Chapter 6 of 7

All 7 chapters
  1. 1Exponential functions, readunderstood
  2. 2Logarithms, readunderstood
  3. 3Laws of logarithms, readunderstood
  4. 4Equations with powers, readunderstood
  5. 5The number e and natural logs, readunderstood
  6. 6Growth and decay models, readunderstood
  7. 7Straight lines from logs, readunderstood

M5.2.6 Chapter 6

Growth and decay models

A herd of goats grows by 8% every year. How do we write a rule for its size, and when does such a rule stop being true?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Model a goat herd growing by 8% a year with an exponential
  • Find a half life from a decay model
  • Say when an exponential model stops making sense

A herd of 50 goats grows by 8% each year. The size after one year is 50 × 1.08, because 100% + 8% = 108% = 1.08. After two years it is 50 × 1.08 × 1.08. After t years it is 50 multiplied by 1.08, t times. This is a power. The model is N = 50 × 1.08 to the power t, where N is the number of goats and t is the time in years.

The general rule has a starting value A, a rate r for each time unit, and a time t. The size is y = A(1 + r) to the power t. With r = 0.08 the factor 1 + r is 1.08, a growth factor. For a fall of 20% each year, r = −0.2 and the factor is 0.8. A factor below 1 means decay, and a factor above 1 means growth.

Worked example

A herd of 50 goats grows by 8% each year. Find its size after 5 years.

  1. The growth factor each year is 1 + 0.08 = 1.08.
  2. After t years the size is N = 50 × 1.08 to the power t.
  3. Put t = 5. To 4 decimal places, 1.08 to the power 5 = 1.4693.
  4. N = 50 × 1.4693 = 73.47. A herd has a whole number of goats, so the size is about 73.

Answer: About 73 goats.

A graph with the x axis years and the y axis goats. A curve rises from the point (0, 50) through the point (5, 73), named (5, 73), and on to about 108 at 10 years, getting steeper each year.(0, 50)(5, 73)246810204060801001200yearsgoats
The herd N = 50 × 1.08 to the power t. At t = 5 it is 73.47 goats, about 73.

The same growth can be written with e. Papers print e with a raised kt. This book cannot raise a letter, so it writes exp(kt) for the same thing. In general exp(x) means eˣ, so exp(kt) is e to the power kt. Every positive number b equals exp(ln b). So 1 + r = exp(k), where k = ln (1 + r). Then (1 + r) to the power t is exp(kt), and the model becomes y = A exp(kt). The number k is the continuous growth rate. Note that k is not r: for 8% growth, k = ln 1.08, which is a little below 0.08.

A table with the heads rate r, factor and k. Rows: 2% and 1.02 and 0.0198; 8% and 1.08 and 0.0770; 10% and 1.1 and 0.0953; 25% and 1.25 and 0.2231; 50% and 1.5 and 0.4055.rate rfactork2%1.020.01988%1.080.077010%1.10.095325%1.250.223150%1.50.4055
The continuous rate k = ln (1 + r) is always a little below r. The gap grows as r grows.

Worked example

Write N = 50 × 1.08 to the power t in the form N = A exp(kt).

  1. Write 1.08 as exp(ln 1.08).
  2. Then 1.08 to the power t is exp(t ln 1.08).
  3. Use ln 1.08 = 0.0770 to 3 significant figures. So N = 50 × exp(0.0770t).
  4. So A = 50, the starting size, and k = 0.0770.

Answer: A = 50 and k = 0.0770.

Two times matter. The doubling time T is when the size is twice the start. Then A exp(kT) = 2A, so exp(kT) = 2 and T = ln 2 ÷ k. For the herd, T = 0.6931 ÷ 0.0770 = 9.0 years. The doubling time does not depend on A, so a herd of 5 or 5 000 doubles in the same time. For decay, k is negative. The half-life is when half is left: exp(kT) = 1/2 and T = ln 2 ÷ |k|, where |k| is k without its minus sign.

