Unit 3, lesson 44

Exponential Growth and Decay Models

How to build and read exponential models from an initial value and a growth factor, how discrete and continuous rates compare, where half-life and doubling time come from, and when such a model stops being believable.

  • Theory part 87
  • Practice part 88
  • Models

Money in a savings account, a colony of bacteria and the carbon-14 in an old bone have one thing in common. Over equal stretches of time each is multiplied by the same factor. That single fact makes them exponential, and it is enough to answer questions such as when the money doubles or how old the bone is.

Key terms

Growth and decay model

Definition An exponential model describes a quantity QQ at time tt by

Q(t)=a bt,Q(t) = a\,b^{t},

where a>0a > 0, b>0b > 0, b≠1b \ne 1 and tt is measured in a fixed unit of time. It is a growth model when b>1b > 1 and a decay model when 0<b<10 < b < 1. Over every time step of length hh the quantity is multiplied by the same factor,

Q(t+h)Q(t)=bh.\frac{Q(t + h)}{Q(t)} = b^{h}.

In plain words Equal steps in time multiply the quantity by equal factors. Growth means the factor is more than 1, and decay means it is between 0 and 1.

Example A culture starts with 500 bacteria and triples every 4 hours. One hour is a quarter of a tripling period, so after tt hours

Q(t)=500⋅3t/4,Q(t) = 500 \cdot 3^{t/4},

which has the form a bta\,b^t with b=31/4≈1.316b = 3^{1/4} \approx 1.316. After 10 hours the model gives 500⋅32.5≈7794500 \cdot 3^{2.5} \approx 7794 bacteria.

Initial value

Definition The initial value of the model Q(t)=a btQ(t) = a\,b^t is Q(0)=aQ(0) = a, the amount at the moment chosen as t=0t = 0. Moving that moment changes aa but not bb. If the quantity is known at two times t1<t2t_1 < t_2, dividing one value by the other gives

b t2−t1=Q(t2)Q(t1),b^{\,t_2 - t_1} = \frac{Q(t_2)}{Q(t_1)},

which determines bb, and then a=Q(t1) b−t1a = Q(t_1)\,b^{-t_1}.

In plain words The initial value is the height at which the graph crosses the vertical axis. It need not be measured directly, because two data points let you run the model back to t=0t = 0.

Example A quantity is 180 at t=2t = 2 and 1440 at t=5t = 5. Then b3=1440÷180=8b^3 = 1440 \div 180 = 8, so b=2b = 2, and a=180⋅2−2=45a = 180 \cdot 2^{-2} = 45. The model is Q(t)=45⋅2tQ(t) = 45 \cdot 2^t, and indeed 45⋅25=144045 \cdot 2^5 = 1440.

Percent rate and growth factor

Definition Suppose a quantity changes by the same percentage in each unit of time, with the rate rr written as a decimal, r>−1r > -1 and r≠0r \ne 0. Then

Q(t)=a(1+r)t.Q(t) = a(1 + r)^t.

The number b=1+rb = 1 + r is the growth factor per unit of time. A positive rr means growth and a negative rr means decay, so a loss of 15 percent per year has r=−0.15r = -0.15 and b=0.85b = 0.85. In the other direction, a growth factor bb corresponds to the percent rate r=b−1r = b - 1.

In plain words Add the rate to 1 to get the number the quantity is multiplied by each period. A 15 percent loss leaves 85 percent.

Example A car bought for 24000 dollars loses 15 percent of its value each year. After 3 years it is worth

24000⋅0.853=14739 dollars,24000 \cdot 0.85^{3} = 14739 \text{ dollars,}

not the 13200 dollars that three flat losses of 3600 dollars would leave. The bacteria above have the growth factor 31/4≈1.3163^{1/4} \approx 1.316 per hour, so they grow by about 31.6 percent per hour.

Discrete versus continuous rate

Definition A discrete rate rr applies the change once per unit of time, which gives Q(t)=a(1+r)tQ(t) = a(1 + r)^t. A continuous rate k≠0k \ne 0 splits the change into ever smaller steps. Applying the rate k/nk/n a total of nn times per unit of time gives the factor (1+k/n)n(1 + k/n)^n, which tends to eke^k as n→∞n \to \infty for every real kk. For k>0k > 0 this is the continuous compounding limit from the lesson on the natural base e, and the same limit holds for k<0k < 0. So a continuous rate gives the model

Q(t)=a ekt,Q(t) = a\,e^{kt},

with growth when k>0k > 0 and decay when k<0k < 0. The two describe the same function exactly when ek=1+re^k = 1 + r, that is

k=ln⁡(1+r),r=ek−1.k = \ln(1 + r), \qquad r = e^{k} - 1.

