Introduction to exponential functions and models

Adaptive game version

Go to Part A: Introduction to exponential functions and models

This tutorial: Part B: The number e and exponential growth and decay

(This topic is also in Section 2.2 in Finite Mathematics and Applied Calculus)

#[I don't like this new tutorial. Take me back to the older tutorial!][No me gusta este nueve tutorial. ¡Regresame al tutorial más viejo (solo en inglés)!]#

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In the %%partAtut we looked at exponential functions in general. However, there is a famous mathematical quantity called $e$ that is frequently used in the discussion of exponential functions, and this tutorial is devoted to recasting a lot of the Part A tutorial using $e$.

The number e

Suppose we invest &D&1 in the bank for 1 year at 100% interest. If the interest is compounded (that is, added to the account) once a year, then we will have &D&2 at the end of the year.

If it is compounded twice a year, then after 6 months we will have half a year's interest added: $1+\frac{1}{2}$, and then half of this quantity added at the end of the year:

$\displaystyle \left(1 + \frac{1}{2}\right) + \frac{1}{2}\left(1 + \frac{1}{2}\right)$ \t $\displaystyle = \left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{2}\right)$ \gap[40] \t #[Factor out][Factorizar]# $\left(1 + \frac{1}{2}\right)$. \\ \t $\displaystyle =\left(1 + \frac{1}{2}\right)^2$.

If the interest is compounded three times a year, a similar argument will show that we will have

$\displaystyle {}=\left(1 + \frac{1}{3}\right)^3$ dollars at the end of the year and if the interest is compounded $m$ times a year, you will have

This is the compound interest formula for a 100% rate of interest.

#[Compounding more times a year increases the value of $m$. For very large values of $m$, something seems to happen; try experimenting by entering larger and larger values of $m$ (positive whole numbers only, please!).][Componiendo más veces al año aumenta el valor de $m$. Para valores muy grandes de $m$, algo interesante parece suceder; intenta experimentar ingresando valores cada vez mayores de $m$ (solo núímeros enteros positivos, ¡por favor!).]# #[The number to which these quantities are approaching for larger and larger $m$ is Euler's number][El número al que se acercan estas cantidades para $m$ cada vez mayor es el número de Euler]#

$e = 2.71828182845904523536028747135266...$ #[Press the "Show /Hide e" button to see an extremely accurate approximation (to 10,000 decimal places!) of $e$.][Presiona el botón "Mostrar/Ocultar e" para ver una aproximación extremadamente precisa (¡a 10,000 lugares decimales!) De $ e $.]#