Worked example

The mass of a sample is M = 80 exp(−0.035t) grams, where t is the time in days. Find (a) the starting mass, (b) the mass after 20 days, (c) the half-life, (d) the time to fall to 10 g.

  1. (a) Put t = 0. Then e⁰ = 1, so M = 80. The starting mass is 80 g.
  2. (b) Put t = 20. Then M = 80 exp(−0.7) = 80 × 0.4966 = 39.7 g to 3 significant figures.
  3. (c) The half-life is T = ln 2 ÷ 0.035 = 0.6931 ÷ 0.035 = 19.8 days.
  4. (d) We need 80 exp(−0.035t) = 10. Divide by 80: exp(−0.035t) = 1/8.
  5. Take ln of both sides: −0.035t = ln (1/8) = −2.0794. So t = 2.0794 ÷ 0.035 = 59.4 days.
  6. Check: 59.4 days is three half-lives, 3 × 19.8 = 59.4, and 80 → 40 → 20 → 10 halves three times.

Answer: (a) 80 g, (b) 39.7 g, (c) 19.8 days, (d) 59.4 days.

A graph with the x axis days and the y axis mass (g). A curve falls from the point (0, 80). Dots mark the mass after each half-life: A at (19.8, 40), B at (39.6, 20) and C at (59.4, 10). Dashed guide lines from each dot meet the axes.ABC20406080204060800daysmass (g)
The mass halves in every 19.8 days. A is (19.8, 40), B is (39.6, 20) and C is (59.4, 10).

Not every quantity decays to nothing. A hot drink cools towards the temperature of the room, and stops there. Its model has a fixed part and a decaying part: θ = A + B exp(−kt). Here A is the temperature it settles at. The term B exp(−kt) is the part that dies away.

Worked example

A hot drink has temperature θ = 20 + 70 exp(−0.05t) °C, with t in minutes. Find (a) the starting temperature, (b) the temperature it approaches, (c) when it reaches 50 °C.

  1. (a) Put t = 0. Then exp(0) = 1, so θ = 20 + 70 = 90 °C.
  2. (b) As t grows, exp(−0.05t) gets close to 0, so θ gets close to 20. The drink cools towards 20 °C and never goes below it.
  3. (c) Put θ = 50: 20 + 70 exp(−0.05t) = 50. Take 20 from both sides: 70 exp(−0.05t) = 30.
  4. Divide by 70: exp(−0.05t) = 3/7. Take ln of both sides: −0.05t = ln (3/7) = −0.8473.
  5. So t = 0.8473 ÷ 0.05 = 16.9 minutes.

Answer: (a) 90 °C, (b) 20 °C, (c) 16.9 minutes.

A graph with the x axis minutes and the y axis temperature (°C). A curve falls from the point (0, 90), named (0, 90), steeply at first and then more slowly towards a dashed horizontal line named y = 20. The curve passes through the point (16.9, 50), named (16.9, 50), with dashed guide lines to the axes.(0, 90)(16.9, 50)y = 20102030405060204060801000minutestemperature (°C)
The drink starts at 90 degrees and cools towards the dashed line at 20 degrees. It reaches 50 degrees after 16.9 minutes.

Often a model is built from data. Two readings of a quantity are enough to find A and k. Put the first reading at t = 0 to get A. Then use the second reading to get k by taking logs.

Worked example

A population of bacteria is 1 200 at t = 0 and 1 900 when t = 3 hours. It follows N = A exp(kt). Find k, and the doubling time.

  1. At t = 0, N = A, so A = 1 200.
  2. At t = 3, 1 900 = 1 200 exp(3k). So exp(3k) = 1 900 ÷ 1 200 = 1.5833.
  3. Take ln of both sides: 3k = ln 1.5833 = 0.4595.
  4. To 3 significant figures, k = 0.4595 ÷ 3 = 0.153 per hour.
  5. The doubling time is ln 2 ÷ k. Keep k unrounded: T = 3 ln 2 ÷ ln 1.5833 = 2.0794 ÷ 0.4595 = 4.525, which is 4.53 hours to 3 significant figures.

