Interest is the price of using money. A bank pays it to you when you save, and you pay it when you borrow. The way it is calculated decides whether $1,000 turns into $1,500 or $1,628.89 after ten years at the same 5% rate.
There are two methods. Simple interest is always calculated on the original amount. Compound interest also earns interest on the interest already added. This page gives both formulas, a chart you can move to see the gap open up, an explanation of why compounding more often helps only a little, and the rule of 72, a shortcut for estimating how long money takes to double. It is a math explanation, not financial advice.
Key terms
- Principal (P): the starting amount.
- Rate (r): the yearly rate written as a decimal. 5% is $latex 0.05$.
- Time (t): the number of years.
- Amount (A): the total at the end, principal plus interest.
- Interest (I): $latex I = A - P$.
Keep the units matched. If the rate is yearly, time must be in years: six months is $latex t = 0.5$.
Simple interest formula
Simple interest is calculated only on the principal, so every year earns the same amount.
$latex I = P\,r\,t \qquad A = P(1 + rt)$
At 5%, $1,000 earns $50 each year. After 10 years that is $500 of interest, so the amount is $1,500. The balance rises along a straight line because the same amount is added every year. That is an arithmetic sequence.
Compound interest formula
With compound interest, the interest is added to the balance at the end of each period, and the next period's interest is calculated on the bigger balance.
$latex A = P\left(1 + \frac{r}{n}\right)^{nt}$
- $latex n$ is the number of compounding periods per year: 1 for annual, 2 for semiannual, 4 for quarterly, 12 for monthly, 365 for daily.
- $latex r/n$ is the rate for one period.
- $latex nt$ is the total number of periods.
For annual compounding $latex n = 1$ and the formula becomes $latex A = P(1+r)^t$. The balance is multiplied by the same factor $latex 1+r$ every year, which makes it a geometric sequence. This table compares both methods for $1,000 at 5% a year:
| Year | Simple interest | Compound interest (annual) |
|---|---|---|
| 1 | $1,050.00 | $1,050.00 |
| 2 | $1,100.00 | $1,102.50 |
| 3 | $1,150.00 | $1,157.63 |
| 5 | $1,250.00 | $1,276.28 |
| 10 | $1,500.00 | $1,628.89 |
| 20 | $2,000.00 | $2,653.30 |
| 30 | $2,500.00 | $4,321.94 |
The first year is identical. After that the compound balance pulls ahead, because each year's interest is calculated on a larger number.
Race the two methods
Move the sliders. At a very low rate the lines almost overlap. Raise the rate and the years and the shaded gap grows: that gap is interest earned on interest. The bars below the chart show what changes when the same rate compounds more often.
Race: simple interest vs compound interest
Change the numbers and watch the gap open up. Press Race to replay the animation.
Interest earned, by how often it compounds
Why compounding frequency matters (and why it stops mattering)
Compounding more often adds interest to the balance sooner, so that interest starts earning interest sooner. The effect is real but small. This is $1,000 at 6% for 10 years:
| Compounding | Amount after 10 years |
|---|---|
| Annual (n = 1) | $1,790.85 |
| Semiannual (n = 2) | $1,806.11 |
| Quarterly (n = 4) | $1,814.02 |
| Monthly (n = 12) | $1,819.40 |
| Daily (n = 365) | $1,822.03 |
| Continuous | $1,822.12 |
Going from yearly to monthly adds $28.55. Going from monthly to daily adds only $2.63. The numbers squeeze toward a limit, because $latex \left(1+\frac{r}{n}\right)^{n}$ approaches $latex e^{r}$ as $latex n$ grows. That limit is called continuous compounding:
$latex A = Pe^{rt}$
To compare accounts that compound differently, convert each to its effective annual rate, $latex \left(1+\frac{r}{n}\right)^{n}-1$. A rate of 6% compounded monthly has an effective annual rate of 6.17%, slightly above the 6% printed on the label. Use the Compound interest calculator to test your own numbers.
The rule of 72
To estimate how long money takes to double, divide 72 by the yearly rate written as a percent:
$latex \text{years to double} \approx \frac{72}{\text{rate in percent}}$
At 8% the estimate is $latex 72 \div 8 = 9$ years. The exact answer comes from solving $latex P(1+r)^t = 2P$, which gives:
$latex t = \frac{\ln 2}{\ln(1+r)}$
At 8% that is 9.006 years, so the shortcut is off by a few days. The Logarithm calculator computes the logarithms, and the article on applications of logarithms shows where this formula comes from.
Why 72? For small rates $latex \ln(1+r)$ is close to $latex r$, and $latex \ln 2 \approx 0.693$, which gives 69.3 divided by the rate. People use 72 because it fits typical rates between 6% and 10% better and divides evenly by 2, 3, 4, 6, 8, 9 and 12. Here is how the two compare, and a curve you can slide along:
| Rate | Exact years to double | 72 ÷ rate |
|---|---|---|
| 2% | 35.00 | 36.00 |
| 4% | 17.67 | 18.00 |
| 6% | 11.90 | 12.00 |
| 8% | 9.01 | 9.00 |
| 9% | 8.04 | 8.00 |
| 12% | 6.12 | 6.00 |
| 18% | 4.19 | 4.00 |
| 24% | 3.22 | 3.00 |
The rule of 72 against the exact doubling time
Slide the rate. The solid curve is exact, the dashed one is 72 divided by the rate.
