Rule of 72

Date: 01/26/99 at 18:46:22
From: Andrea Vasseur
Subject: The rule of 72

I need to know how the rule of 72 was started, and I need specifics 
on how it works. I understand that you divide 72 by the rate of change, 
but I do not understand how that always leads to the answer.

Date: 01/28/99 at 02:26:30
From: Doctor Schwa
Subject: Re: The rule of 72

We're trying to find when a given amount of money P has doubled if it
grows at an interest rate r (say, .06) with compounding.

So, after t years, the amount you have is

   P * (1 + r)^t

which is supposed to be double the original amount, so

   2P = P (1+r)^t

and dividing by P,

   2 = (1 + r)^t

Now, taking the natural log of both sides,

   ln 2 = t ln(1+r)

Now, ln(2) is about .72 (actually it's .693, a little less than .72)
and ln(1+r) is about r (actually it's a little less; how much less
depends on how big r is). So, by approximating the ln 2 = .72 and the 
ln (1+r) = r, we've overestimated both sides, so roughly the errors 
cancel, and we get .72 = t r and thus t = .72/r.

How far off is the approximation if r = .06?  Let's check. The 
approximation says .72/.06 = 12 years should be the doubling time, and 
the exact calculation ln(2) / ln(1.06) = 11.9 roughly. So we're off by 
about a month out of 12 years. Not bad.

If the interest rates are really small, like .01, then you need to use
a number closer to the true value of .693 for ln 2. To see why, look at 
ln(2) / ln(1.01) = 69.66, which is fairly far off of 72.

And if the interest rates are really big, like .24, you need to use a 
number bigger than 72. To see why, look at ln(2) / ln(1.24) = 3.22, 
when the rule of 72 would give 72/24 = 3.

So in short, the rule of 72 doesn't give an exact answer, but for most 
common interest rates (say, in the 6-9 percent range,) it is accurate 
enough for most purposes. And it has the added convenience that 72 is 
evenly divisible by 6, 8, and 9, so you get nice round numbers for your 
answers.

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/