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Module 7 · L2 of 20 ~40 min ⚡ +95 XP available

Compound Interest in Depth

The formula $A = P(1+r)^n$ is deceptively simple. But what if you need to find the rate that gets you to your goal? Or the exact number of years? In this lesson you'll learn to transpose the formula, use logarithms to unlock the exponent, and discover the Rule of 72, the mental maths shortcut used by investors worldwide.

Today's hook, A bank offers 6% p.a. compound interest. Without any calculation, how many years does it take to double your money? Warren Buffett's answer changed his life.

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Get oriented

Recall what you already know, meet the key ideas and settle the terms.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall, your gut answer first

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You want to double your money. A bank offers 6% p.a. compound interest.

Without calculating exactly guess how many years it will take to double. Then explain what makes this hard to estimate mentally.

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One formula, four variables, own them all

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$A = P(1+r)^n$ has four variables. Know three, find the fourth. Real-world planning demands all four:

"How much do I need to invest?" → solve for $P$
"What rate do I need?" → solve for $r$
"How long will it take?" → solve for $n$

Transposing for $P$ and $r$ uses algebra. Transposing for $n$ uses logarithms because the unknown is trapped in an exponent. The natural logarithm $\ln$ is the key that unlocks it.

Rearrange the compound-interest formula to isolate each variable.

$n = \dfrac{\ln(A/P)}{\ln(1+r)}$, logarithms unlock the exponent

Finding P (present value)

Divide $A$ by $(1+r)^n$. This tells you how much to invest today to reach a future goal.

Finding r (the rate)

Take the $n$th root: $r = \left(\frac{A}{P}\right)^{1/n} - 1$. Use the $x^y$ key or $\sqrt[n]{x}$ on your calculator.

Finding n (using ln)

$\ln$ both sides: $n = \frac{\ln(A/P)}{\ln(1+r)}$. Works with $\log_{10}$ too, the base cancels.

What you'll master

Know

Key facts

How to transpose $A = P(1+r)^n$ for any variable
The logarithmic method for solving exponential equations
The Rule of 72 and its limitations

Understand

Concepts

Why logarithms are needed to isolate an exponent
How small changes in $r$ create large changes in $A$ over long periods
When the Rule of 72 is accurate vs when it breaks down

Can do

Skills

Transpose the compound interest formula for $P$, $r$, or $n$
Use natural logarithms to find exact doubling or tripling times
Estimate doubling times mentally using the Rule of 72
Verify calculator results using approximation

Key terms

TranspositionRearranging a formula to make a different variable the subject.

Present value ($P$)The current worth of a future sum of money, discounted at a given rate.

Natural logarithm ($\ln$)The logarithm with base $e \approx 2.718$, used to solve exponential equations.

Doubling timeThe number of periods required for an investment to double in value.

Rule of 72An approximation: doubling time $\approx 72 \div \text{annual rate (\%)}$.

Effective annual rateThe actual annual return after accounting for compounding frequency.

Transpose for P and r

Make principal or rate the subject of the compound interest formula.

Transposing for P and r

core concept

The standard compound interest formula has four variables. If you know any three, you can find the fourth. This is the mathematics behind goal-setting.

Finding the Principal ($P$)

Divide both sides by $(1+r)^n$:

$$P = \dfrac{A}{(1 + r)^n}$$

Example: You need $30,000 in 5 years. The account pays 4% p.a. annually. How much do you deposit today?

$P = \dfrac{30{,}000}{(1.04)^5} = \dfrac{30{,}000}{1.21665} = \$24{,}657.81$

Finding the Rate ($r$)

Take the $n$th root of both sides, then subtract 1:

$$r = \left(\dfrac{A}{P}\right)^{1/n} - 1$$

Example: An investment grew from $8,000 to $12,000 in 6 years. What annual rate was earned?

$r = \left(\dfrac{12{,}000}{8{,}000}\right)^{1/6} - 1 = (1.5)^{0.1\overline{6}} - 1 = 1.0699 - 1 = 6.99\%$

Real-world anchor, House prices. A house bought for $\$850,000$ in 2014 is now worth $\$1,200,000$. Using the rate formula: $r = (1{,}200{,}000/850{,}000)^{1/10} - 1 = (1.4118)^{0.1} - 1 \approx 3.46\%$ p.a. compound growth, that is the annualised capital growth rate.

Find P: $P = A \div (1+r)^n$, this is called the "present value".; Find r: $r = (A/P)^{1/n} - 1$, take the $n$th root of the ratio, then subtract 1.

