[PDF] exponential to log conversion calculator
[PDF] exponential to natural log calculator
[PDF] exponentielle de 0 7
[PDF] exponentielle de 0 valeur
[PDF] exponentielle de x u003d0
[PDF] exponentielle traduction en arabe
[PDF] exponents and logarithms worksheet
[PDF] exponents surds and logarithms pdf
[PDF] exposition art contemporain paris janvier 2019
[PDF] exposition art paris avril 2019
[PDF] exposition art paris mai 2019
[PDF] exposition art paris mars 2019
[PDF] exposition cite architecture paris
[PDF] exposition de peinture paris 2020
[PDF] exposition mode paris avril 2019
58Chapter 2. Nonlinear Functions (LECTURE NOTES 4)
2.4 Exponential Functions
An exponential function is given by
f(x) =ax wherexis any real number,a >0 anda?= 1. If basea=e≈2.718, the exponential function becomes the (natural) exponential function,f(x) =ex. Related to this, as mgets larger, (1 +1m)mapproaches e. Exponential functions are used in financial formulas. If principal (present value) amountPis invested at interest raterper year over timet,simple interest,I, is I=Prt. IfPis invested at interest raterper year, compoundedmtimes per year fortyears,compound amountis A=P?
1 +rm?
mt. If interest rateris compoundedcontinuously, compound amount aftertyears is
A=Pert.
Exercise 2.4 (Exponential functions)
1.Properties of exponential functions.5
5-5 -5x y (e)(a) (b)(c)(d)
Figure 2.11 (Various Exponential Functions)
(Use WINDOW -5 5 1 -5 5 1 1) (a) Exponential functionf(x) = 2xcorresponds to graph (i)(a)(ii)(b)(iii)(c)(iv)(d)(v)(e) (b) Exponential functionf(x) = 3xcorresponds to graph (i)(a)(ii)(b)(iii)(c)(iv)(d)(v)(e) f(x) = 3xincreases more rapidly thanf(x) = 2x. Section 4. Exponential Functions (LECTURE NOTES 4)59 (c) Exponential functionf(x) = 0.5xcorresponds to graph (i)(a)(ii)(b)(iii)(c)(iv)(d)(v)(e) f(x) = 0.5x=?1 2? x=1x2x=12x= 2-x=f(-x), which is reflection iny-axis tof(x) = 2x. (d) Exponential functionf(x) = 2-x-4corresponds to graph (i)(a)(ii)(b)(iii)(c)(iv)(d)(v)(e) f(x) = 2-x-4= 2-(x+4)=f(-(x+ 4)), which a left translation 4 units fromf(-x) = 2-x. (e) Exponential functionf(x) =-2x+ 2 corresponds to graph (i)(a)(ii)(b)(iii)(c)(iv)(d)(v)(e) -f(x) =-2xis a reflection off(x) = 2xinx-axis and-f(x) + 2 is a translation up 2 units.
2.Exponential function applications: radioactive decay.Quantity (in ounces)
present at timet(in years) is
Q(t) = 500(5-0.2t)
(a) Quantity left int= 8 years is approximately (i)28(ii)38(iii)48ounces.
Calculator: 500?(5?(-0.2?8))
(b) Length of time until quantity reduces to 10 ounces: (i)12.12(ii)12.34(iii)12.67years. Calculator: Type 500?(5?(-0.2?X)) into Y=, set WINDOW 0 30 1 0 500 1 1, GRAPH, TRACE, arrow close to Y= 10. Later, we will use logarithms to solve this sort of question.
3.Compound Interest:A=P?1 +r
m? mt. (a) If $700 is invested at 11% interest compoundedyearly(or annually), cal- culate its value after 8 years.
A=P?1 +r
m? mt= 700?1 +0.111?
1(8)=1513.18/1613.18/1713.18
(b) If $700 is invested at 11% interest compoundedmonthly, calculate its value after 8 years.
A=P?1 +r
m? mt= 700?1 +0.1112? (12)8=1580.88/1680.88/1780.88
4.Compound interest: related questions.
(a)Interest rate,r? i. IfA= 700,P= 15,t= 10 years, interest compounded yearly
SinceA=P?1 +r
m? mt, then 700 = 15?1 +r1?