In calculus kk turns out to be the relative rate of change, the rate of change of QQ divided by QQ, as in the lesson on exponential growth and decay.

In plain words A continuous rate of 5 percent per year adds slightly more than 5 percent over a year, because the growth keeps feeding itself during the year. For small rates the two numbers are close but not equal.

Example A continuous rate k=0.05k = 0.05 per year gives the annual factor e0.05≈1.05127e^{0.05} \approx 1.05127, an increase of about 5.127 percent per year. Starting from 1000, after 10 years the continuous model gives 1000e0.5≈1648.721000e^{0.5} \approx 1648.72, while 5 percent applied once a year gives 1000⋅1.0510≈1628.891000 \cdot 1.05^{10} \approx 1628.89.

Half-life

Definition For a decay model Q(t)=a btQ(t) = a\,b^t with 0<b<10 < b < 1, the half-life is the time T>0T > 0 with Q(t+T)=12Q(t)Q(t + T) = \frac{1}{2}Q(t) for every tt. It is the solution of bT=12b^T = \frac{1}{2}, namely

T=ln⁡12ln⁡b=ln⁡2−ln⁡b,T = \frac{\ln \frac{1}{2}}{\ln b} = \frac{\ln 2}{-\ln b},

and for Q(t)=a ektQ(t) = a\,e^{kt} with k<0k < 0 it is T=ln⁡2/(−k)T = \ln 2 / (-k). The half-life depends neither on aa nor on tt, and the model can be written as

Q(t)=a(12)t/T.Q(t) = a\left(\frac{1}{2}\right)^{t/T}.

In plain words Whatever amount you start with, half of it is left after one half-life, a quarter after two, an eighth after three, and so on.

Example Carbon-14 has a half-life of about 5730 years. A bone that has 30 percent of the carbon-14 it had when the animal died satisfies (12)t/5730=0.3\left(\frac{1}{2}\right)^{t/5730} = 0.3. Taking ln⁡\ln of both sides,

t=5730⋅ln⁡0.3ln⁡0.5≈9953 years.t = 5730 \cdot \frac{\ln 0.3}{\ln 0.5} \approx 9953 \text{ years.}

The half-life itself is rounded, so about 9950 years is a fair answer.

Doubling time

Definition For a growth model Q(t)=a btQ(t) = a\,b^t with b>1b > 1, the doubling time is the time D>0D > 0 with Q(t+D)=2Q(t)Q(t + D) = 2Q(t) for every tt, namely

D=ln⁡2ln⁡b,D = \frac{\ln 2}{\ln b},

and for Q(t)=a ektQ(t) = a\,e^{kt} with k>0k > 0 it is D=ln⁡2/kD = \ln 2 / k. The model can be written as Q(t)=a⋅2t/DQ(t) = a \cdot 2^{t/D}.

In plain words The doubling time is the same no matter when you start counting. For a growth of a few percent per period, dividing 70 by the percent rate gives a good estimate, because ln⁡2≈0.693\ln 2 \approx 0.693 and ln⁡(1+r)\ln(1 + r) is close to rr when rr is small.

Example At 3 percent growth per year,

D=ln⁡2ln⁡1.03≈23.45 years,D = \frac{\ln 2}{\ln 1.03} \approx 23.45 \text{ years,}

and the estimate 70÷3≈23.370 \div 3 \approx 23.3 years is close. At a continuous rate of 3 percent the doubling time is ln⁡2/0.03≈23.1\ln 2 / 0.03 \approx 23.1 years.

Compound interest

Definition A principal P>0P > 0 at a nominal annual rate r>0r > 0, compounded nn times per year with nn a positive integer, grows after tt years to

A(t)=P(1+rn)nt,A(t) = P\left(1 + \frac{r}{n}\right)^{nt},

and to A(t)=PertA(t) = Pe^{rt} under continuous compounding, which is the model a ekta\,e^{kt} with a=Pa = P and continuous rate k=rk = r. These are growth models with annual growth factors (1+rn)n\left(1 + \frac{r}{n}\right)^n and ere^r. Interest compounded nn times a year is credited only at the end of each period, so the formula gives the actual balance only when ntnt is a whole number of periods.

In plain words Compound interest is exponential growth at a rate set by the bank. Questions about how long something takes lead to exponential equations, solved with logarithms.

Example How long does 5000 dollars at 4 percent compounded monthly take to grow to 8000 dollars? Setting 5000(1+0.0412)12t=80005000\left(1 + \frac{0.04}{12}\right)^{12t} = 8000 and taking ln⁡\ln of both sides gives

12t=ln⁡1.6ln⁡(1+0.0412)≈141.2.12t = \frac{\ln 1.6}{\ln\left(1 + \frac{0.04}{12}\right)} \approx 141.2.

