Exponential growth and decay

1. Overview

Exponential growth and decay are mathematical models that describe how quantities change over time when the rate of change is proportional to the current amount. This means the larger the quantity, the faster it grows (or decays). You'll use these models to solve real-world problems involving populations, investments, depreciation, and more. The key is understanding the exponential formula and applying it correctly.

Key Definitions

• Exponential Growth: When a quantity increases by a constant percentage rate over equal time periods.
• Exponential Decay: When a quantity decreases by a constant percentage rate over equal time periods.
• Multiplier: The decimal value used to increase or decrease the initial amount (e.g., a 5% increase has a multiplier of 1.05).
• Initial Amount ($P$): The starting value before any growth or decay has occurred.
• Rate ($r$): The percentage increase or decrease per time period.

Core Content

Note: The IGCSE syllabus classifies Exponential Growth and Decay as a Supplement (Extended) topic. However, a basic understanding of percentage increases and decreases is required at the Core level.

Extended Content (Extended Only)

In exponential change, the amount of change depends on the current value. As the value grows, the amount added each time also increases (Growth). As the value shrinks, the amount lost each time decreases (Decay). This leads to the characteristic curved shape of exponential graphs.

Method: Using the General Formula

To calculate the final amount after $n$ time periods:

$\qquad \qquad \bf{A = P \left(1 \pm \frac{r}{100}\right)^n}$

• Use $+$ for Growth
• Use $-$ for Decay

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Worked Example 1 — Exponential Growth (Population)

The population of a town is currently 12,000. It is increasing at a rate of 3.5% per year. What will the population be in 8 years? Give your answer to the nearest whole number.

Step-by-Step Working:

1. State the question: Find the population after 8 years, given an initial population of 12,000 and a growth rate of 3.5% per year.
2. Identify the variables:
   • Initial amount ($P$) = $12,000$
   • Rate ($r$) = $3.5$
   • Time ($n$) = $8$
3. Determine the multiplier:
   • Since it is growth, use $(1 + \frac{r}{100})$
   • Multiplier $= 1 + \frac{3.5}{100}$
   • Multiplier $= 1 + 0.035$
   • Multiplier $= 1.035$
4. Substitute into the formula:
   • $A = P \left(1 + \frac{r}{100}\right)^n$
   • $A = 12,000 \times (1.035)^8$
5. Calculate:
   • $A = 12,000 \times 1.316809...$
   • $A = 15,801.71...$
6. Round to the nearest whole number:
   • $A = 15,802$
7. Final Answer:
   • $\bf{15,802}$

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Worked Example 2 — Exponential Decay (Value of Equipment)

A printing company buys a new machine for $75,000. The machine's value depreciates at a rate of 8% per year. What will be the value of the machine after 6 years? Give your answer to the nearest dollar.

Step-by-Step Working:

1. State the question: Find the value of the machine after 6 years, given an initial value of $75,000 and a depreciation rate of 8% per year.
2. Identify the variables:
   • Initial amount ($P$) = $75,000$
   • Rate ($r$) = $8$
   • Time ($n$) = $6$
3. Determine the multiplier:
   • Since it is decay, use $(1 - \frac{r}{100})$
   • Multiplier $= 1 - \frac{8}{100}$
   • Multiplier $= 1 - 0.08$
   • Multiplier $= 0.92$
4. Substitute into the formula:
   • $A = P \left(1 - \frac{r}{100}\right)^n$
   • $A = 75,000 \times (0.92)^6$
5. Calculate:
   • $A = 75,000 \times 0.606355...$
   • $A = 45,476.66...$
6. Round to the nearest dollar:
   • $A = 45,477$
7. Final Answer:
   • $\bf{$45,477}$

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Worked Example 3 — Finding the Rate (Growth)

The number of bacteria in a petri dish increased from 500 to 1500 in 4 hours. Assuming exponential growth, calculate the percentage increase per hour.