1(10)or (1 +r)10=70015or taking tenth root of both sides,1 +r=?700
15?
1/10orr=?70015?
1/10-1≈0.15/0.39/0.47.
Calculator: (700/15)?(0.1)-1
60Chapter 2. Nonlinear Functions (LECTURE NOTES 4)
ii. IfA= 700,P= 15,t= 10 years, interest compounded monthly
SinceA=P?1 +r
m? mt, 700 = 15?1 +r12?
12(10)or?1 +r12?
120=70015or taking 120th root of both sides,1 +r
12=?70015?
1/120orr= 12?
?70015?
1/120-1?
≈0.15/0.39/0.47.
Calculator: 12?((700/15)?(1/120)-1)
(b)Principal,P? i. IfA= 700,t= 5 years,r= 0.08 interest compounded yearly
SinceA=P?1 +r
m? mt, 700 =P?1 +0.081? 1(5) orP= 700(1 + 0.08)-5≈476.41/500.00/528.89.
Calculator: 700?1.08?(-5)
ii. IfA= 700,t= 5 years,r= 0.08 interest compounded monthly
SinceA=P?1 +r
m? mt, 700 =P?1 +0.0812? 12(5) orP= 700?1 +0.08 12? -60≈469.85/499.00/518.89.
Calculator: 700?(1 + 0.08/12)?(-60)
5.Compound interest (continuously):A=Pert
(a) If $700 is invested at 11% interest compounded continuously, calculate its value after 8 years.
A=Pert= 700e0.11(8)=1687.63/1967.36/2267.36.
(b) An amount $700 invested at 11% interestcompounded annually($1613.18) isless/greater than $700 invested at 11% interestcompounded monthly($1680.88) isless/greater than $700 invested at 11% interestcompounded continuously($1687.63) after 8 years.
2.5 Logarithmic Functions
Logarithmic functions are related to exponential functions. Assumea >0,a?= 1 and x >0 y= logaxif and only ifay=x(oray-x= 0) where "log ax" is read "logarithm ofxto the basea". If basea=e, the logarithmic function becomes the (natural) logarithmic function,f(x) = logex= lnx; if base a= 10, the logarithmic function becomes the (common) logarithmic function,f(x) = log
10x= log10x. For any positivex,yanda,a?= 1, and any real numberr,
logaxy= logax+ logay
Section 5. Logarithmic Functions (LECTURE NOTES 4)61
loga?x
y?= log ax-logay
logaxr=rlogax
Also, log
aa= 1, loga1 = 0 and logaar=r. The change-of-base theorem for loga- rithms log ax=logbx logba=lnxlna and change-of-base theorem for exponentials is a x=e(lna)x
Exercise 2.5 (Logarithmic Functions)
1.Graphs of Logarithmic Functions.Consider the following graphs of various
logarithm functions. 5 5-5 -5x y (a) (b) (c) (d)
Figure 2.12 (Various Logarithmic Functions)
(Calculator: Use WINDOW -5 5 1 -5 5 1 1) (a) Logarithm functionf(x) = log2xcorresponds to graph (i)(a)(ii)(b)(iii)(c)(iv)(d) Use change-of-base theorem: forf(x) = log2x, type ln(X)/ln(2) into Y=, then GRAPH or your calculator may have MATH logBASE 2 X ENTER (b) Logarithm functionf(x) = log3xcorresponds to graph (i)(a)(ii)(b)(iii)(c)(iv)(d)
Type ln(X)/ln(3) into Y=, then GRAPH
(c) Logarithm functionf(x) = log0.9xcorresponds to graph (i)(a)(ii)(b)(iii)(c)(iv)(d)
Type ln(X)/ln(0.9) into Y=, then GRAPH
62Chapter 2. Nonlinear Functions (LECTURE NOTES 4)
(d) Logarithm functionf(x) = log0.1xcorresponds to graph (i)(a)(ii)(b)(iii)(c)(iv)(d)
Type ln(X)/ln(0.1) into Y=, then GRAPH
(e) In general, thex-intercept off(x) = logaxis (i)0(ii)1(iii)2.