2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427427466391932003059921817413596629043572900334295260595630738132328627943490763233829880753195251019011573834187930702154089149934884167509244761460668082264800168477411853742345442437107539077744992069551702761838606261331384583000752044933826560297606737113200709328709127443747047230696977209310141692836819025515108657463772111252389784425056953696770785449969967946864454905987931636889230098793127736178215424999229576351482208269895193668033182528869398496465105820939239829488793320362509443117301238197068416140397019837679320683282376464804295311802328782509819455815301756717361332069811250996181881593041690351598888519345807273866738589422879228499892086805825749279610484198444363463244968487560233624827041978623209002160990235304369941849146314093431738143640546253152096183690888707016768396424378140592714563549061303107208510383750510115747704171898610687396965521267154688957035035402123407849819334321068170121005627880235193033224745015853904730419957777093503660416997329725088687696640355570716226844716256079882651787134195124665201030592123667719432527867539855894489697096409754591856956380236370162112047742722836489613422516445078182442352948636372141740238893441247963574370263755294448337998016125492278509257782562092622648326277933386566481627725164019105900491644998289315056604725802778631864155195653244258698294695930801915298721172556347546396447910145904090586298496791287406870504895858671747985466775757320568128845920541334053922000113786300945560688166740016984205580403363795376452030402432256613527836951177883863874439662532249850654995886234281899707733276171783928034946501434558897071942586398772754710962953741521115136835062752602326484728703920764310059584116612054529703023647254929666938115137322753645098889031360205724817658511806303644281231496550704751025446501172721155519486685080036853228183152196003735625279449515828418829478761085263981395599006737648292244375287184624578036192981971399147564488262603903381441823262515097482798777996437308997038886778227138360577297882412561190717663946507063304527954661855096666185664709711344474016070462621568071748187784437143698821855967095910259686200235371858874856965220005031173439207321139080329363447972735595527734907178379342163701205005451326383544000186323991490705479778056697853358048966906295119432473099587655236812859041383241160722602998330535370876138939639177957454016137223618789365260538155841587186925538606164779834025435128439612946035291332594279490433729908573158029095863138268329147711639633709240031689458636060645845925126994655724839186564209752685082307544254599376917041977780085362730941710163434907696423722294352366125572508814779223151974778060569672538017180776360346245927877846585065605078084421152969752189087401966090665180351650179250461950136658543663271254963990854914420001457476081930221206602433009641270489439039717719518069908699860663658323227870937650226014929101151717763594460202324930028040186772391028809786660565118326004368850881715723866984224220102495055188169480322100251542649463981287367765892768816359831247788652014117411091360116499507662907794364600585194199856016264790761532103872755712699251827568798930276176114616254935649590379804583818232336861201624373656984670378585330527583333793990752166069238053369887956513728559388349989470741618155012539706464817194670834819721448889879067650379590366967249499254527903372963616265897603949857674139735944102374432970935547798262961459144293645142861715858733974679189757121195618738578364475844842355558105002561149239151889309946342841393608038309166281881150371528496705974162562823609216807515017772538740256425347087908913729172282861151591568372524163077225440633787593105982676094420326192428531701878177296023541306067213604600038966109364709514141718577701418060644363681546444005331608778314317444081194942297559931401188868331483280270655383300469329011574414756313999722170380461709289457909627166226074071874997535921275608441473782330327033016823719364800217328573493594756433412994302485023573221459784328264142168487872167336701061509424345698440187331281010794512722373788612605816566805371439612788873252737389039289050686532413806279602593038772769778379286840932536588073398845721874602100531148335132385004782716937621800490479559795929059165547050577751430817511269898518840871856402603530558373783242292418562564425502267215598027401261797192804713960068916382866527700975276706977703643926022437284184088325184877047263844037953016690546593746161932384036389313136432713768884102681121989127522305625675625470172508634976536728860596675274086862740791285657699631378975303466061666980421826772456053066077389962421834085988207186468262321508028828635974683965435885668550377313129658797581050121491620765676995065971534476347032085321560367482860837865680307306265763346977429563464371670939719306087696349532884683361303882943104080029687386911706666614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%%Q: That is for &D&1 invested for one year. What if we invest an amount $\$P$ for $t$ years at an interest rate of $r$, compounded continuously, what will be the accumulated amount $A$ at the end of that period?
%%A: #[If you look at the %%compoundInteresttut you will find that the expression $\left(1 + \frac{1}{m}\right)^m$, for &D&1 invested for one year at 100% interest compounded $m$ times a year comes from a more general formula: ][Si a el %%compoundInteresttut, encontrarás que la expresión $\left(1 + \frac{1}{m}\right)^m$, para &D&1 invertido durante un año al 100% de interés compuesto $m$ veces por año proviene de una fórmula mas general:]#

#[Compound interest formula][Formula para interés compuesto]#
#[If an investment earns interest an annual interest rate of $r$ compounded $m$ times per year for $t$ years, then the value of the investment after $t$ years will be ][Si una inversión gana intereses a una tasa anual de $r$ pero se compone $m$ veces por año, entonces el valor de la inversión después de $t$ años será]#

$P\left(1+\frac{r}{m}\right)^{mt}\qquad$ \t #[Invest $P$ for $t$ years at $r$% compounded $m$ times per year.][Invertir $P$ para $t$ al $r$% compuesto $m$ veces por año.]#

#[We can use a little algebra rewrite this expression to show the the expression leading to $e$ more clearly:][Podemos usar un poco de álgebra para reescribir esta expresión para mostrar la expresión que conduce a $e$ mas claramente:]#

$P\left(1+\frac{r}{m}\right)^{mt}$ \t ${}= P\left(1+\frac{1}{m/r}\right)^{\frac{m}{r}.rt}$ \\ \t ${}=P\left(1+\frac{1}{M}\right)^{M.rt}$ \t #[Write $\frac{m}{r}$ as $M$.][Escribe $\frac{m}{r}$ como $M$.]# \\ \t ${}=P\left[\left(1+\frac{1}{M}\right)^{M}\right]^{rt}$ \t #[Law of exponents:][Ley de exponentes:]# $b^{Mq} = (b^M)^q$

