Mathematics Advanced • Year 12 • Module 7 • Lesson 2

Compound Interest in Depth

Build fluency transposing A = P(1 + r)ⁿ for P, r, n — including the logarithmic method and the Rule of 72.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Starting from A = P(1 + r)ⁿ, write the rearranged formula that solves for each variable:

Solve for P: P = ________________________________

Solve for r: r = ________________________________

Solve for n (using natural logs): n = ________________________________

Q1.2 State the Rule of 72 in one sentence, and use it to estimate the doubling time for an annual rate of 8%.

Q1.3 Why must you use logarithms when the unknown is the exponent n? Answer in one line.

Stuck? Revisit lesson § Formula Reference and § Logarithms.

2. Worked example — house growth rate and time-to-target

Follow each line. Every step has a short reason.

Problem. A house grew in value from $850,000 to $1,200,000 over 8 years (compound growth). (a) Find the annual rate r. (b) Hence find how many years from purchase until it is worth $1,500,000 at the same rate.

Step 1 — Choose the right transposition for r.

r = (A/P)^1/n − 1

Reason: the unknown is the base of the exponent; take the nth root.

Step 2 — Substitute and simplify.

r = (1,200,000 / 850,000)^1/8 − 1 = (1.4118)^0.125 − 1

r = 1.0439 − 1 = 0.0439 ≈ 4.39% p.a.

Step 3 — For (b), the unknown is the exponent n — use logarithms.

n = ln(A/P) / ln(1 + r)

Reason: ln of an exponent brings the exponent down as a multiplier.

Step 4 — Substitute (P = 850,000, A = 1,500,000, r = 0.0439).

n = ln(1,500,000 / 850,000) / ln(1.0439) = ln(1.7647) / ln(1.0439)

n = 0.5681 / 0.04298 ≈ 13.2 years from original purchase (i.e., 5.2 more years from year 8).

Conclusion. (a) r ≈ 4.39% p.a.; (b) about 13.2 years from purchase.

3. Faded example — find the rate that grew $15,000 to $22,000 in 5 years

Fill in each blank line. 4 marks

Step 1 — Identify which formula to use:

The unknown is ______ , so use the transposition r = ________________________.

Step 2 — Substitute the given values P = 15,000, A = 22,000, n = 5:

r = ( ______ / ______ )^1/______ − 1

Step 3 — Evaluate the ratio inside the bracket:

r = ( ______________ )^0.2 − 1

Step 4 — Evaluate the power and subtract 1:

r = ______________ − 1 = ______________ = ______________ % p.a.

Conclusion. The annual rate is r ≈ ______________ % p.a.

Stuck? Revisit lesson § Transposing — Finding the Rate.

4. Graduated practice — transpose the formula

For each, identify which variable is unknown, write the correct transposed formula, substitute and evaluate.

Foundation — single-transposition substitution (4 questions)

Q	Given	Unknown & final value
4.1 1	A = 20,000, r = 5% p.a., n = 6 years. Find P.
4.2 1	A = 30,000, P = 20,000, n = 8 years. Find r.
4.3 1	P = 5,000, r = 5% p.a., A = 10,000. Find n (years).
4.4 1	Use Rule of 72 to estimate doubling time at 9% p.a.

Standard — typical HSC difficulty (6 questions)

Show working: write the transposed formula and a substitution line for each.

4.5 An investor needs $40,000 in 6 years. The available rate is 5.2% p.a. compounded annually. How much must be invested today? 2 marks

4.6 An investment of $9,000 grew to $14,500 in 7 years (compound). Find the annual rate to two decimal places. 2 marks

4.7 How many years does it take $6,000 to double at 4.5% p.a. compound interest? Use logarithms, then verify with the Rule of 72. 2 marks

4.8 A credit card charges 18% p.a. compounded daily. How many days does an unpaid $2,000 balance take to grow to $2,500? (r_daily = 0.18/365; find n in days.) 2 marks

4.9 An asset is to grow from $12,000 to $30,000 over 12 years (compound, annual). Find r to two decimal places. 2 marks

4.10 A bond promises to pay $50,000 in 8 years. If the prevailing discount rate is 4.5% p.a. compounded annually, what is the fair price (present value) today? 2 marks

Extension — combine concepts (2 questions)

4.11 $5,000 is invested at 6% p.a. compounded monthly. How many full months until the balance first exceeds $10,000? 3 marks

4.12 Show algebraically that the Rule of 72 comes from approximating ln(2) ≈ 0.72 and ln(1 + r) ≈ r. Hence explain why the rule is most accurate around r ≈ 8%. 3 marks

Stuck on 4.12? Start with n = ln(2) / ln(1 + r), then replace each log with its small-r approximation.