2.Logarithmic function inverse of exponential function
5 5-5 -5 x y y = log xy = 3 x 3 Figure 2.13 (Logarithmic Versus Exponential Function) (a) Functionsy= log3xandy= 3xreflections of one another through (i)y-axis(ii)x-axis(iii)45odegree line.
See figure above.
(b) Inverse function of natural logarithmic functiony= lnxis (i)10x(ii)ex(iii)5x (c) Inverse function of common logarithmic functiony= log10xis (i)10x(ii)ex(iii)5x (d) Sincey= logaxmeansx=ay, theny= log0.5xmeans x= (circle one ormore!) (i)(0.5)y(ii)?1 2? y(iii)1y2y(iv)12y(v)2-y (e) 10 = log
0.2xmeans
x= (one ormore!) (i)(0.2)y(ii)?1 5?
10(iii)110510(iv)1510(v)5-10
(f) 2 = log
0.1xmeans
x= (circle one) (i)2-2(ii)7-0.1(iii)10-2(iv)12-10 (g) Write log
416384 = 7 in exponential form
(i)74= 16384(ii)47= 16384(iii)163844= 7 (h) Write log100000 = 5 in exponential form (i)10000010= 5(ii)105= 100000(iii)510= 100000 (i) Write 5
2= 25 in logarithmic form
(i)log525 = 2(ii)log255 = 2(iii)log225 = 5 Section 5. Logarithmic Functions (LECTURE NOTES 4)63 (j) Write 4 -2=1
16in logarithmic form
(i)log41
16=-2(ii)log-24 =116(iii)log-2116= 4
3.Properties of logarithms.
(a) log bx+ logby-logbz= (i)logbxy z(ii)logbxzy(ii)logbzxy. (b) 5log bx-4logby= (i)logb5x
4y(ii)logbx5y4(iii)logby4x5.
(c) 2log b(x+ 3)-4logb(x-3) = (i)logb2(x+3) (d) log aa= (i)0(ii)1(iii)e. (Hint: log aa=ymeansay=aand soy= ?) (e) log aak= (i)0(ii)1(iii)k. (Hint: log aak=ymeansay=akand soy= ?) (f) log a1 = (i)0(ii)1(iii)k. (Hint: log a1 =ymeansay= 1 and soy= ?) (g) log a-3 = (i)0(ii)1(iii)undefined. (Hint: log a-3 =ymeansay=-3,a >0, and soy= ?) (h) lnx+ lny= (i)lnxy(ii)lnxz y(iii)lnzxy. (i) 5lnx-4lny= (i)ln5x
4y(ii)lnx5y4(iii)lny4x5.
(j) lne= (i)0(ii)1(iii)e (Hint: lne= logee=ymeansey=eand soy= ?) (k) 2log(x+ 3)-4log(x-3) = (i)log102(x+3) (l) log10 = (i)0(ii)1(iii)e (Hint: log
1010 =ymeans 10y= 10 and soy= ?)
4.Working with logarithmic functions.Let
log a2 = 0.245 loga7 = 0.404 (a) log a14 = (circle one or more) (i)loga(2·7)(ii)loga2 + loga7(iii)0.245 + 0.404 = 0.649 (b) log a2
7= (circle one or more)
(i)loga2 + loga7(ii)loga2-loga7(iii)0.245-0.404 =-0.159 (c) log a8
7= (circle one or more)
(i)loga8-loga7 (ii)loga23-loga7 (iii)3loga2-loga7 (iv)3(0.245)-0.404 = 0.331
64Chapter 2. Nonlinear Functions (LECTURE NOTES 4)
(d) log a49 = (circle one or more) (i)loga72(ii)2loga7(iii)2(0.404) = 0.808 (e) log a49a= (circle one or more) (i)loga72a(ii)2loga7 + logaa(iii)2(0.404) + 1 = 1.808 (f) log a3⎷ a= (circle one or more) (i)logaa1/3(ii)13logaa(iii)13(1) =13 (g) log a5⎷ a= (circle one) (i)13(ii)15(iii)17 (h) log a77= (circle one) (i)7(0.404) = 2.828 (ii)7(7) = 49 (iii)77= 823543 (i) log a64 = (circle one) (i)6(0.245) = 1.47(ii)2(6) = 12(iii)26= 64
5.Change-of-base for logarithms and exponentials.Evaluate to 3 decimal points
of accuracy if necessary. (a) log
11345 = (i)2.437(ii)2.447(iii)2.457
Since log
ax=lnx lna, log11345 =ln 345ln 11 (b) log
1.1345 = (i)61.321(ii)61.301(iii)61.311
log
1.1345 =ln 345
ln 1.1 (c) Write 9 x-4using basee: (i)e(ln 9)(x+4)(ii)e(ln9)(x-4)(ii)e(9)(x-4)
Hint:ax=e(lna)(x)
6.Logarithm and exponential equations.
