Using the Rules of 69, 70, & 72

Doubling Time Shortcuts

The "Rule of 69," "Rule of 70," and "Rule of 72" refer to shortcuts for estimating the doubling time of a process that increases at a constant rate. You can use the rules to compute the doubling time of a bacterial colony growing in a petri dish, the number of periods it takes for money in an account to double, or any other real-world growth scenario.

If the growth rate is R% and you select the "Rule of N" (N = 69, 70, or 72), then it requires approximately N/R periods for the quantity to grow to twice its original size.

Below we explain how to choose the appropriate value of N and why the method yields accurate estimates.

When to Use N = 69

Use N = 69 when dealing with any continuous* growth process, or when you have a periodic growth process and the rate is between 0% and 0.5%.

Example 1: A culture of cells grows at a continuous rate of 6.9% per day. In how many days will the number of cells double? Solution: Simply calculate 69/6.9 = 10, so the answer is about 10 days.

Example 2: A credit card has an annual rate of 4.8%, with interest compounded monthly. How many months does it take for the credit card balance to double if you never make any payments? Solution: Since the annual periodic rate is 4.8%, the monthly periodic rate is 4.8/12 = 0.4%. Therefore, it will take around 69/0.4 = 172.5 months for the balance to double. (Approximately 14 and a half years)

When to Use N = 70

Use N = 70 to approximate the doubling time when the periodic rate is between 0.5% and 4.9%.

Example 3: You invest money in a savings account that earns 3.5% interest compounded yearly. How long must you wait for your money to double if you never add or subtract funds? Solution: Calculate 70/3.5 = 20, so the answer is roughly 20 years.

When to Use N = 72

Use N = 72 when the growth rate is between 4.9% and 11%.

Example 4: Since 2007, a business has grown annually by 8%. At this rate, when will the business double? Solution: We compute 72/8 = 9, so the business will double in size in about 9 years. This means in 2016 it will be roughly twice as large as it was in 2007.

Why the Method Works

If you start with a quantity Q and it grows at a rate of R%, the doubling time T can be found by solving this equation:

2Q = Q(1 + R/100)^T

The exact solution is T = [Ln 2]/[Ln(1 + R/100)], where Ln is the natural logarithm function, and Ln 2 ≈ 0.693147. For small values of R between 0 and 0.5, the function Ln(1 + R/100) is approximately equal to (R/100). Thus,

[Ln 2]/[Ln(1 + R/100)] ≈ 0.693147/(R/100) ≈ 69/R.

When R is between 0.5 and 4.9, the function Ln(1 + R/100) ≈ 0.99(R/100). Thus,

[Ln 2]/[Ln(1 + R/100)] ≈ 0.693147/[0.99(R/100)] ≈ 70/R.

And finally, when R is between 4.9 and 11, the function Ln(1 + R/100) is roughly equal to 0.962(R/100). And so we get,

[Ln 2]/[Ln(1 + R/100)] ≈ 0.693147/[0.962(R/100)] ≈ 72/R.

If you need a shortcut rule for percentages greater than 11% but less than 20%, you must use a higher value of N, such as 75. For instance, if a process grows by 15% each year, then the doubling time is roughly 75/15 = 5 years. (The actual value is 4.9595 years.) If you used N = 69, 70, or 72 in this case, you would get a value too low.

For continuous growth, the doubling time T comes from solving the equation 2Q = Qe^(R/100)T, which has a solution of [Ln 2]/[R/100]. Continuous growth means that the period of compounding is instantaneous, or so small that it is effectively continuous.

Refinements

The function Ln(1 + R/100) can be better approximated by the function R/100 - R²/20000, for R values between 0 and 11. This gives a better shortcut estimation formula for the doubling time:

[Ln 2]/[Ln(1 + R/100)] ≈ 0.693/[R/100 - R²/20000] ≈ 13860/(200R - R²).

Example 5: A savings account accrues interest at a rate of 5% per year. The time it takes for a person's money to double is approximately

13860/(200*5 - 5*5) = 13860/(975) = 14.2154 years.

(Compare this to the actual value of 14.2067.)

© Had2Know 2010