Answer: k = 0.153 per hour, and the population doubles every 4.53 hours.

Now ask when a model stops making sense. A model is a rule that fits the data we have. It holds near those data. Far from them it can give nonsense. Exponential growth is the clearest case, because it never slows. In real life food, space, disease and other limits slow the growth, and the curve flattens.

Worked example

The herd model N = 50 × 1.08 to the power t was fitted to counts over the first 5 years. Find N after 100 years. A pasture feeds at most 400 goats. When does the model first pass 400?

  1. After 100 years, N = 50 × 1.08 to the power 100 = 109 988 goats.
  2. That is far more than any pasture can feed. The model is being used far outside its data, which cover only 5 years.
  3. For 400 goats, solve 50 × 1.08 to the power t = 400. Divide by 50: 1.08 to the power t = 8.
  4. Take ln of both sides: t ln 1.08 = ln 8. So t = 2.0794 ÷ 0.0770 = 27.0 years.
  5. After about 27 years the model says 400 goats. From then on it should not be trusted, because the pasture is full.

Answer: N = 109 988 after 100 years, and N reaches 400 after 27.0 years.

A graph with the x axis years and the y axis goats. A curve rises from the point (0, 50), getting steeper. A dashed horizontal line named y = 400 marks the most goats the pasture feeds. The curve meets it at the point (27, 400), named (27, 400).(27, 400)y = 40081624322004006000yearsgoats
The model passes the pasture limit of 400 goats after about 27 years.

See it three ways

With things

The herd starts at 50 goats. Each year it is multiplied by 1.08. After 1 year it is 54, and after 2 years it is 58.32.

As a picture

Three boxes in a row joined by arrows. The boxes hold 50, 54 and 58.32. Each arrow is labelled × 1.08.50× 1.0854× 1.0858.32

In words

Each year the herd is multiplied by the same factor, 1.08. Equal steps of time multiply the size by equal factors. This is the mark of an exponential.

In symbols

N = 50 × 1.08 to the power t. For t = 2 this gives 50 × 1.08 × 1.08 = 58.32.

Try it

A half-life with beans

You need: 64 dried beans, a pen, a cup.

  1. Mark one side of each bean with a dot. Count 64 beans.
  2. Shake them in the cup and throw them on a table. Take away every bean that lands dot upwards. Count what is left.
  3. Throw the beans that are left, take away the dot-up ones, and count again. Do this 5 times.

What you will see: About half of the beans are taken away each throw, so the number left falls like 64, 32, 16, 8, 4, 2. Your numbers will differ a little by chance. One throw is the half-life.

If you cannot try it, this is what happens.
A table with the heads throw and beans left. Rows: 0 and 64; 1 and 32; 2 and 16; 3 and 8; 4 and 4; 5 and 2.throwbeans left064132216384452

A wrong idea many people have

Growth by 8% a year means N = 50 × 8 to the power t, or N = 50 × 0.08 to the power t.

Why it feels right: The number 8 is in the question, and 0.08 is how a calculator wants 8%.

The test that breaks it: Put t = 1. With 8 the herd is 50 × 8 = 400 after one year, which is far more than 8% growth. With 0.08 the herd is 50 × 0.08 = 4, which has shrunk. The right answer after one year is 54.

What is true: The factor is 1 + 0.08 = 1.08: the whole herd (100%) plus 8% more. Catch the slip by putting t = 1 into your model and checking it gives 8% more than the start.

A wrong idea many people have

For 8% growth, k = 0.08 in N = A exp(kt).

Why it feels right: Both r and k are called a rate, and 0.08 is the number in the question.

The test that breaks it: With k = 0.08, after one year the factor is exp(0.08) = 1.0833, which is 8.33% growth, not 8%. The right value is k = ln 1.08 = 0.0770, and exp(0.0770) = 1.08.

What is true: The rate r goes with the form (1 + r) to the power t, and k = ln (1 + r) goes with the form exp(kt). They are close for small r but not equal. Convert with k = ln (1 + r).

A wrong idea many people have

After two half-lives everything has gone.

