Growth, decay and compound interest

What the specification requires

• The 2026 specification states that candidates may be required to "form and solve exponential equations relating to compound interest, growth and decay, etc".
• These are the contexts where exponential equations meet real life, and they are the most predictable applied questions on the paper.

The general growth and decay model

$$A = A_0 \times r^{\,t}$$

• Each part:

  • $A_0$ — the initial amount, the value when $t = 0$
  • $r$ — the multiplier applied each period
  • $t$ — the number of periods
  • $A$ — the amount after $t$ periods

• The multiplier is what makes the model work:

  • Growth of $p$% — multiplier $= 1 + \dfrac{p}{100}$
  • Decay of $p$% — multiplier $= 1 - \dfrac{p}{100}$

• A 7% annual increase gives $r = 1.07$; a 7% annual decrease gives $r = 0.93$.

• Note that $r > 1$ means growth and $0 < r < 1$ means decay. A multiplier can never be negative.

Compound interest

$$A = P(1 + i)^n$$

• Each part:

  • $P$ — the principal, the amount invested or borrowed
  • $i$ — the interest rate per period, as a decimal
  • $n$ — the number of periods
  • $A$ — the final amount

• The interest is added to the balance each period, so the next period's interest is calculated on a larger amount. That is what "compound" means.

• Interest earned $= A - P$, not $A$. Read which one the question wants.

Matching the rate to the period

• If the compounding is not annual, both the rate and the number of periods change.

  • Monthly: $i = \dfrac{\text{annual rate}}{12}$, and $n = 12 \times$ years
  • Quarterly: divide by 4, multiply the years by 4
  • Weekly: divide by 52, multiply by 52

• Worked through — $5000 at 6% p.a. compounded monthly for 3 years:

  • $i = \dfrac{0.06}{12} = 0.005$ and $n = 36$
  • $$A = 5000(1.005)^{36} = 5983.40$$

• Compare with annual compounding: $5000(1.06)^3 = 5955.08$. More frequent compounding earns more, which is a useful sense-check.

Depreciation

• Depreciation is decay applied to the value of an asset:

$$A = P(1 - i)^n$$

• A car worth $28 000 depreciating at 15% per year is worth $28000(0.85)^n$ after $n$ years.
• Note that depreciation never reaches zero in this model — it approaches it. The value after 10 years is $28000(0.85)^{10} = \$5514$, still positive.

Solving for the time

• When the unknown is $t$ or $n$, the equation is exponential and needs logarithms:

  • Isolate the power term.
  • Take logs of both sides.
  • Bring the exponent down.
  • Solve.

• Worked through — how long for $2000 at 4.5% p.a. to reach $3000?

  • $3000 = 2000(1.045)^n$, so $(1.045)^n = 1.5$
  • $$n\log(1.045) = \log 1.5$$
  • $n = \dfrac{\log 1.5}{\log 1.045} = 9.21$ years

• Round up in context. If interest is paid annually, the balance only reaches $3000 after the 10th payment, so the answer is 10 years — the decimal 9.21 does not correspond to any moment when interest is credited.

Solving for the rate

• If the rate is unknown, take the $n$th root rather than a logarithm:
  • $A = P r^n$ gives $r = \sqrt[n]{\dfrac{A}{P}} = \left(\dfrac{A}{P}\right)^{1/n}$
  • Then the percentage rate is $100(r - 1)$.

New Zealand contexts

• Inflation: the Reserve Bank of New Zealand targets 1–3% annual inflation. At 2%, an item costing $25 today is modelled at $25(1.02)^t$ dollars in $t$ years.
• KiwiSaver: a balance of $18 000 growing at 5% per year reaches $18000(1.05)^{20} = \$47\,760$ after 20 years.
• Population: Stats NZ figures for regional growth are commonly modelled this way over short periods.