#[So, as $m$ gets larger an larger, so does $M$, and the quantity in square brackets appraches $e$. We therefore have the following:][Así, a medida que $m$ crece cada vez más, también lo hace $M$, y la cantidad en corchetes se aproxima a $e$. Por lo tanto, tenemos lo siguiente:]#

The number e and continuous compounding

The number e and continuous compounding

#[The number $e$ is the limiting value of the quantities $\left(1 + \dfrac{1}{m}\right)^m$ as $m$ gets larger and larger and has the value 2.71828182845904523536….][El número $e$ es el valor límite de las cantidades $\left(1 + \dfrac{1}{m}\right)^m$ a medida que $m$ se hace cada vez más grande y tiene el valor 2,71828182845904523536….]#

#[If $\$P$ is invested at an annual interest rate $r$ compounded continuously, the accumulated amount after $t$ years is][Si $\$P$ se invierte a una tasa de interés anual $r$ compuesta continuamente, la cantidad acumulada después de $t$ años es]#

$A(t) = Pe^{rt}.\qquad$ P*e^(r*t) #[or][o]# P*EXP(r*t)

Natural logarithm

The logarithm with base $e$ is called the natural logarithm and is often written as $\ln$. Thus, for instance, $\ln 4$ means $\log_e 4$.

%%Examples

1. If &D&500 is invested at an annual rate of 4% compounded continuously, then at the end of 15 years the investment will be worth

$A(15)^{ }$\t $= 500e^{(0.04)(15)}\qquad$\t 500*e^(0.04*15) %%or 500*EXP(0.04*15) \\ \t $\approx \$911.06$

2. If &D&1 is invested at an annual rate of 100% compounded continuously, then at the end of $x$ years the investment will be worth

$A(x)= e^{x}$ #[dollars][dólares]#.

#[One for you][Uno para t]#

3. How long to invest
If you invest &D&1,000 ar 5% interest compounded continuously, how many years will it take for the investment to reach one million dollars? Putting all this information in the formula gives

$1{,}000{,}000 = 1{,}000e^{0.05t}\qquad$\t #[The unknown is $t$.][Lo desconocido es $t$.]# \\ $1{,}000 =e^{0.05t}\qquad$\t #[Divide both sides by 1,000.][Dividir ambos lados por 1,000.]# \\ $0.05t = \ln 1{,}000 \approx 6.90776$ \t #[Rewrite in logarithmic form.][Reescribir en forma logarítmica.]# \\ $t \approx \dfrac{6.90776}{0.05} \approx 138$ #[years][años]#

#[One for you][Uno para t]#

The number e and exponential functions

If we write the continuous compounding formula $A(t) = Pe^{rt}$ as $A(t) = P(e^r)^t$, we see that $A(t)$ is an exponential function of $t$, where the base is $b = e^r$, so we have really not introduced a new kind of function. In fact, every exponential function $f(t) = Ab^t$ can be written in this way:

$Ab^t$ \t ${}= A(e^{\ln b})^t \qquad$ \t #[When you raise $e$ to $\log_e b$ you obtain $b$.][Cuando elevas $e$ al $\log_e b$ obtienes $b$.]# \t #[(See the logarithm identities in the %%logstut.)][(Ve los identidades logarítmicas en el %%logstut.)]# \\ \t ${}=Ae^{(\ln b)t}$, \t #[Laws of exponents][Leyes de los exponentes]#

#[which has the form][que tiene la forma]# $Ae^{rt}$, #[where][donde]# $r = \ln b$

Exponential functions expressed in terms of e

#[We can write any exponential function in the following form:][Podemos escribir cualquier función exponencial de la siguiente forma:]#

$f(x) = Ae^{rx}$,

#[where $A$ and $r$ are constants. If $r$ is positive, $f$ models exponential growth; if $r$ is negative, $f$ models exponential decay.][donde $A$ y $r$ son constantes. Si $r$ es positivo, $f$ modela un crecimiento exponencial; si $r$ es negativo, $f$ modela el decaimiento exponencial.]# #[If $r$ is positive, it is called the growth constant. If $r$ is negative, its absolute value is called the decay constant.][Si $r$ es positivo, se denomina constante de crecimiento. Si $r$ es negativo, su valor absoluto se denomina constante de desintegración.]#

