Here’s a handy rule of thumb for calculating compound interest:

For example, how long would it take for your money to double if it grew at a yearly rate of 5%? $72/5 = 14.4$ years. And, sure enough, $1.05^{14.4} = 2.01$!

So let’s say you’re trying to figure out how much money you’ll have saved for retirement if you save now $100,000 and it grows at an annual rate of 5% for the next 30 years? It will double every 14.4 years, so in 28.8 years it will double twice, so it will be a little more than$400,000. How’d we do? $100,000 * 1.05^{30} = 432,194$. Pretty good for something you can do in your head!

Probably obvious, but this trick can also convert between a “doubling time” and an interest rate. If I tell you that your money will double in 10 years, you know the interest rate is about $72/10 = 7.2\%$. And, sure enough, $1.072^{10} = 2.004$. Very close!

### How does it work?

What’s the equation we’re trying to solve?

• r is the interest rate
• y is the doubling time (the number of years it takes to double)
• and 2 is because we want our money to double

We need to turn an exponent into multiplication/division, which usually means taking the log of both sides. Let’s try it:

So far, everything we’ve done is exact. Now it’s time to make a few approximations:

The first one is trivial, you can just check it with a calculator. Why, though, is the second one true?

You can think about it this way. $e^0 = 1$. And the derivative of $e^x$ at 0 is also 1. So, if you zoom in really close around 0, it looks like a straight line with a y-intercept of 1 and a slope of 1. Which means, for small values of r, $1+r \approx e^r$. And if we take the natural log of both sides, we get $ln(1+r) \approx r$.

So, using what we have so far, we can say:

This works just fine, especially for really small values of r. For example, how long would it take for your money to double at a 1% interest rate? 69.3 years right? $1.01^{69.3} = 1.99$. Close!

So where does 72 come from?

Well, $ln(1+r) \approx r$ gets to be a worse approximation as $r$ gets large. In particular, $r$ is an overestimate for $ln(1+r)$. So, when we divide by $r$ in $\frac{0.693}{r}$, we’re dividing by something that’s too large. For “normal” interest rate values - say, 8% - $r$ is about 4% bigger than $ln(1+r)$. So, to adjust for that fact, we can just make numerator 4% bigger as well. What’s $0.693 * 1.04$? 0.72!