Why it feels right: One half-life takes away half, and two halves make a whole.

The test that breaks it: Take 80 g. After one half-life, 40 g is left. After the second half-life, half of 40 g goes, and 20 g is left. After the third, 10 g.

What is true: Each half-life halves what is left, not what you started with. The amount gets smaller and smaller but never reaches 0. After n half-lives the fraction left is (1/2) to the power n.

A wrong idea many people have

If the model fits the first 5 years well, it will fit 100 years.

Why it feels right: A good fit looks like proof that the rule is the real one.

The test that breaks it: The herd model fitted to 5 years gives 109 988 goats after 100 years. No pasture can feed that many, so the model is wrong there.

What is true: A model is trusted near the data it was built from. Far outside them, things the model ignores, such as food, space and disease, matter. Always say how far a model is used, and ask if the rate can really stay the same.

Key words

growth factor
the number you multiply by each time step, 1 + r. A factor above 1 is growth, and below 1 is decay.
doubling time
the time for a growing quantity to become twice as big. T = ln 2 ÷ k.
half-life
the time for a decaying quantity to fall to half. T = ln 2 ÷ |k|.
model
a rule that fits data and predicts new values, such as N = 50 × 1.08 to the power t.

Back to your guess

The answer is C: About 108 goats, because each year the herd is multiplied by 1.08.

The herd starts at 50 and each year it grows by 8% of its size at that time. So the number added grows too. After 10 years the size is 50 × 1.08 ten times, which is about 108. Adding 4 goats a year would give 90, and 540 comes from multiplying by 10 instead of using a power.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A population of 200 grows by 5% each year. After x years it is P. Which rule is right?

    Tap your answer.

    1. Not this one. You used 5 as the factor. A factor of 5 means 500% each year. Growth of 5% is a factor 1 + 0.05. Go to M5.2.6. Try another.

    2. Not this one. A factor of 0.05 shrinks the population to 5% of itself each year. Growth of 5% needs 105%, which is 1.05. Go to M5.2.6. Try another.

    3. Not this one. That adds a fixed 1.05 each year, which is a straight line. Growth by a percentage multiplies, so x must be a power. Go to M5.2.6. Try another.

    4. Yes, that is right. Each year the population is multiplied by 1 + 0.05 = 1.05. After x years it has been multiplied x times, so P = 200 × 1.05 to the power x.

    See the answer

    P = 200 × 1.05ˣ

    Each year the population is multiplied by 1 + 0.05 = 1.05. After x years it has been multiplied x times, so P = 200 × 1.05 to the power x.

    • If you chose A: You used 5 as the factor. A factor of 5 means 500% each year. Growth of 5% is a factor 1 + 0.05. Go to M5.2.6.
    • If you chose B: A factor of 0.05 shrinks the population to 5% of itself each year. Growth of 5% needs 105%, which is 1.05. Go to M5.2.6.
    • If you chose C: That adds a fixed 1.05 each year, which is a straight line. Growth by a percentage multiplies, so x must be a power. Go to M5.2.6.

  2. A quantity falls by 12% each hour. By what number is it multiplied each hour?

    Tap your answer.

    1. Yes, that is right. A fall of 12% leaves 100% − 12% = 88%, which is the factor 0.88.

    2. Not this one. A factor of 1.12 is growth of 12%. For a fall, take the 12% away from 100%. Go to M5.2.6. Try another.

    3. Not this one. That would leave only 12% of the quantity. A fall of 12% leaves 88%. Go to M5.2.6. Try another.

    4. Not this one. Multiplying by 12 would make the quantity twelve times bigger. Go to M5.2.6. Try another.

    See the answer

    0.88

    A fall of 12% leaves 100% − 12% = 88%, which is the factor 0.88.

    • If you chose B: A factor of 1.12 is growth of 12%. For a fall, take the 12% away from 100%. Go to M5.2.6.
    • If you chose C: That would leave only 12% of the quantity. A fall of 12% leaves 88%. Go to M5.2.6.
    • If you chose D: Multiplying by 12 would make the quantity twelve times bigger. Go to M5.2.6.