%%Examples

1. $f(x)$ \t ${}= 100e^{0.15x}\qquad $ \t #[Exponential growth with growth constant 0.15.][Crecimeinto exponencial con constante de crecimiento 0.15]# \\ 2. $f(t) $ \t ${}=Ae^{-0.00012101t}\qquad$ \t #[Exponential decay of carbon 14 with decay constant $0.00012101$][Desintegración exponencial del carbono 14 con constante de desintegración $-0.00012101$]# \\ 3. $f(t) $ \t ${}=100e^{0.15t}= 100(e^{0.15})^t\qquad$ \\ \t ${}= 100(1.1618)^t$ \t #[Converting $Ae^{rt}$ to the form $Ab^t$][Convertiendo $Ae^{rt}$ a la forma $Ab^t$]#

Half-life and doubling time

#[We have seen that in exponential functions $f(t)$, every increase of $t$ by the same fixed amount results in an increase of $f(t)$ by the same fixed multiple, so if $Q(t) = Q_0e^{kt}$ is an exponential growth functions and doubles after a certain time $t= t_d$, then it will continue to double for each subsequent inrcease of time by that same amount $t_d$. So, we call $t_d$ the doubling time..][Hemos visto que en funciones exponenciales $f(t) = Ab^t $, cada aumento de $t$ por la misma cantidad fija da como resultado un aumento de $f(t)$ por el mismo múltiplo fijo, entonces si $f(t)$ es una función de crecimiento exponencial y se duplica después de un cierto tiempo $t = t_d $, entonces continuará duplicándose para cada aumento subsiguiente de tiempo en la misma cantidad $t_d$. Entonces, llamamos $t_d$ el tiempo de duplicación .]#

#[To calculate the doubling time, we use the fact that $Q(t + t_d)$ must be twice $Q(t)$:][Para calcular el tiempo de duplicación, usamos el hecho que $Q(t + t_d)$ debe ser el doble de $Q(t)$:]#

\t $Q(t+t_d) = 2Q(t)$ \\ \t $Q_0e^{k(t+t_d)} = 2Q_0e^{kt}$ \\ \t $Q_0e^{kt}e^{kt_d} = 2Q_0e^{kt} \qquad $ \t #[Laws of exponents][Leyes de los exponentes]# \\ \t $e^{kt_d} = 2$ \t #[Cancel the][Cancela la]# $Q_0e^{kt}$. \\ \t $kt_d = \ln 2$ \t #[Logarithmic form][Forma logarítmico]# \\ \t $t_d = \dfrac{\ln 2}{k}$ \t #[Formula for doubling time][Formula para tiempo de duplicación]#

#[Similarly, if $Q(t)$ is exponential decay $Q(t) = Q_0e^{-kt}$, then there is a fixed half-life $t_h$, the time it takes for the quantity to halve in size. A similar calculationto that above shows that][De manera similar, si $Q(t)$ es un decaimiento exponencial $Q(t) = Q_0e^{-kt}$, entonces hay una vida media $t_h$ fija, el tiempo que tarda la cantidad a reducir a la mitad. Un cálculo similar al anterior muestra que]#

\\ \t $t_h = \dfrac{\ln 2}{k}$ \t #[Formula for half-life][Formula para tiempo medio]#

#[Multilying b $k$ gives the following.][Multiplicar por $k$ da el siguiente.]#

Half-life and doubling time

For a quantity undergoing exponential growth, $Q(t)=Q_0e^{kt}$, the value of $Q$ doubles every $t_d$ units of time, where #

$t_dk = \ln2$. \t \gap[40] Doubling time × Growth constant = ln 2

For a quantity undergoing exponential decay, $Q(t)=Q_0e^{-kt}$, the value of $Q$ halves every $t_h$ units of time, where the half-life an decay constant are related by #

$t_hk = \ln2$. \t \gap[40] Half-life × Decay constant = ln 2

Now try some of the exercises in Section 2.2 in Finite Mathematics and Applied Calculus. or move ahead by pressing "Next tutorial" on the sidebar.