Interest is the price of money over time. The difference between simple and compound interest lies in which amount the rate applies to in each period.

Notation: PP is the principal (initial amount), rr the rate per period (as a decimal), tt the number of periods, II the interest and A=P+IA = P + I the final amount.

Simple interest

The rate always applies to the initial principal, so each period earns the same interest:

I=P⋅r⋅tA=P (1+r t)I = P \cdot r \cdot t \qquad A = P\,(1 + r\,t)

Worked example: $1,000 at 2% per month for 12 months

  1. Decimal rate: r=0.02r = 0.02.
  2. Interest: I=1000⋅0.02⋅12=240I = 1000 \cdot 0.02 \cdot 12 = 240.
  3. Amount: A=1000+240=1240A = 1000 + 240 = 1240.

Compound interest

The rate applies to the previous period’s amount, so interest earns interest:

A=P (1+r)tI=A−PA = P\,(1 + r)^t \qquad I = A - P

Worked example: the same $1,000 at 2% per month for 12 months

  1. Monthly factor: 1.021.02.
  2. Amount: A=1000⋅1.0212≈1000⋅1.26824≈1268.24A = 1000 \cdot 1.02^{12} \approx 1000 \cdot 1.26824 \approx 1268.24.
  3. Interest: I≈268.24I \approx 268.24, which is $28.24 more than with simple interest.

Side by side

Principal of $1,000 at 10% per period:

PeriodSimpleCompound
01,0001,000
11,1001,100
21,2001,210
31,3001,331
102,0002,593.74

Simple interest grows as an arithmetic sequence (adds 100 each period); compound interest grows as a geometric sequence (multiplies by 1.1). They match in the first period; after that, the gap keeps widening.

Equivalent rates

Two rates are equivalent when they take the same principal to the same amount over the same term. With compound interest:

(1+ryear)=(1+rmonth)12(1 + r_{\text{year}}) = (1 + r_{\text{month}})^{12}

Worked example: what annual rate is 1% per month?

  1. (1.01)12≈1.1268(1.01)^{12} \approx 1.1268.
  2. ryear≈0.1268=12.68%r_{\text{year}} \approx 0.1268 = 12.68\% per year.

Dividing an annual rate by 12 is only correct for simple interest.

Finding the term with logarithms

When time is the unknown, take logarithms of both sides.

Worked example: how long does it take to double money at 10% per year?

  1. P (1.1)t=2P⇒1.1t=2P\,(1.1)^t = 2P \Rightarrow 1.1^t = 2.
  2. t=log⁡2log⁡1.1≈0.30100.0414≈7.27t = \dfrac{\log 2}{\log 1.1} \approx \dfrac{0.3010}{0.0414} \approx 7.27 years.

The “rule of 72” gives a quick estimate: 72÷10=7.272 \div 10 = 7.2 years.

Common mistakes

  1. Plugging the rate in as a percentage (22 instead of 0.020.02).
  2. Mixing units: an annual rate with a term in months.
  3. Converting compound rates by dividing or multiplying by 12.

Frequently asked questions

Which yields more, simple or compound interest?

At the same rate and for more than one period, compound interest yields more. For less than one period, simple interest gives a slightly larger amount.

Is 1% per month the same as 12% per year?

Only with simple interest. With compound interest, 1% per month is equivalent to 1.01¹² − 1 ≈ 12.68% per year.

How long does it take to double money at compound interest?

Solve (1 + r)^t = 2, that is, t = log 2 / log(1 + r). At 10% per period, t ≈ 7.27 periods.

Where does compound interest show up in everyday life?

In investments with reinvested returns, overdrafts, credit card debt and the installments of loans and mortgages.