Despite what you may have heard, Einstein probably never said that compound interest was the greatest force in the universe. It is, however, an interesting beastie.

The Mathematical Ninja likes quick fixes. The Mathematical Ninja LOVES estimating powers of e, but he loves quick-and-dirty estimates even more.

Especially to one particular question: how long will it take my money to double?

For example, say you have £1,000 to invest at 8%, compounded annually. How long will it take to double? The Mathematical Ninja says, without hesitation, nine years. In fact, $(1.08)^9 = 1.99900$, which is pretty close. At 6%, it would take twelve years.

The Mathematical Ninja is using something called The Rule of 72: to double your money at an annual rate of $r$%, it takes $72 ÷ r$ years. Very handy rule. But why does it work?

You're trying to solve $\left(1 + \frac{r}{100}\right)^k=2$, where $k$ is the number of years. With an unknown power, there's only one thing for it: $\log$s.

We've got $k \ln \left(1 + \frac{r}{100}\right) = \ln(2)$, so:

$$k = \frac{\ln(2) }{\ln\left(1 + \frac{r}{100}\right)}$$.

Now, if you know your expansions, you know that $\ln \left(1 + x\right) \simeq x$, so that gives us something like $k \simeq \ln(2) \div \frac{r}{100}$, or $k \simeq \frac{100 \ln(2)}{r}$.

$100 \ln 2$ is, of course, about 69.3 -- which isn't too far from 72, but it's not quite right. The difference comes from the approximation being, well, approximate; $\ln \left(1 + x\right) \simeq x - \frac{1}{2}x^2$, meaning you have to adjust $k$ slightly upwards in all cases. 72 is generally not far off.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.