  3. A quantity decays as M = A exp(−kt), with k above 0. Which is its half-life?

    Tap your answer.

    1. Not this one. You left out the ln. Halving means exp(−kT) = 1/2, and the log of that brings in ln 2. Go to M5.2.6. Try another.

    2. Yes, that is right. We need exp(−kT) = 1/2. Take ln of both sides: −kT = −ln 2. So T = ln 2 ÷ k.

    3. Not this one. The ln belongs to the 2, not to k. Halving gives exp(−kT) = 1/2, so kT = ln 2. Go to M5.2.6. Try another.

    4. Not this one. You turned the fraction upside down. A bigger k means faster decay, so the half-life must get smaller as k gets bigger. Go to M5.2.6. Try another.

    See the answer

    ln 2 ÷ k

    We need exp(−kT) = 1/2. Take ln of both sides: −kT = −ln 2. So T = ln 2 ÷ k.

    • If you chose A: You left out the ln. Halving means exp(−kT) = 1/2, and the log of that brings in ln 2. Go to M5.2.6.
    • If you chose C: The ln belongs to the 2, not to k. Halving gives exp(−kT) = 1/2, so kT = ln 2. Go to M5.2.6.
    • If you chose D: You turned the fraction upside down. A bigger k means faster decay, so the half-life must get smaller as k gets bigger. Go to M5.2.6.

  4. A herd model N = 50 × 1.08 to the power t was fitted to counts over the first 5 years. Which prediction is the least trustworthy?

    Tap your answer.

    1. Not this one. Year 3 lies inside the 5 years of data, so the model has been tested there. Go to M5.2.6. Try another.

    2. Not this one. Year 4.5 lies inside the 5 years of data, so it is a safe prediction. Go to M5.2.6. Try another.

    3. Not this one. Year 1 lies inside the 5 years of data, so it is a safe prediction. Go to M5.2.6. Try another.

    4. Yes, that is right. A model is trusted near its data. The time of 80 years is far outside the 5 years of data, and things such as food and space will have changed the growth.

    See the answer

    The size after 80 years

    A model is trusted near its data. The time of 80 years is far outside the 5 years of data, and things such as food and space will have changed the growth.

    • If you chose A: Year 3 lies inside the 5 years of data, so the model has been tested there. Go to M5.2.6.
    • If you chose B: Year 4.5 lies inside the 5 years of data, so it is a safe prediction. Go to M5.2.6.
    • If you chose C: Year 1 lies inside the 5 years of data, so it is a safe prediction. Go to M5.2.6.

  5. A population grows by 5% each year. In the form A exp(kt), what is k, to 3 significant figures?

    Tap your answer.

    1. Yes, that is right. The factor is 1 + 0.05 = 1.05, and exp(k) = 1.05, so k = ln 1.05 = 0.0488 to 3 significant figures.

    2. Not this one. That is the growth factor 1 + r, which goes with the form (1 + r) to the power t. For exp(kt) you need its natural log. Go to M5.2.6. Try another.

    3. Not this one. The ln of 1.05 is small, near 0.05. A value of 0.488 would be growth of 63% a year. Go to M5.2.6. Try another.

    4. Not this one. You used r as k. They are close for small r, but k = ln (1 + r), not r. Go to M5.2.6. Try another.

    See the answer

    0.0488

    The factor is 1 + 0.05 = 1.05, and exp(k) = 1.05, so k = ln 1.05 = 0.0488 to 3 significant figures.

    • If you chose B: That is the growth factor 1 + r, which goes with the form (1 + r) to the power t. For exp(kt) you need its natural log. Go to M5.2.6.
    • If you chose C: The ln of 1.05 is small, near 0.05. A value of 0.488 would be growth of 63% a year. Go to M5.2.6.
    • If you chose D: You used r as k. They are close for small r, but k = ln (1 + r), not r. Go to M5.2.6.

  6. From an earlier chapter

    Solve ln x = 2. Which is x, to 3 significant figures?