Pause, copy the transposition formulas: present value $P = A \div (1+r)^n$ and rate $r = (A/P)^{1/n} - 1$ (take the $n$th root of the ratio, then subtract 1) into your book.

Did you get this? True or false: to find the present value $P$, you divide $A$ by $(1+r)^n$.

Work the examples

Follow three worked problems that solve for each variable in turn.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · FIND P (PRESENT VALUE)

You need $40,000 in 6 years. The best rate you can find is 5.2% p.a. compounded annually. How much must you invest today?

$P = \dfrac{A}{(1+r)^n} = \dfrac{40{,}000}{(1.052)^6}$

Identify $A = 40000$, $r = 0.052$, $n = 6$. Use the transposed formula.

PROBLEM 2 · FIND r (THE RATE)

An investment of $\$9,000$ grew to $\$14,500$ over 7 years with annual compounding. Find the annual interest rate.

$r = \left(\dfrac{14{,}500}{9{,}000}\right)^{1/7} - 1$

Use $r = (A/P)^{1/n} - 1$ with $A = 14500$, $P = 9000$, $n = 7$.

PROBLEM 3 · FIND n (LOGARITHMS)

How many years will it take $\$5,000$ to grow to $\$10,000$ at 5% p.a. compound interest?

$n = \dfrac{\ln(A/P)}{\ln(1+r)} = \dfrac{\ln(10{,}000/5{,}000)}{\ln(1.05)}$

Use the logarithm formula. We need $n$ such that $(1.05)^n = 2$.

Quick check: An investment doubles at 8% p.a. compound interest. Which expression gives the exact number of years?

The Rule of 72 · the investor's mental shortcut

The Rule of 72

mental maths

We just saw the exact formula $n = \ln(2)/\ln(1+r)$ for the doubling time. That raises a question: in practice, is there a quick mental estimate, one that doesn't require a calculator, that gives a reasonable answer for common rates? This card answers it → the Rule of 72: doubling time $\approx 72 \div \text{rate (\%)}$, accurate for rates 4–12%, derived from $\ln 2 \approx 0.693$ with an adjustment for the fact that $\ln(1+r) < r$.

Investors and bankers use a powerful mental shortcut:

$$\text{Doubling time (years)} \approx \dfrac{72}{\text{annual rate (\%)}}$$

Examples at a glance:

At 6% p.a.: doubling time $\approx 72 \div 6 = 12$ years (exact: 11.9 years)
At 8% p.a.: doubling time $\approx 72 \div 8 = 9$ years (exact: 9.0 years)
At 10% p.a.: doubling time $\approx 72 \div 10 = 7.2$ years (exact: 7.27 years)
At 4% p.a.: doubling time $\approx 72 \div 4 = 18$ years (exact: 17.67 years)

The Rule of 72 is accurate for rates between 4% and 12%. It works because $\ln(2) \approx 0.693$ and $\ln(1+r) \approx r$ for small $r$, giving $n \approx 0.693/r$. Using 72 instead of 69.3 makes mental division easier and corrects for $\ln(1+r) < r$.

Warren Buffett's 20% track record. Berkshire Hathaway has achieved roughly 20% compounded annual returns since 1965. Rule of 72: $72 \div 20 = 3.6$ years to double. Over 58 years, $\$1,000$ invested in 1965 would have doubled about 16 times, growing to over $\$36$ million. That is the mathematical reality behind the phrase "compound interest is the eighth wonder of the world."

Rule of 72: Doubling time $\approx 72 \div \text{rate (\%)}$. Only use for quick estimates; always verify with the exact logarithm formula for exam answers.; The rule is derived from $\ln(2) \approx 0.693 \approx 0.72$ adjusted for the fact that $\ln(1+r) < r$.

Pause, copy the Rule of 72: doubling time $\approx 72 \div \text{rate (\%)}$, for quick estimation only; always verify with the exact formula $n = \ln(2)/\ln(1+r)$ in the HSC, into your book.

Fill the gap: Using the Rule of 72, an investment at 9% p.a. doubles in approximately years. The exact answer using logarithms is $n = \ln(2) \div \ln(1.09) \approx$ years.

Avoid the traps

Fix the index and rounding slips that cost marks.

Common errors · the 3 traps that cost marks

Trap 01

Using division instead of logarithms to find n

The formula $n = (A-P)/(Pr)$ finds $n$ for simple interest only. For compound, the exponent makes the relationship non-linear, you must use $n = \ln(A/P) \div \ln(1+r)$.