(a) Solve log
981 =yfory.
i.True/False9y= 81 ii.True/False(32)y= 34 iii.True/False32y= 34 iv. and soy= (i)4(ii)6. (iii)2 (b) Solve log q6 =1
2forq.
i.True/Falseq1/2= 6 ii.True/False⎷ q=⎷36 iii. and soq= (i)6(ii)⎷
6. (iii)36
(c) Solve log(x+ 5)(x+ 1) = 1 forx. i.True/False101= (x+ 5)(x+ 1) =x2+ 6x+ 5 ii.True/False0 =x2+ 6x-5 = (x-3)(x-2) iii. and sox= (i)1,3(ii)2,4. (iii)2,3 (d) Solve lnx+ ln2x=-2 forx. Section 5. Logarithmic Functions (LECTURE NOTES 4)65 i.True/Falseln2x2=-2 ii.True/Falsee-2= 2x2 iii.True/False1
2e-2=x2
iv. and sox= (i)1 ⎷2e(ii)12e2(iii)2⎷e. (e) Solve 21
2x= 27 forx.
i.True/Falseln212x= ln27 ii.True/False(2x)ln21 = ln27 iii.True/False2x=ln27 ln21 iv. and sox= (circle one or more) (i)1
2ln 27ln 21(ii)0.541(iii)0.675.
(f) Solve 21
2x= 27 forx.
i.True/Falselog10212x= log1027 ii.True/False(2x)log1021 = log1027 iii.True/False2x=log1027 log1021 iv. and sox= (circle one or more) (i)1 2log
1027log1021(ii)0.541(iii)0.675.
(g) Solvee2x= 27 forx. i.True/Falselne2x= ln27 ii.True/False(2x)lne= ln27 iii.True/False2x=lne ln27 iv. and sox= (circle one or more) (i)1
2·lne(ii)12·1ln 27(iii)0.164
(h) Solvee-0.02x= 12 forx. x= (circle one or more) (i)-1
0.02·lne(ii)-ln 120.02(iii)-16.423
Since lne-0.02x= ln12, then-0.02x= ln12, and sox=ln 12 -0.02=?
7.Compound interest: number of interest periods,n=mt
(a) IfA= 700,P= 15,r= 0.08 interest compounded yearly
SinceA=P?1 +r
m? mt, 700 = 15?1 +0.081? mtor (1 + 0.08)mt=70015or taking natural logs of both sides,ln(1 + 0.08)mt= ln700
15ormtln(1 + 0.08) = ln70015
orn=mt=ln700
15ln1.08≈48/50/52.
Calculator: ln(700/15)/ln(1.08)
(b) IfA= 700,P= 15,r= 0.08 interest compounded monthly
SinceA=P?1 +r
m? mt, 700 = 15?1 +0.0812? mtor?1 +0.0812? mt=70015or taking natural logs of both sides,ln?1 +0.08 12? mt= ln70015or 12tln?1 +0.0812?= ln70015 orn= 12t=ln700 15 ln(1+0.0812)≈563/578/589.
Calculator: ln(700/15)/ln(1 + 0.08/12)
66Chapter 2. Nonlinear Functions (LECTURE NOTES 4)
(c)Doubling time for compound interest: rule of 70, 72.True/FalseIfm= 1, thenA=P?1 +r m? mt=P(1 +r)t, and time to double principal whenP= 1 is given by 2 = (1 +r)tor t=ln2 ln(1 +r)≈ln2r≈0.693r,ifris small r <0.12, doubling time ist≈72 100r.
