# The Rule of 72 (Investing)

## Introduction

When investing in anything, the investor always wants to know “when do I get my money back?”.  This question elicits the following math problem:

Given an interest rate i, when will my principle P double in value (that is, I have made my money back)?

First off, this question is not quite well defined – we are not told how often the interest is compounded (or if it is compounded), but we will work with that.

The rule of 72 is a very convenient little trick for those interested in this rule. The rule states:

Divide 72 by the interest rate.  This will give you approximately the number of years it will take to double your money.

## Example

Lets say I have a savings account that gives me 3% interest annually.  Then, dividing 72 / 3 to get 24, we get that it would take approximately 24 years for my money to double.  In an investment that earns 10% annually, it would take approximately 72 / 10 = 7.2 years.  Again, notice the exact answer will change depending on how often the money is compounded.

## Why does this work?

The general compound interest formula to find the accumulated value $A$ of an investment of principle $P$ with interest rate $i$ calculated $m$ times per year for $n$ interest periods is

$\displaystyle A = P\left(1 + \frac{i}{100m}\right)^n$.

In this situation, $A = 2P$, $i$ is given, $n$ is what we are looking for, and $m$ is what is not well-defined.  Lets assume first off $m$ is 1, that is, that interest is calculated once per year.  Notice first that since $A = 2P$, and $P \neq 0$, we can divide by $P$ on both sides and it cancels out completely (thus the time it takes has nothing to do with the amount of money you start with).  So choosing $m = 1$ we get:

$\displaystyle 2 = \left(1 + \frac{i}{100}\right)^n$.

We can now solve for $n$ based on $i$:

$\displaystyle 2 = \left(1 + \frac{i}{100}\right)^n$

$\displaystyle \ln 2 = n \ln \left(1 + \frac{i}{100}\right)$

$\displaystyle n = \frac{\ln 2}{\ln \left(1 + \frac{i}{100}\right)}$

Now to see how this relates to the more elementary formula 72/i, lets run some numbers:

So we can see that this is a good estimate (within a fifth of a year) when i is between 5 and 20%.  However, when we look at compounding the interest monthly (m = 12), we get a slightly different picture:

Here we can see that it is just as good an estimate (again, within a fifth of a year) from 12% up to 100% (where it is exact!).

Going one further to compounding interest continually (that is, the limit as interest is calculated more and more often–more than every second–the formula for which is $A = Pe^{\frac{i}{100}t}$ ), we get the following:

We see that this is best for higher interest rates: 15% and up.

## Conclusion

So this rule does work and is a good quick measure, but depending on how often the interest is calculated, it may be better or worse.  When interest is calculated yearly, this estimate is good for small interest rates (5% – 20%), and as interest is calculated more and more often, the estimate gets better for higher interest rates (good for 12% to 100% and higher for monthly and over 15% for continually).  Overall, this is a good quick estimate to use for those who want an answer quick, or who don’t want to have to use logs to find the exact value.  For small interest rates (1-4%) this technique is never that good, but in practice, doing this technique with a small value, say 3%, you get 24 years, where the actual is closer to 23 years…either way, it’s a LOOOOONG time to double your money.  Hope you found this helpful / interesting!

Rob

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### 3 comments to The Rule of 72 (Investing)

• erauqssidlroweht

My gr 11 math teacher taught us this when we spent a week doing the entire consumer math curriculum. It is immensely useful.

• Jason

This is really cool Rob!

One interesting thing I can add is that e can be defined as the limit of (1+1/x)^x as x approaches infinity.

Define a function y = f(x)= (1+1/x)^x

Now, take the natural logarithm of both sides,

ln(y) = ln[(1+1/x)^x]

• Jason

= xln(1+1/x) (by property of logarithms)
= ln(1+1/x)/(1/x)

But this is an intermediate form (0/0) so L’Hospital’s rule can be applied.

Thus, by L’Hospital’s rule,

lim(x->inf)ln(y)= lim(x->inf)[[1/(1+1/x)]*(-1/x^2)]/(-1/x^2)
= lim(x->inf)[(1/(1+1/x)]
= 1

But e^ln(y) = y = f(x)

So lim(x->inf)f(x) = lim(x->inf)[e^ln(y)]
= e^[lim(x->inf)ln(y)] (since e^x is continuous)
= e^1
= e

So lim(x->inf)[(1+1/x)^x] = e

e is such a beautiful number!