Interest arrives once a month, so the balance first passes 8000 dollars at the end of month 142, which is after about 11.8 years.

Logarithmic scales

Definition A logarithmic scale places a positive quantity xx at the position log⁡x\log x, so each unit step along the scale multiplies the quantity by 1010. A semi-log graph has a logarithmic vertical axis. For an exponential model

log⁡Q(t)=log⁡a+tlog⁡b,\log Q(t) = \log a + t \log b,

which is a linear function of tt. So on a semi-log graph an exponential model is a straight line with slope log⁡b\log b and vertical intercept log⁡a\log a. The pH scale, sound levels in decibels and earthquake magnitudes are logarithmic scales too. Each uses a constant multiple of a logarithm, such as −log⁡x-\log x for pH or 10log⁡II010 \log \frac{I}{I_0} for decibels, so on each of them a fixed step stands for a fixed ratio.

In plain words On a logarithmic scale equal distances mean equal ratios, not equal differences. That is why exponential growth, which bends sharply upward on an ordinary graph, becomes a straight line.

Example The bacteria from the first example triple every 4 hours, so their common logarithm goes up by the same amount every 4 hours.

tt in hoursQ(t)Q(t)log⁡Q(t)\log Q(t)
005005002.6992.699
44150015003.1763.176
88450045003.6533.653
121213500135004.1304.130

The values of QQ jump by 1000, then 3000, then 9000, while log⁡Q\log Q rises by log⁡3≈0.477\log 3 \approx 0.477 each time. On a semi-log graph the model is the line with slope 14log⁡3≈0.119\frac{1}{4}\log 3 \approx 0.119 per hour.

Interpreting model assumptions

Definition An exponential model assumes that the quantity changes by the same percentage in every time step of equal length, for as long as the model is used. The growth factor stays fixed, nothing limits the size of the quantity, and in a discrete model the change happens once per period. A model is fitted to data from a limited range of times, and its predictions outside that range, called extrapolations, are only as good as these assumptions.

In plain words Ask what would have to stay constant for the model to hold, and for how long that is believable. Exponential growth always outruns food, space and money eventually.

Example The bacteria model predicts that after one week, which is 168 hours, there are

500⋅3168/4=500⋅342≈5.5×1022\begin{aligned} 500 \cdot 3^{168/4} &= 500 \cdot 3^{42} \\[0.4em] &\approx 5.5 \times 10^{22} \end{aligned}

bacteria. At roughly 10−1210^{-12} grams each, that is around 55 million kilograms of bacteria from one small culture, so the assumption of unlimited food and space broke down long before. The logistic model builds such a limit into the formula.

Check yourself

  1. A radioactive substance has a half-life of 8 days. Starting with 200 milligrams, how much is left after 20 days, and when are 10 milligrams left?
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The amount after tt days is

Q(t)=200(12)t/8,Q(t) = 200\left(\frac{1}{2}\right)^{t/8},

so after 20 days Q(20)=200⋅2−2.5≈35.4Q(20) = 200 \cdot 2^{-2.5} \approx 35.4 milligrams. Setting Q(t)=10Q(t) = 10 gives (12)t/8=0.05\left(\frac{1}{2}\right)^{t/8} = 0.05, so

t=8⋅ln⁡0.05ln⁡0.5=8ln⁡20ln⁡2≈34.6 days.\begin{aligned} t &= 8 \cdot \frac{\ln 0.05}{\ln 0.5} = \frac{8 \ln 20}{\ln 2} \\[0.8em] &\approx 34.6 \text{ days.} \end{aligned}
  1. A population grows continuously at 2 percent per year. Find its doubling time and the percent by which it grows in one year.
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The doubling time is

D=ln⁡20.02≈34.7 years.D = \frac{\ln 2}{0.02} \approx 34.7 \text{ years.}

In one year the population is multiplied by e0.02≈1.0202e^{0.02} \approx 1.0202, an increase of about 2.02 percent.

  1. The sound level of a sound with intensity II is L=10log⁡II0L = 10 \log \frac{I}{I_0} decibels, where I0I_0 is a fixed reference intensity. How many times as intense as a 60 decibel sound is a 90 decibel sound?
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Let I1I_1 be the intensity of the 90 decibel sound and I2I_2 that of the 60 decibel sound. Subtracting the two levels gives

10log⁡I1I0−10log⁡I2I0=10log⁡I1I2=90−60=30,\begin{aligned} &10 \log \frac{I_1}{I_0} - 10 \log \frac{I_2}{I_0} \\[0.8em] &\quad = 10 \log \frac{I_1}{I_2} = 90 - 60 = 30, \end{aligned}

so the logarithm of the ratio is 33 and the ratio is 103=100010^3 = 1000.

How this lesson connects