Exponential Growth and Decay Modeling

• If the exact same percentage change happens $n$ times, raise the multiplier to the power of $n$. This creates an exponential relationship, unlike "simple" change models which just add or subtract a fixed amount and do not use powers.

The General Exponential Model

• Bacteria populations can double every hour, leading to massive numbers from just a few starting cells.

• This type of repeated, proportional change is modeled using the general formula for exponential growth and exponential decay:

$$y = a \times b^x$$

• $y$ is the final amount.

• $a$ is the initial value (the starting quantity at time $x=0$).

• $b$ is the multiplier (the constant scale factor).

• $x$ is the number of time periods (years, hours, days).

Constructing the Formula:

• To build a specific model, substitute the starting quantity for $a$ and the growth/decay factor for $b$.

• Example 1 (Growth): A population of 500 bacteria triples every hour. The initial value is $a = 500$ and the multiplier is $b = 3$. The formula is $y = 500 \times 3^x$.

Financial Growth and Decay (Compound Interest & Depreciation)

• When you leave money in a savings account, you earn interest on your interest, making your money grow faster over time.

• This is called compound interest. Conversely, items like cars or laptops lose value over successive years, which is known as depreciation.

• Both scenarios use the standard financial formula:

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

• $A$ is the final amount.

• $P$ is the principal (the original amount invested or initial value).

• $r$ is the percentage rate of change.

• $n$ is the number of time periods.

• If you are asked to find the accumulated interest rather than the total amount, you must perform a final step: $Interest = A - P$.

Worked Example: Compound Interest

A bank offers a savings account with 4.2% compound interest per year. If Maya invests £3,400, calculate the total value of her investment after 5 years and interpret the result.

Step 1: Identify the values for the formula $A = P(1 + \frac{r}{100})^n$.

• $P = 3400$

• $r = 4.2$

• $n = 5$

Step 2: Substitute the values into the equation to create your multiplier.

• $A = 3400 \times \left(1 + \frac{4.2}{100}\right)^5$

Step 3: Calculate the final amount, rounding to 2 decimal places for currency.

• $A = 4176.438...$

• $A = £4,176.44$

Step 4: Interpret the final value in the context of the problem.

• After 5 years, Maya's initial investment has grown to £4,176.44, meaning her money appreciated and she earned £776.44 in accumulated interest.

Solving for Time in Real-World Models

• Sometimes scientists need to know exactly when a decaying radioactive substance will drop below a safe threshold.

• In the OCR GCSE exam, you are not expected to use logarithms to solve for an unknown time power ($n$ or $t$).

• Instead, you must use trial and improvement by substituting whole-number values for $n$ into your calculator until you cross the target value.

• When modeling populations (such as town sizes or bird flocks), always round your final interpretation to the nearest whole number, as you cannot have fractions of living things.

Worked Example: Trial and Improvement

A population of 8,000 birds decreases by 12% each year. The model is given by $P = 8000 \times 0.88^t$. Determine the number of full years it takes for the population to fall below 5,000.

Step 1: Use trial and improvement by testing values for $t$.

• Try $t = 3$: $P = 8000 \times 0.88^3 = 5451.7...$

Step 2: The value is still above 5,000, so increase $t$.

• Try $t = 4$: $P = 8000 \times 0.88^4 = 4797.5...$

Step 3: Interpret the result based on the threshold.

• At 3 years, the population is ~5,452. At 4 years, it is ~4,798.

• It takes 4 full years for the population to fall below 5,000.