    Tap your answer.

    1. Not this one. You worked out 2 × e. But ln x = 2 means x = e², which is e multiplied by itself. Go to M5.2.5. Try another.

    2. Not this one. That is ln 2. The equation says ln x = 2, so x is the number whose ln is 2. Go to M5.2.5. Try another.

    3. Yes, that is right. Change to index form. ln x = 2 means x = e², which is 7.389 to 3 decimal places and 7.39 to 3 significant figures.

    4. Not this one. You took the square root of 2. The equation ln x = 2 is in log form. Its index form is x = exp(2). Go to M5.2.5. Try another.

    See the answer

    7.39

    Change to index form. ln x = 2 means x = e², which is 7.389 to 3 decimal places and 7.39 to 3 significant figures.

    • If you chose A: You worked out 2 × e. But ln x = 2 means x = e², which is e multiplied by itself. Go to M5.2.5.
    • If you chose B: That is ln 2. The equation says ln x = 2, so x is the number whose ln is 2. Go to M5.2.5.
    • If you chose D: You took the square root of 2. The equation ln x = 2 is in log form. Its index form is x = exp(2). Go to M5.2.5.

  7. A herd of 120 goats grows by 6% each year. Find its size after 8 years, to the nearest whole number.

    See the answer

    191

    The factor is 1.06. After 8 years, N = 120 × 1.06 to the power 8 = 120 × 1.5938 = 191.3. So the herd is about 191 goats.

  8. A sample has a half-life of 12 days. Its mass is M = A exp(−kt), with t in days. Find k, to 3 significant figures.

    See the answer

    0.0578

    After one half-life the mass is halved, so exp(−12k) = 1/2. Take ln of both sides: −12k = −ln 2, so k = ln 2 ÷ 12 = 0.6931 ÷ 12 = 0.0578 to 3 significant figures.

  9. A culture has 500 cells at t = 0 and 2 000 cells when t = 5 hours. It follows N = A exp(kt). Find k, to 3 significant figures.

    See the answer

    0.277

    At t = 0, A = 500. At t = 5, 2 000 = 500 exp(5k), so exp(5k) = 4. Take ln: 5k = ln 4 = 1.3863. So k = 0.2773, which is 0.277 to 3 significant figures.

  10. A herd starts at 50 goats and grows by 8% each year. After how many whole years does it first have more than 200 goats?

    See the answer

    19

    We need 50 × 1.08 to the power t > 200, so 1.08 to the power t > 4. Take ln: t > ln 4 ÷ ln 1.08 = 18.01. The smallest whole number is 19. Check: 50 × 1.08¹⁸ = 199.8, which is not above 200, and 50 × 1.08¹⁹ = 215.8.

  11. A population follows N = A exp(kt). N = 500 when t = 2 and N = 2 000 when t = 5. Find k to 3 significant figures, then A to the nearest whole number. Type k, then A.

    See the answer

    0.462, 198

    Divide the readings: exp(5k) ÷ exp(2k) = 2 000 ÷ 500 = 4, so exp(3k) = 4. Then 3k = ln 4 = 1.3863 and k = 0.4621. Put t = 2: 500 = A exp(2 × 0.4621) = A × 2.5198, so A = 198.4.

  12. A drink cools so that its temperature is θ = 18 + 62 exp(−0.04t) °C, with t in minutes. Find θ when t = 10, to 3 significant figures.

    See the answer

    59.6

    Put t = 10: θ = 18 + 62 exp(−0.4) = 18 + 62 × 0.6703 = 18 + 41.56 = 59.6 °C. The temperature approaches 18 °C in the long run.

Keep it

Copy this into your exercise book.

Growth by r: y = A(1 + r) to the power t. Also y = A exp(kt), with k = ln (1 + r). Doubling time ln 2 ÷ k. Half-life ln 2 ÷ |k|. Trust a model near its data only.

Draw: A decay curve with dots at each half-life, the mass halving every time.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

No internet tomorrow? Save the whole book as one file (909 KB). It opens on any phone.