P g e 30 4.2 Eponentil Functions 1. Properties of Eponents: (i) (iii) (iv) (v) (vi) 1 If 1, 0 1, nd 1, then E1. Solve the following eqution 4 3. 1 2 89 8(2 ) 7 Definition: The eponentil function with se ( 0, 1) is defined the eqution. 2. Properties of eponentil function (i) Grphs The eponentil function is one-to-one function. (iii) Domin: (, ), Rnge: ( 0, ) (iv) no -intercept, -intercept is 1. (v) Horizontl smptote is -is or =0. E2. Find domin, rnge, -intercept nd -intercept for 3 9
P g e 31 Emple 3: Find n eponentil function of the form f() = - + c tht hs the given horizontl smptote nd -intercept nd psses through point P. = 60; -intercept 409; P (1, 147.25) 4.3 Nturl Eponentil Function 1 e lim 1 n n e function. n 2.718 is clled the nturl eponentil E1. Find domin for ech of the following functions: (i) (iii) e 3 e 3 1 1 e 1 35 n 1 n 1 10 2.5937424601 100 2.7048138294 1000 2.7169239322 10000 2.7181459268 100000 2.7182682372 1000000 2.7182804693 10000000 2.7182816926 n Emple 2: Simplif the epression. (Do not epnd the epression in the denomintor.)
P g e 32 Applictions: Emple 3: Assume tht the nnul interest rte r= 12% is clculted ever month. Tht mens tht the nnul interest rte is compounded monthl. Wht is the mount in the ccount fter 1 month, 2 months, 3 months, 1 er, 2 ers, 3 ers, nd t ers, if the initil investment is $1000? Compound Interest Formul 1: A P 1 where P= principl r = nnul interest rte epressed s deciml n= numer of interest period per er t = numer of ers P is invested A = mount fter t ers nt =numer of interest period in t ers Emple 4: $2400 is invested t rte of 5.5% per er compounded monthl. Find the principl fter 3 ers nd 4 month. r n nt Emple 5. Suppose tht $10,000 is invested in n ccount tht drws 8% nnul interest. Clculte the mount in this ccount fter 10 ers if the interest is compounded () Annull () (c) Seminnull Monthl (d) Dil
P g e 33 Compound Interest Formul 2: A where P= principl, or initil investment rt Pe r = nnul interest rte epressed s deciml tht is compounded continuousl t = numer of ers P is invested A = mount fter t ers Effective Annul Yield The effective ield (or effective nnul interest rte) for n investment is the simple interest rte tht would ield t the end of one er the sme mount s is ielded the compounded rte tht is ctull pplied. Emple 6: Find Effective Annul Yield for the following ccounts: () 4% nnul interest rte compounded qurterl. () 4% nnul interest rte compounded continuousl. 4.4 Logrithmic Functions Definition: The epression (red s se of ) mens the eponent to which must e rised to ield. 3. Equivlent Forms Logrithmic form Eponentil form Emple: Determine the per nnum interest rte r required for n investment with continuous compound interest to ield n effective rte of 5.5%.
P g e 34 Definition: The rithmic function with se ( 0, 1) is defined the eqution. The common rithmic function is defined 10. The nturl rithmic function is defined e ln. 4. Properties of Logrithmic Functions (i) Grph The rithmic function is one-to-one function. (iii) Domin: ( 0, ), Rnge: (, ) (iv) no -intercept, -intercept is 1. (v) Verticl smptote is -is or =0. E1. Find domin for ech of the following functions: (i) ln 10 2 ln 10 2 (iii) 3 e 5 E2. Find the domin nd -intercept for f ( ) 2 ln(3 7) E3.You invest $5,650 t 8% per nnum compounded continuousl. Determine the time T (in ers) for our investment to e worth $8,240. E4. Mone is invested t interest rte r, compounded continuousl. Epress the time required for the mone to qudruple, s function of r.
P g e 35 4.5 Properties of Logrithms 5. Properties of Logrithms If u nd w denote positive rel numers, then (i) (iii) ( uw) u w c u c u u u w w E1. Write ( 23) 2(1 ) 6(1 ) s one rithm. E2. Epress f ( ) 4 7 4 (35) without rithms. 35 82 E3.. Epress f ( ) (81 27 ) without rithms. 3 E 4: Epress in terms of rithms of,, or z. E5: Evlute given tht nd
P g e 36 4.6 Eponentil nd Logrithmic Equtions Chnge of Bse Formul : If u 0 nd if nd re positive rel numers different from1, then u u Proof: E1. Solve ech of the following equtions: (i) E2. Use common rithms to solve for in terms of : E3. Use common rithms to solve for :
P g e 37 E4. Solve ech of the following equtions: (i) (iii) (iv) E3. Solve the following eqution:
P g e 38 Applictions Nturl Growth/Dec Model: where is the initil quntit. If k>0, P(t) is n eponentil incresing function of t nd k is clled growth constnt. If k<0, P(t) is nturl declining quntit nd k is clled dec constnt. For nturl declining quntit, we often use Hlf-life to descrie its propert. Definition: Hlf-life is the length of time it tkes for quntit to decrese to the hlf of its initil vlue. Emple 1: A certin rdioctive sustnce with hlf-life of 2400 ers is used in estimting the ge of relics. Prt 1: Find the dec constnt. Your nswer must e EXACT. Answers which included deciml points ANYWHERE will e mrked incorrect. Enter n EXACT, smolic nswer (such s 3/2, not 1.5, or ln(2), not 0.69 Dec constnt= Prt 2: If certin relic contins 28.7% of the originl mount of the sustnce, determine the ge T of the relic in ers. Your nswer must e correct to the nerest whole numer. WARNING: Do not round off vlues during intermedite clcultions. You m prefer to enter smolic nswer. Emple 2: 90% of rdioctive mteril remins fter 30 ds. Prt 1: Find the dec constnt. IMPORTANT: Deciml nswers will e mrked incorrect-- deciml point cn't e ANYWHERE in our nswer. Enter n EXACT, smolic nswer (such s 3/2, not 1.5, or ln(2), not 0.69) Dec constnt= Prt 2: Find the time T (in ds) fter the initil mesurement when 44% of the originl mount of rdioctive mteril remins. You m enter smolic nswer or round our nswer to the nerest whole numer. WARNING: Do not round off vlues during intermedite clcultions, s error m compound to give ou n inccurte nswer. T=
P g e 39 Emple 3:The grph shown elow is the grph of n eponentil growth curve, P = P 0 e kt Prt 1: Find the growth constnt, k. Your nswer must e ect, tht is, smolic (NO deciml plces nwhere). Prt 2: Find P 0. Your nswer must e correct to one deciml plce. (Use the ect form of k. to insure our nswer is ccurte.) Emple 4: Two popultions of cteri re growing eponentill in seprte petri dishes. The popultion in the first dish hs growth constnt 0.16 nd initil popultion 1000. The popultion in the second dish hs growth constnt 0.08 nd initil popultion 6000. Prt 1. Find the time t which the two popultions hve equl size. Your nswer must e EXACT--deciml points ANYWHERE will e mrked incorrect (so ou must convert those growth constnts). Time t which the popultions re equl= Prt 2. Wht is the common vlue of the two popultions t tht time? Either enter n ect epression or round to the nerest whole numer. Popultion= Emple 5: Chemists use numer denoted ph to descrie quntittivel the cidit of sicit of solutions. B definition, ph = -[H + ], where [H + ] is the hdrogen ion concentrtion in moles per liter. Mn solutions hve ph etween 4 nd 13. Find the corresponding rnge of [H + ]. < [H + ] <