Compound Interest and Finance

Money grows exponentially when it earns interest on both the original principal and the accumulated interest. This page covers the mathematics of compound interest — the most directly useful application of exponential functions you will encounter. Whether you are saving for retirement, comparing loan offers, or valuing an investment, these formulas are the tools.

The Compound Interest Formula

When interest is compounded $n$ times per year at annual rate $r$, the balance after $t$ years is:

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

where:

• $A$ = future amount (what you will have)
• $P$ = principal (what you start with)
• $r$ = annual interest rate (as a decimal)
• $n$ = number of compounding periods per year
• $t$ = time in years

Common compounding frequencies:

Example 1: You invest $5,000 at 6 percent annual interest compounded monthly. What is the balance after 10 years?

$A = 5000\left(1 + \frac{0.06}{12}\right)^{12 \times 10} = 5000(1.005)^{120}$

$(1.005)^{120} = e^{120\ln(1.005)} = e^{120 \times 0.004988} = e^{0.5985} = 1.8194$

$A = 5000 \times 1.8194 = 9{,}097$

After 10 years, the balance is approximately $9,097.

Example 2: Compare the balance of $10,000 at 5 percent for 20 years, compounded annually vs. quarterly vs. daily.

More frequent compounding yields more money, but the gains diminish as $n$ increases.

Continuous Compounding

As $n \to \infty$, the compound interest formula approaches:

$A = Pe^{rt}$

This is continuous compounding — interest is added at every instant. The derivation uses the limit definition of $e$:

$\lim_{n \to \infty} \left(1 + \frac{r}{n}\right)^{nt} = e^{rt}$

Example 3: The $10,000 at 5 percent for 20 years, compounded continuously:

$A = 10000e^{0.05 \times 20} = 10000e^1 = 10000 \times 2.7183 = 27{,}183$

Notice this is $27,183 — very close to the daily compounding result of $27,181. Continuous compounding is the theoretical upper limit, but daily compounding essentially reaches it.

Effective Annual Rate (EAR)

Different compounding frequencies with the same nominal rate produce different actual returns. The effective annual rate (EAR) standardizes the comparison:

$\text{EAR} = \left(1 + \frac{r}{n}\right)^n - 1$

For continuous compounding:

$\text{EAR} = e^r - 1$

Example 4: A bank offers 4.8 percent compounded monthly. Another offers 4.85 percent compounded annually. Which is better?

Bank 1: $\text{EAR} = \left(1 + \frac{0.048}{12}\right)^{12} - 1 = (1.004)^{12} - 1 = 1.04907 - 1 = 0.04907$

Bank 2: $\text{EAR} = 0.0485$ (already annual)

Bank 1’s EAR is 4.907 percent, which beats Bank 2’s 4.85 percent despite the lower nominal rate. Monthly compounding makes up the difference.

Present Value

Present value answers: “How much do I need to invest today to have a specific amount in the future?”

Rearrange the compound interest formula:

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

For continuous compounding:

$P = Ae^{-rt}$

Example 5: How much must you invest today at 7 percent compounded quarterly to have $50,000 in 15 years?

$P = \frac{50000}{(1 + 0.07/4)^{60}} = \frac{50000}{(1.0175)^{60}}$

$(1.0175)^{60} = e^{60 \times 0.01735} = e^{1.041} = 2.832$

$P = \frac{50000}{2.832} = 17{,}655$

You need to invest approximately $17,655 today.

Example 6: What is the present value of $100,000 due in 20 years if the discount rate is 5 percent, compounded continuously?

$P = 100000 \times e^{-0.05 \times 20} = 100000 \times e^{-1} = 100000 \times 0.3679 = 36{,}788$

The $100,000 future payment is worth $36,788 in today’s dollars.

The Rule of 72

The Rule of 72 is a quick estimate for doubling time:

$T_d \approx \frac{72}{r}$

where $r$ is the annual interest rate as a whole number (e.g., use 8 for 8 percent, not 0.08).

Example 7: At 8 percent annual growth, how long to double?

$T_d \approx \frac{72}{8} = 9 \text{ years}$

The exact answer is $\frac{\ln 2}{0.08} = 8.66$ years. The Rule of 72 gives 9 years — close enough for quick mental math.

Why 72? The number 72 works because it has many divisors (2, 3, 4, 6, 8, 9, 12) making mental division easy, and it closely approximates $100 \ln 2 \approx 69.3$ with an adjustment for compound (not continuous) growth.

The Rule of 72 is most accurate for rates between 4 and 12 percent. For very low or very high rates, the “Rule of 69.3” (using 69.3 instead of 72) is more precise but harder to compute mentally.

Comparing Investment Options

When comparing investments with different terms, rates, and compounding, always convert to a common measure. The two most useful comparisons:

1. Effective annual rate (EAR) — for comparing rates
2. Future value at a common time horizon — for comparing outcomes

Example 8: Three investment options for $20,000 over 5 years:

Option A has the highest EAR and future value, even though its nominal rate (5.2 percent) is lower than Option B’s (5.3 percent). More frequent compounding overcomes the rate difference.

Solving for Time and Rate

Finding Time

“How long until $8,000 grows to $20,000 at 6 percent compounded semiannually?”

$20000 = 8000(1.03)^{2t}$

$2.5 = (1.03)^{2t}$

$\ln(2.5) = 2t \cdot \ln(1.03)$

$t = \frac{\ln(2.5)}{2\ln(1.03)} = \frac{0.9163}{2 \times 0.02956} = \frac{0.9163}{0.05912} = 15.50 \text{ years}$

Finding Rate

“What annual rate, compounded quarterly, turns $15,000 into $25,000 in 8 years?”

$25000 = 15000\left(1 + \frac{r}{4}\right)^{32}$

$\frac{5}{3} = \left(1 + \frac{r}{4}\right)^{32}$

$\left(\frac{5}{3}\right)^{1/32} = 1 + \frac{r}{4}$

$1.01609 = 1 + \frac{r}{4}$

$r = 4 \times 0.01609 = 0.06436$

The required rate is approximately 6.44 percent.

Real-World Application: Comparing Loan Offers

A car buyer is choosing between two loan offers for $30,000:

• Offer A: 4.5 percent APR compounded monthly for 60 months
• Offer B: 4.2 percent APR compounded daily for 60 months

Which loan costs less in total interest?

The total amount repaid depends on the monthly payment formula (which involves annuities, a topic for a finance course). But we can compare using EAR:

Offer A: $\text{EAR} = (1 + 0.045/12)^{12} - 1 = (1.00375)^{12} - 1 = 0.04594$, or about 4.594 percent.

Offer B: $\text{EAR} = (1 + 0.042/365)^{365} - 1 = (1.0001151)^{365} - 1 = 0.04289$, or about 4.289 percent.

Offer B has the lower effective rate and will cost less in total interest.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Compound interest follows $A = P(1 + r/n)^{nt}$; continuous compounding follows $A = Pe^{rt}$
• More frequent compounding yields more growth, but with diminishing returns — daily and continuous compounding give nearly identical results
• The effective annual rate (EAR) is the true comparison tool: $\text{EAR} = (1 + r/n)^n - 1$
• Present value $P = A/(1+r/n)^{nt}$ tells you what a future sum is worth today
• The Rule of 72 gives a quick doubling-time estimate: $T_d \approx 72 / r$ (where $r$ is the rate as a whole number)
• To find unknown $t$ or $r$, take logarithms of both sides of the compound interest equation

Return to College Algebra for more topics in this section.