Trap 02

Forgetting to subtract 1 when finding r

The formula gives $(A/P)^{1/n} = 1 + r$, so $r = (A/P)^{1/n} - 1$. If you stop at $r = (A/P)^{1/n}$ you get a number slightly over 1, that is $1+r$, not the rate.

Trap 03

Using Rule of 72 as an exact answer

The Rule of 72 is an approximation for mental estimates only. Exam questions asking for the exact number of years always require logarithms. Quote the Rule of 72 estimate, then verify with $n = \ln(A/P) \div \ln(1+r)$.

Match each goal to the correct transposed formula:

Drill it and revisit

Run the drill, then return to your opening answer and name what has changed.

Quick-fire practice · 5 transpositions

Find $P$: You need $\$40{,}000$ in 6 years at 5.2% p.a. compounded annually.

Find $r$: $\$9{,}000$ grew to $\$14{,}500$ in 7 years. Annual rate?

Find $n$ (exact): $\$6{,}000$ doubles at 4.5% p.a. compound interest. How many years?

A credit card charges 18% p.a. compounded daily. How many days for an unpaid $\$2{,}000$ to reach $\$2{,}500$? (daily rate $= 0.18/365$)

A house grew from $\$850{,}000$ to $\$1{,}200{,}000$ in 8 years. Find the annual compound growth rate.

Revisit your thinking

At 6% p.a., the exact doubling time is:

$n = \dfrac{\ln(2)}{\ln(1.06)} = \dfrac{0.6931}{0.05827} = 11.9$ years

The Rule of 72 gives $72 \div 6 = 12$ years, remarkably close.

The mental difficulty comes from the exponent: our brains think linearly, so we intuitively expect $6\% \times 10 = 60\%$ growth in 10 years. The actual growth is $(1.06)^{10} - 1 = 79\%$, nearly 20 percentage points higher. Exponential growth consistently outruns linear intuition.

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Practise compound interest

Answer the question bank, then write full short-answer responses.

Multiple choice

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Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 43 marks

Q1. How much must be invested today at 4.8% p.a. compound interest (annual) to have $25,000 at the end of 5 years? Show all working. (3 marks)

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ApplyBand 43 marks

Q2. An investment of $\$12,000$ grew to $\$18,000$ in 4 years with annual compounding. (a) Find the annual interest rate. (b) Verify using the Rule of 72 and comment on the accuracy. (3 marks)

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AnalyseBand 54 marks

Q3. (a) Find the exact number of years for $6,000 to double at 7.2% p.a. compound interest. (b) Use the Rule of 72 to estimate the answer. (c) Explain why the Rule of 72 gives such a precise estimate at this particular rate. (4 marks)

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📖 Comprehensive answers (click to reveal)

Drill 1: $P = 40000/(1.052)^6 = 40000/1.3549 = \$29{,}524.88$ · 2: $r = (14500/9000)^{1/7} - 1 = 7.02\%$ · 3: $n = \ln(2)/\ln(1.045) = 0.6931/0.04402 = 15.75$ years; Rule of 72: $72/4.5 = 16$ years · 4: daily rate $= 0.000493$; $n = \ln(1.25)/\ln(1.000493) = 0.2231/0.000493 \approx 452.5$ days · 5: $r = (1{,}200{,}000/850{,}000)^{1/8} - 1 = 4.39\%$ p.a.

Q1 (3 marks): $P = 25000/(1.048)^5 = 25000/1.26471 = \$19{,}768.25$ [2]. They need to deposit $\$19{,}768.25$ today [1].

Q2 (3 marks): (a) $r = (18000/12000)^{1/4} - 1 = (1.5)^{0.25} - 1 = 1.10668 - 1 = 0.1067 = 10.67\%$ p.a. [2]. (b) Doubling time at 10.67%: $72 \div 10.67 \approx 6.75$ years. At 10.67%, doubling takes about 6.9 years exactly, the Rule of 72 is slightly off at this rate [1].

Q3 (4 marks): (a) $n = \ln(2)/\ln(1.072) = 0.6931/0.06952 = 9.97 \approx 10$ years [2]. (b) $72 \div 7.2 = 10$ years [1]. (c) The Rule of 72 uses 72 rather than $100\ln(2) \approx 69.3$ to correct for $\ln(1+r) < r$. At $r = 7.2\%$ these two corrections cancel almost perfectly, the rule happens to be exact here because $100\ln(2)/\ln(1.072) \approx 72$ [1].

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Module overview · Maths Advanced · Checkpoint 1