2.6 Applications: Growth and Decay; Mathemat-
ics of Finance Fory0amount present at timet= 0, let amount present at timetbe y=y0ekt. Ifk >0, thenkis agrowth constantandyis an exponential growth function (used in bacterial growth, for example); ifk <0, thenkis adecay constantandyis an exponential decay function (used in radioactive decay, for example). In addition to thisunboundedmodel, thelimited growth modelis given by y=L-(L-y0)ekt wherek <0 andLis a limit to growth.
Also,effective rate for compound interestis
r E=? a+r m? m-1 which becomesrE=er-1 if interest is compounded continuously. Exercise 6.6 (Applications: Growth and Decay; Mathematicsof Finance)
1.Biological growth function,kandtknown.How many cells will there be after
10 hours, if there are an initial 5000 cells and growth ratek= 0.02?
(a) Theyiny=y0ektdescribes the (i)cell growth rate(ii)initial cell count(iii)cell count (b) They0describes initial cell count given by (i)4(ii)10( (iii)5000 (c) so att= 10,y= 5000e0.02(10)≈(i)6004(ii)6053(iii)6107 Section 6. Applications: Growth and Decay; Mathematics of Finance(LECTURE NOTES 4)67
2.Population decay,kandtknown.How many people will there be after 10 years,
if there are an initial population of 5000 people and the population decays at a rate ofk=-0.02? (a) Theyiny=y0ektdescribes the (i)population decay rate(ii)initial population(iii)population. (b) They0describes initial population given by (i)-0.02(ii)10(iii)5000. (c) att= 10,y= 5000e-0.02(10)≈(i)4094(ii)4154(iii)4254.
3.Biological growth,kunknown andtknown.If cells in a bacterial culture divide
every 4 hours (cell countdoublesevery 4 hours), how many cells will there be after 10 hours, if there are an initial 5000 cells? (a) Sincey0= 5000,y= (i)5000e5000t(ii)5000ekt(iii)kekt (b) (i)True(ii)FalseAtt= 0, initial cell count isP= 5000ek(0)= 5000. (c) Since cell countdoubleseveryt= 4 hours, att= 4, four hours after initial count of 5000 cells, there must be (i)2500(ii)25000(iii)10000cells.
In other words,y= 5000ek(4)= 10000.
(d) Rewrite 5000ek(4)= 10000 ase4k= 2. Taking natural logarithms of both sides, lne4k= ln2 or 4k= ln2 and sok≈(i)0.173(ii)0.456 (e) Att= 10,y= 5000e0.173(10)≈(i)28,204(ii)29,204(iii)30,204
4.Population growth,kunknown andtknown.A population doubles every 25
years. If the population was 10,000 in 2000, what will it be in 2100? (a)y=y0ekt= (i)10000e5000t(ii)10000ekt(iii)kekt. (b) (i)True(ii)FalseAtt= 0, initial population isy= 10000ek(0)= 10000. (c) Since populationdoubleseveryt= 25 years, att= 25, 25 years after initial count of 10000, there must be (i)2500(ii)25000(iii)20000people.
In other words,P= 10000ek(25)= 20000.
(d) Rewrite 10000ek(25)= 20000 ase25k= 2. Taking natural logarithms of both sides, 25k= ln2 and sok=ln2
25≈(i)0.0173(ii)0.0277.
Noticek=ln 2
25=0.693125≈69.312500≈702500and so population doubling is an example of therule of 70.
(e) So, att= 2100-2000 = 100, y= 10000e0.0277(100)≈(i)158,587(ii)159,587(iii)200,587.
5.Radioactive decay,kunknown andtknown.If the half-life of blueberrium is
745 hours, how much of 23 grams of blueberrium will be left after 1045 hours?
(a) Sincey0= 23,P= (i)23e23t(ii)23e-kt(iii)ke-kt. (b) (i)True(iii)FalseAtt= 0,P= 23e-k(0)= 23.quotesdbs_dbs5.pdfusesText_10
[PDF] Changing from Exponential to Logarithmic Form