Step-by-Step Working:

1. State the question: Find the percentage increase per hour, given an initial amount of 500, a final amount of 1500, and a time period of 4 hours.
2. Identify the variables:
   • Final amount ($A$) = $1500$
   • Initial amount ($P$) = $500$
   • Time ($n$) = $4$
   • Rate ($r$) = ? (what we are trying to find)
3. Substitute into the formula:
   • $A = P \left(1 + \frac{r}{100}\right)^n$
   • $1500 = 500 \left(1 + \frac{r}{100}\right)^4$
4. Divide both sides by 500:
   • $\frac{1500}{500} = \left(1 + \frac{r}{100}\right)^4$
   • $3 = \left(1 + \frac{r}{100}\right)^4$
5. Take the fourth root of both sides:
   • $\sqrt[4]{3} = 1 + \frac{r}{100}$
   • $1.31607... = 1 + \frac{r}{100}$
6. Subtract 1 from both sides:
   • $1.31607... - 1 = \frac{r}{100}$
   • $0.31607... = \frac{r}{100}$
7. Multiply both sides by 100:
   • $0.31607... \times 100 = r$
   • $31.607... = r$
8. Round to three significant figures:
   • $r = 31.6$
9. Final Answer:
   • $\bf{31.6%}$

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Worked Example 4 — Finding the Time (Decay)

A radioactive substance decays at a rate of 7% per day. How many days will it take for the substance to decay to half of its original mass?

Step-by-Step Working:

1. State the question: Find the number of days it takes for a substance to decay to half its original mass, given a decay rate of 7% per day.
2. Identify the variables:
   • Initial amount ($P$) = $P$ (we can assume any value, as we are looking for half of it)
   • Final amount ($A$) = $0.5P$ (half of the initial amount)
   • Rate ($r$) = $7$
   • Time ($n$) = ? (what we are trying to find)
3. Substitute into the formula:
   • $A = P \left(1 - \frac{r}{100}\right)^n$
   • $0.5P = P \left(1 - \frac{7}{100}\right)^n$
4. Divide both sides by P:
   • $0.5 = \left(1 - \frac{7}{100}\right)^n$
   • $0.5 = (1 - 0.07)^n$
   • $0.5 = (0.93)^n$
5. Use Trial and Improvement (or Table Mode on Calculator):
   • Try $n = 9$: $(0.93)^9 = 0.5217...$ (Too high)
   • Try $n = 10$: $(0.93)^{10} = 0.4852...$ (Too low)
   • Try $n = 9.5$: $(0.93)^{9.5} = 0.5031...$ (Closer)
   • Try $n = 9.9$: $(0.93)^{9.9} = 0.4887...$ (Very close)
6. Approximate Answer:
   • $n \approx 9.9$
7. Final Answer:
   • $\bf{9.9 \text{ days}}$ (approximately)

Key Equations

The Exponential Formula:

$\qquad \qquad \bf{A = P\left(1 \pm \frac{r}{100}\right)^n}$

(Note: Use + for growth, - for decay)

Symbol	Meaning	Typical Units
$A$	Final Amount	People, Dollars ($), Grams, etc.
$P$	Principal (Initial) Amount	Same as $A$
$r$	Percentage Rate	% per year/day/hour
$n$	Number of time periods	Years, Months, Days, etc.

Formula Sheet Info: This formula is not usually provided on the IGCSE formula sheet. You must memorize it.

Common Mistakes to Avoid

• ❌ Wrong: Calculating simple percentage increase/decrease repeatedly instead of using the exponential formula. This only works for the first time period.
  • ✓ Right: Use the formula $A = P(1 \pm \frac{r}{100})^n$ to account for the compounding effect.
• ❌ Wrong: Using the rate, $r$, as a decimal directly in the formula without dividing by 100. For example, using 5 instead of 0.05 for a 5% rate.
  • ✓ Right: Always divide the percentage rate by 100 before using it in the formula: $\frac{r}{100}$.
• ❌ Wrong: Confusing growth and decay and using the wrong sign in the formula.
  • ✓ Right: Double-check whether the quantity is increasing (growth, use "+") or decreasing (decay, use "-").
• ❌ Wrong: Rounding intermediate calculations, leading to inaccurate final answers.
  • ✓ Right: Keep all the digits in your calculator until you reach the final step, then round to the required degree of accuracy.

Exam Tips

• Calculator Tip: Use the power button ($x^y$ or $x^\square$) on your calculator. You do not need to multiply the number repeatedly.
• Command Words: "Calculate the value after..." means find $A$. "Find the percentage increase..." means you may need to find the difference between $A$ and $P$ first.
• Finding $n$: If a question asks "How many years until the population exceeds X?", use the Table Mode on your calculator or use Trial and Improvement by plugging in different values for $n$.
• Contexts: Be prepared for biological contexts (bacteria doubling), financial contexts (compound interest), and physics (radioactive half-life/decay).
• Typical Values: In decay questions involving value, the answer will almost never be negative. It will simply get closer and closer to zero.