5. Self-check the easy 3

Tick once you have checked your method.

For 4.1 (find P) I divided A by (1 + r)ⁿ, not multiplied.

For 4.2 (find r) I took the nth root then subtracted 1 — not the other way around.

For 4.3 (find n) I used n = ln(A/P) / ln(1 + r), not (A − P)/(P × r).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Transpositions

P = A / (1 + r)ⁿ. r = (A/P)^1/n − 1. n = ln(A/P) / ln(1 + r).

Q1.2 — Rule of 72

Doubling time in years ≈ 72 ÷ (annual rate as a percentage). At 8% p.a.: 72 ÷ 8 = 9 years (exact is 9.006).

Q1.3 — Why logarithms?

Because logarithms bring an exponent down as a multiplier: ln(bⁿ) = n · ln(b), letting us isolate n by ordinary division.

Q3 — Faded example: rate that grew $15,000 to $22,000 in 5 years

Unknown is r. r = (A/P)^1/n − 1.
r = (22,000 / 15,000)^1/5 − 1 = (1.4667)^0.2 − 1 = 1.0794 − 1 = 0.0794 = 7.94% p.a.

Q4.1 — Find P given A = 20,000, r = 5%, n = 6

P = 20,000 / (1.05)⁶ = 20,000 / 1.340096 = $14,924.31.

Q4.2 — Find r given A = 30,000, P = 20,000, n = 8

r = (30,000/20,000)^1/8 − 1 = (1.5)^0.125 − 1 = 1.05199 − 1 = 0.05199 ≈ 5.20% p.a.

Q4.3 — Find n given P = 5,000, A = 10,000, r = 5%

n = ln(10,000/5,000) / ln(1.05) = ln 2 / ln 1.05 = 0.6931 / 0.04879 = 14.21 years.

Q4.4 — Rule of 72 at 9%

72 ÷ 9 = 8 years (exact: 8.04).

Q4.5 — P for $40,000 in 6 yr at 5.2%

P = 40,000 / (1.052)⁶ = 40,000 / 1.354807 = $29,524.88.

Q4.6 — r for $9,000 → $14,500 in 7 yr

r = (14,500/9,000)^1/7 − 1 = (1.6111)^0.1429 − 1 = 1.07016 − 1 = 0.0702 = 7.02% p.a.

Q4.7 — Time for $6,000 to double at 4.5%

n = ln 2 / ln 1.045 = 0.6931 / 0.04402 = 15.75 years. Rule of 72: 72 ÷ 4.5 = 16 years (1.6% error — within acceptable Rule-of-72 range).

Q4.8 — Days for $2,000 → $2,500 at 18% p.a. daily

r_daily = 0.18/365 = 0.000493. n = ln(2,500/2,000) / ln(1.000493) = ln 1.25 / ln 1.000493 = 0.22314 / 0.000493 = 452.5 days ≈ 1 year 88 days.

Q4.9 — r for $12,000 → $30,000 in 12 yr

r = (30,000/12,000)^1/12 − 1 = (2.5)^0.0833 − 1 = 1.07930 − 1 = 0.0793 = 7.93% p.a.

Q4.10 — Present value of $50,000 in 8 yr at 4.5%

P = 50,000 / (1.045)⁸ = 50,000 / 1.422101 = $35,159.88.

Q4.11 — Months for $5,000 to exceed $10,000 at 6% p.a. monthly

Monthly rate r = 0.06/12 = 0.005. Solve 5,000(1.005)ⁿ > 10,000 ⇒ (1.005)ⁿ > 2 ⇒ n > ln 2 / ln 1.005 = 0.6931 / 0.004988 = 138.98. The first integer satisfying the inequality is n = 139 months (≈ 11 years 7 months).

Q4.12 — Derivation of the Rule of 72

Start with n = ln 2 / ln(1 + r). For small r, ln(1 + r) ≈ r (first-order Taylor expansion), and ln 2 ≈ 0.6931. So n ≈ 0.6931 / r. Multiplying numerator and denominator by 100 to convert r to a percentage: n ≈ 69.31 / (100r). Replacing 69.31 by 72 makes mental division easier and also partially corrects for the under-estimate ln(1 + r) < r. The two errors — using 72 instead of 69.31 (too big) and approximating ln(1+r) by r (too small) — cancel near r ≈ 8%, which is why the rule is most accurate there.