tking logrithms w find tht log = log ( b ) = log + log = log + log b W cn scl th -is so tht Y b is Now our rltionship is of th form stright lin in th (, Y )-pln Y = log ; th Y-is is clld logrithmic Y = log b + log, which is ( log ) Y = b + log log grdint is log b Sinc this is just stright lin, w cn us known points to find th grdint log b nd th Y-is intrcpt log From ths w cn sil find th vlus of nd b, nd hnc spcif th qution EXAMPLES = b Th rltionship btwn two vribls, nd, is of th form = b An primnt to tst this rltionship producd th dt shown in th grph, whr log is plottd ginst Find th vlus of nd b W nd to obtin stright lin qution: log = log b (tking logs of both sids) = b log = log + log b log = log + log b i Y = log b + log log From th grph, th Y-is intrcpt is log = ; so ( 7,5) = hsnuknt Pg 7 HSN

13 Highr Mthmtics Unit Eponntils nd Logrithms Using th grdint formul: log b 5 = 7 = 7 b = 7 Th rsults from n primnt wr notd s follows: Th rltionship btwn ths dt cn b writtn in th form = b Find th vlus of nd b, nd stt th formul for in trms of W nd to obtin stright lin qution: log = log b (tking logs of both sids) = b log = log + log b log = log + log b i Y = log b + log W cn find th grdint log b (nd hnc b), using two points on th lin: using (, ) nd ( 8, ), So log = 7 + log log b So b = 8 = 7 (to dp) 7 = = 8 (to dp) Now w cn work out log (nd hnc ) b substituting point into this qution: using (, ), Thrfor = log log = nd = so = 7 + log log = 7 so = = 9 (to dp) 9 = 97 (to dp) Not Dpnding on th points usd, slightl diffrnt vlus for nd b m b obtind hsnuknt Pg 7 HSN

14 Highr Mthmtics Unit Eponntils nd Logrithms Equtions in th form = b Anothr common rltionship is =, whr, > In this cs, th rltionship cn b rprsntd b stright lin if w chng both s to logrithmic ons EXAMPLE Th rsults from n primnt wr notd s follows: Th rltionship btwn ths dt cn b writtn in th form = Find th vlus of nd b, nd stt th formul for in trms of W nd to obtin stright lin qution: = log = log (tking logs of both sids) log = log + log log = log + b log i Y = bx + log b log log b b W cn find th grdint b using two points on th lin: using ( 7, ) nd ( 85, ), So log 59 log log = + b b = 85 7 = 59 (to dp) Now w cn work out b substituting point into this qution: using ( 7, ), = log log = 59 7 = = = (to dp) b Thrfor 59 = hsnuknt Pg 7 HSN

15 Highr Mthmtics Unit Eponntils nd Logrithms 7 Grph Trnsformtions Grph trnsformtions wr covrd in Unit utcom Functions nd Grphs, but w will now look in mor dtil t ppling trnsformtions to grphs of ponntil nd logrithmic functions EXAMPLES Shown blow is th grph of = f ( ) whr f ( ) = log = f ( ) ( 9,) () Stt th vlu of (b) Sktch th grph of = f ( + ) + () = log 9 = = (sinc 9) (b) Th grph shifts two units to th lft, nd on unit upwrds: = f ( + ) + = ( 7,) (,) Shown blow is prt of th grph of = log5 = f ( ) Sktch th grph of = log5 ( ) = log 5 ( ) = log5 ( ) = log5 = log 5 So rflct in th -is ( 5,) ( 5, ) hsnuknt Pg 7 HSN

16 Highr Mthmtics Unit Eponntils nd Logrithms Th digrm shows th grph of = = (, ) n sprt digrms, sktch th grphs of: () = ; (b) = () Rflct in th -is: (b) = = = (, ) So scl th grph from () b in th -dirction: (,8 ) = = hsnuknt Pg 7 HSN