\(A = P(1 + \frac{r}{100})^n\)

Where:

• \(A\) = final amount
• \(P\) = principal (original amount)
• \(r\) = percentage increase per period (as a number)
• \(n\) = number of periods

Example:

A city's population of 50,000 increases by 2% annually. What will the population be after 5 years?

Using the compound formula:

\(A = 50,000 \times (1 + \frac{2}{100})^5\)

\(A = 50,000 \times (1.02)^5\)

\(A = 50,000 \times 1.1041\)

\(A = 55,205\)

The city's population will be approximately 55,205 after 5 years.

\(r = \left(\left(\frac{A}{P}\right)^{\frac{1}{n}} - 1\right) \times 100\%\)

Example:

A house value increased from $200,000 to $250,000 over 5 years. What was the average annual percentage increase?

\(r = \left(\left(\frac{250,000}{200,000}\right)^{\frac{1}{5}} - 1\right) \times 100\%\)

\(r = \left((1.25)^{0.2} - 1\right) \times 100\%\)

\(r = \left(1.0456 - 1\right) \times 100\%\)

\(r = 0.0456 \times 100\%\)

\(r = 4.56\%\)

The average annual percentage increase was approximately 4.56%.

\(\text{EAR} = \left(1 + \frac{r}{100 \times m}\right)^m - 1\)

Where:

• \(\text{EAR}\) = effective annual rate (as a decimal)
• \(r\) = nominal annual percentage increase
• \(m\) = number of compounding periods per year

Example:

If an investment grows at a rate of 8% per year, compounded quarterly, what is the effective annual rate?

\(\text{EAR} = \left(1 + \frac{8}{100 \times 4}\right)^4 - 1\)

\(\text{EAR} = (1 + 0.02)^4 - 1\)

\(\text{EAR} = 1.0824 - 1\)

\(\text{EAR} = 0.0824 = 8.24\%\)

The effective annual rate is 8.24%.

\(A = Pe^{rt}\)

Where:

• \(A\) = final amount
• \(P\) = principal (original amount)
• \(r\) = rate (as a decimal)
• \(t\) = time in years
• \(e\) = Euler's number (approximately 2.71828)

Example:

If $5,000 is invested at 6% annual interest, compounded continuously, how much will it be worth after 10 years?

\(A = 5,000 \times e^{0.06 \times 10}\)

\(A = 5,000 \times e^{0.6}\)

\(A = 5,000 \times 1.8221\)

\(A \approx 9,110.50\)

The investment will be worth approximately $9,110.50 after 10 years.

\(\text{Years to double} \approx \frac{72}{\text{Annual Percentage Increase}}\)

Example:

If your money grows at 8% per year, approximately how long will it take to double?

Years to double ≈ 72 ÷ 8 = 9 years

Let's verify this with the exact formula:

Using \(A = P(1 + \frac{r}{100})^n\) and setting \(A = 2P\):

\(2P = P(1 + \frac{8}{100})^n\)

\(2 = (1.08)^n\)

\(\log(2) = n \times \log(1.08)\)

\(n = \frac{\log(2)}{\log(1.08)} = \frac{0.301}{0.0334} \approx 9.01\)

It would take about 9.01 years, so the Rule of 72 gives a very close approximation!

Mathematical Derivation:

The Rule of 72 comes from the approximation that \(\ln(2) \approx 0.693\), and when we use the natural logarithm to solve the doubling time equation:

\(T = \frac{\ln(2)}{\ln(1+r)} \approx \frac{0.693}{r}\) (for small values of r)

Multiplying by 100 (to use the percentage instead of the decimal), we get approximately 70, which is rounded to 72 for easier mental calculations with common interest rates.