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Cambridge IGCSE0580

Exponential growth and decay

Mathematics 0580 Chapter Notes

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Exponential growth and decay
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In this chapter

  1. 1Exponential growth
  2. 2Exponential decay
  3. 3Finding the number of periods
  4. 4Finding the rate of growth or decay
  5. 5Formulas with a starting value and a multiplier

1. Exponential growth

Exponential growth is a repeated percentage increase. The amount is multiplied by the same multiplier each time, so the rise gets bigger every year. Population change and compound interest both work this way.

For a rise of r% per period, the multiplier is 1 + r/100. A rise of 4.4% uses 1.044. For n periods, use the multiplier as a power.

final amount = original × (1 + r/100)ⁿ

Examiner insight

“The most common error was to treat the growth as 5 × 2.5 per cent.”

Worked example 12 marks

Tom invests char"24 2500 at a rate of 4.4% per year compound interest. Calculate the value of the investment at the end of 6 years.

  1. 1

    multiplier = 1 + 4.4/100 = 1.044, so value = 2500 × 1.044⁶ [1]

  2. 2

    = char"24 3237 [1]

2. Exponential decay

Exponential decay is a repeated percentage decrease. Depreciation of a car is an example. The amount is multiplied by the same multiplier each time, but the multiplier is less than 1.

For a fall of r% per period, the multiplier is 1 - r/100. A fall of 5% uses 0.95. Each fall is a percentage of the new value, so the amounts lost get smaller.

final amount = original × (1 - r/100)ⁿ

Examiner insight

“Where errors were made it tended to be by applying a 10 per cent increase instead of decrease, or by using a simple decrease of $250 each year.” “Some candidates gave the answer 73.7, the percentage remaining, rather than the overall percentage decrease.”

Worked example 13 marks

A scientist records the mass of a radioactive substance at the same time each day. On 10 January, the scientist records a mass of 150 grams. Each day, the mass of the substance decreases by 5%. Find the date on which the scientist first records a mass of less than 96 grams.

  1. 1

    multiplier = 1 - 5/100 = 0.95, so mass after n days = 150 × 0.95ⁿ [1]

  2. 2

    n = 8: 150 × 0.95⁸ = 99.5 (not less than 96). n = 9: 150 × 0.95⁹ = 94.5 (less than 96) [1]

  3. 3

    9 days after 10 January is 19 January [1]

3. Finding the number of periods

When the question asks how long, the power n is the unknown. The syllabus does not require logarithms, so use trial and improvement: try values of n until the amount crosses the target.

Show the value for the last n that fails and the first n that works. Then read the question: "complete years" means a whole number of periods.

Examiner insight

“Candidates should always write down their supporting methods as fully as possible.”

Worked example 13 marks

Mia invests char"24 3000 in an account that pays compound interest at a rate of 3.2% per year. Find the number of complete years it takes for the value of Mia's investment to become greater than char"24 4800.

  1. 1

    value after n years = 3000 × 1.032ⁿ [1]

  2. 2

    n = 14: char"24 4662.70 (too small). n = 15: char"24 4811.90 (greater than char"24 4800) [1]

  3. 3

    15 years [1]

4. Finding the rate of growth or decay

To find an unknown rate, write the exponential equation, then undo the power with an nth root. Divide both sides by the original amount first, then take the root.

The root is the multiplier. Keep its full calculator value until the end, then subtract 1 and multiply by 100 to give r.

(1 + r/100)ⁿ = final/original

Examiner insight

“Candidates who only gave the cube root as 1.10 were unable to achieve a 3 significant figure final answer.” “There were many attempts without any exponential statement, such as finding the overall percentage change over 3 years and dividing this by 3.”

Worked example 13 marks

Ryan invests some money. The value of the investment increases exponentially at a rate of r% per year. At the end of 6 years, the value of the original investment has increased by 19.4%. Find the value of r.

  1. 1

    an increase of 19.4% has multiplier 1.194, so (1 + r/100)⁶ = 1.194 [1]

  2. 2

    1 + r/100 = 1.194^(1/6) = 1.02999… [1]

  3. 3

    r = 3.00 to 3 significant figures [1]

5. Formulas with a starting value and a multiplier

An exponential relationship can be written y = ab^(x). Here a is the starting value, because b⁰ = 1. The number b is the multiplier for each step of x.

Put in x = 0 to find a. Put in a second point to get an equation in b alone. Do not solve as simultaneous equations. Give b to 3 significant figures when it is not exact.

y = ab^(x) b > 1 for growth, 0 < b < 1 for decay

Examiner insight

“Some candidates gave the 2 significant figure value 0.95 for b rather than 3 significant figures as required.” “Some candidates set up two equations and attempted to solve as a pair of simultaneous equations which was not appropriate in this case.”

Worked example 13 marks

The temperature, T °C, of a liquid is given by the formula T = ab^(x) where x is the number of minutes after 2 pm. T = 80 at 2 pm. T = 20 at 2.30 pm. Find the value of a and the value of b.

  1. 1

    at 2 pm, x = 0: 80 = ab⁰ = a, so a = 80 [1]

  2. 2

    at 2.30 pm, x = 30: 20 = 80b³⁰, so b³⁰ = 1/4 [1]

  3. 3

    b = (1/4)^(1/30) = 0.955 [1]

Recap

  • Exponential growth and decay are a repeated percentage change: multiply by the same multiplier each time.
  • Extended: growth multiplier 1 + r/100, decay multiplier 1 - r/100. Final amount = original × multiplierⁿ.
  • Extended: for the number of periods, use trial and improvement and show the values either side of the target.
  • Extended: for the rate, take the nth root of final ÷ original, then subtract 1 and multiply by 100.
  • Extended: in y = ab^(x), a is the value at x = 0 and b is the multiplier. Give b to 3 significant figures.
  • Never find the rate by dividing the overall change by the number of years.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Do not use a calculator. A town has a population of 20 000. The population grows by 10% each year. Find the population after 2 years.2 marks
  2. Do not use a calculator. A car is worth char"24 8000. Its value decreases by 25% each year. Find its value after 2 years.2 marks
  3. Do not use a calculator. A culture has 500 bacteria at 09 00. The number doubles every hour. Find the number of bacteria at 13 00.2 marks
  4. A machine is bought for char"24 12 000. Its value decreases by 6% each year. Write down the multiplier for one year. Find the value of the machine after 4 years, to the nearest dollar.3 marks
  5. Rahim invests char"24 600 at 2.5% per year compound interest. Find the interest he earns in 7 years, to the nearest cent.3 marks
  6. A substance loses 3% of its mass each day. Find the percentage decrease in its mass after 10 days. Give your answer to 3 significant figures.3 marks
  7. A pond plant covers 4 m². Each week its area grows by 20%. Find the number of complete weeks until the area is greater than 20 m².3 marks
  8. A painting rises in value from char"24 4000 to char"24 5300 in 5 years. The value increases exponentially at a rate of r% per year. Find r to 3 significant figures.3 marks
  9. Do not use a calculator. The value, char"24 V, of a machine is V = ab^(t), where t is the number of years after 2020. The value is char"24 6400 in 2020 and char"24 2700 in 2023. Find the value of the machine in 2024. Find also the percentage by which the value falls each year.5 marks
  10. A lake holds 48 000 fish of species A and 30 000 fish of species B. Each year the number of species A decreases by 2.5% and the number of species B increases by 4%. Find the number of complete years until there are more fish of species B than of species A. Find the percentage decrease in the number of species A over these years, to 3 significant figures.5 marks

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