Model exponential growth and decay using $y = A \cdot r^t$ or $y = A e^{kt}$, including problems involving population growth, radioactive decay, depreciation and continuous compound interest

What this dot point is asking

QCAA wants you to model real-world growth and decay scenarios with exponential functions, choosing between $y = A r^t$ (discrete or growth factor form) and $y = A e^{kt}$ (continuous form), and to solve for any of the parameters $A$, $r$, $k$ or $t$ from worded conditions.

The two standard forms

Discrete growth factor form. $y = A r^t$ where:

• IMATH_7 is the initial value at $t = 0$.
• IMATH_9 is the per-time-unit multiplier.
• IMATH_10 for growth, $0 < r < 1$ for decay.
• IMATH_12 is time in the units that match $r$.

Continuous form. $y = A e^{kt}$ where:

• IMATH_15 is initial value.
• IMATH_16 is the continuous growth rate ($k > 0$ growth, $k < 0$ decay).
• Used when growth is compounded continuously (continuously compounded interest, radioactive decay in mathematically clean form).

The two forms convert via $r = e^k$ or $k = \ln r$.

Building the model from a scenario

Identify $A$ from the initial value. Identify $r$ or $k$ from a single additional condition (typically "after $T$ time units the value is $V$").

For percentage growth at rate $p\%$ per period: $r = 1 + p/100$. For percentage decay: $r = 1 - p/100$ (assuming $p < 100$).

For half-life $T_{1/2}$: $r^{T_{1/2}} = 1/2$, so $r = (1/2)^{1/T_{1/2}}$.

For doubling time $T_d$: $r^{T_d} = 2$, so $r = 2^{1/T_d}$.

Continuous compound interest

$A(t) = P e^{rt}$ where $P$ is the principal, $r$ is the annual interest rate (as a decimal, continuously compounded), $t$ is time in years.

For discrete annual compounding the model is $A = P(1 + r)^t$; for $n$ times per year it is $A = P(1 + r/n)^{nt}$. As $n \to \infty$, this approaches $A = P e^{rt}$.

Worked example

A population doubles every $7$ years. The current population is $1500$. Find (a) the population after $20$ years and (b) the time to reach $10\,000$.

Half-life-style model: $P(t) = 1500 \cdot 2^{t/7}$.

(a) $P(20) = 1500 \cdot 2^{20/7} = 1500 \cdot 2^{2.857} = 1500 \cdot 7.24 = 10\,860$.

(b) Solve $10\,000 = 1500 \cdot 2^{t/7}$.

$2^{t/7} = 6.667$

$t/7 = \log_2 6.667 = \ln 6.667 / \ln 2 = 1.897 / 0.693 = 2.737$

$t = 19.2$ years.

Common traps

Confusing $A$ with the answer to the question. $A$ is the initial value, not the value at the time being asked about.

Using $r =$ percentage rate. $r$ is a growth factor (a number near $1$), not the percentage. A $5\%$ annual growth means $r = 1.05$.

Forgetting that decay $r < 1$. A decay model with $r > 1$ models growth, not decay.

Mixing time units. If $r$ is per year, $t$ must be in years. If you switch units, recompute $r$.

How this appears in IA1 and EA

IA1. Building a discrete model from a scenario and predicting a value at a stated time.

EA Paper 1. Multiple choice on identifying the growth factor or the doubling time.

EA Paper 2. A multi-part contextual problem: build the model, predict a value, solve for the time at which a target is met. Often combined with calculus in Year 12 to find an instantaneous rate of change.

In one sentence

Exponential growth and decay are modelled by $y = A r^t$ (initial value $A$, growth factor $r > 1$ or decay factor $0 < r < 1$) or equivalently $y = A e^{kt}$ with $r = e^k$, and worded scenarios provide enough information to identify $A$ and $r$ (or $k$) from the initial value plus one further condition.