The shortcut is best between 6% and 10%. At very high rates it underestimates the time (24% gives 3.00 years by the rule and 3.22 exactly). Always plug in the rate as a number of percent, 8 and not 0.08.
Worked examples
EXAMPLE 1
Find the interest and the amount for $2,000 at 5% simple interest for 3 years.
Use $latex I = Prt$: $latex I = 2000 \times 0.05 \times 3 = 300$.
The amount is $latex A = 2000 + 300 = 2{,}300$. So the interest is $300 and the amount is $2,300.
EXAMPLE 2
Same money, but compounded annually: $2,000 at 5% for 3 years.
$latex A = 2000(1+0.05)^3 = 2000 \times 1.157625 = 2{,}315.25$.
The interest is $315.25, which is $15.25 more than simple interest paid on the same money. The gap is small after 3 years and grows quickly after that.
EXAMPLE 3
Find the amount for $5,000 at 6% a year compounded monthly for 2 years.
Here $latex n = 12$, so the rate per month is $latex 0.06/12 = 0.005$ and the number of periods is $latex nt = 24$.
$latex A = 5000(1+0.005)^{24} = 5{,}635.8$. The amount is $5,635.80.
EXAMPLE 4
How long does money take to double at 8% compounded yearly?
Solve $latex 1.08^t = 2$. Taking logarithms, $latex t = \ln 2 / \ln 1.08 = 9.006$.
It takes about 9 years. The rule of 72 gives $latex 72/8 = 9$, which is very close.
Common mistakes
- Writing the rate as a whole number. Use 0.05 in the formulas, not 5. The rule of 72 is the one place where you use 5.
- Mixing time units. A monthly rate needs t in months. An annual rate needs t in years.
- Confusing n and t. The exponent is $latex nt$, the number of periods, not t alone.
- Forgetting to subtract the principal. The formulas for $latex A$ give the total. The interest is $latex A - P$.
- Rounding early. Keep every digit of $latex (1+r/n)$ until the last step.
- Expecting daily compounding to be a big upgrade. Past monthly, the gains are tiny.
Solved exercises
EXERCISE 1
Find the interest and the amount for $1,500 at 4% simple interest for 5 years.
Solution
$latex I = 1500 \times 0.04 \times 5 = 300$, so the interest is $300 and the amount is $latex 1500 + 300 = 1800$, that is $1,800.
EXERCISE 2
How many years does $800 need at 3% simple interest to earn $120 of interest?
Solution
From $latex I = Prt$ we get $latex t = \frac{I}{Pr} = \frac{120}{800 \times 0.03} = 5$. It takes 5 years.
EXERCISE 3
Find the amount for $3,000 at 7% compounded annually for 4 years.
Solution
$latex A = 3000(1.07)^4 = 3000 \times 1.31079601 = 3{,}932.39$. The amount is $3,932.39.
EXERCISE 4
Find the amount for $2,500 at 8% a year compounded quarterly for 3 years.
Solution
$latex n = 4$, $latex nt = 12$, rate per quarter $latex 0.08/4 = 0.02$. Then $latex A = 2500(1.02)^{12} = 3{,}170.6$. The amount is $3,170.60.
EXERCISE 5
(a) Use the rule of 72 to estimate the time to double at 6%. (b) What yearly rate doubles money in 9 years? Compare with the exact values.
Solution
(a) $latex 72/6 = 12$ years. Exactly, $latex \ln 2/\ln 1.06 = 11.9$ years.
(b) The rule gives $latex 72/9 = 8$%. Exactly, $latex 1+r = 2^{1/9}$, so $latex r = 0.0801$, about 8.01%.
EXERCISE 6
Compare $10,000 at 6% for 30 years with simple interest and with annual compounding.
Solution
Simple: $latex A = 10000(1 + 0.06 \times 30) = 28{,}000$. Compound: $latex A = 10000(1.06)^{30} = 57{,}434.91$.
The compound result is $57,434.91, more than double the $28,000 from simple interest.
Frequently asked questions
Which is better, simple or compound interest?
When you earn interest, compound is better, since once the first period has passed it always pays more. When you pay interest on a loan, the same property works against you. Check which method the contract uses.
Does compounding daily beat monthly by much?
No. In the table above, going from monthly to daily adds $2.63 on $1,000 over ten years. The first jump, from yearly to a few times a year, is the one that counts.
Is the rule of 72 exact?
No, it is an estimate that works best for rates of about 6% to 10%. For exact answers use $latex t = \ln 2/\ln(1+r)$.
What about regular deposits every month?
Those follow a different formula, which adds up a geometric series. This page covers a single deposit that is left alone.
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