exponential equation help?

Question

According to the rule of 72, an investment at r% interest compounded continuously will double in approximately 72/r years. Show that a more accurate doubling time is 69.3/r years.
I am getting confused because if the formula is P(t)=Pe^(rt)
wouldn't (72/r)/(r) cancel out? But then it doesn't work.

Answer 1

You are approaching the question in the wrong way: why start with 72, when you already know that 69.3 is more accurate?

Think of it this way:

Suppose you make an initial investment P at t=0 (time = year 0) and interest r%. Let P(t) be your total amount at year t, so we can say that P(0)=P. Since we are investing at r%, after one year our total investment would be (100+r)% of P, or P(1+.01r). After the next year our total investment would be (100+r)% of P(1+.01r), or P(1+.01r)^2. And so on. We can summarize this as follows:

P(0) = P
P(1) = P*(1+.01r)
P(2) = P*(1+.01r)^2
...
and we can generalize it to:
P(t) = P*(1+.01r)^t, where t is time measured in years.

This is our formula. Now, the question is, at what time T does our investment double? In other words, for what value T does P(T)=2*P? Well, we can plug in the formula for P(t):

P(T) = 2*P
P*(1+.01r)^T = 2*P
(1+.01r)^T = 2
ln[(1+.01r)^T] = ln2 = .693
T*ln(1+.01r) = .693

For small values of x, ln(1+x) is very close to x. You can easily show this with limits. Since .01r is a small number, we know that ln(1+.01r) is very close to .01r.

Thus, we have:

T*.01r = .693
T = .693/.01r = 69.3/r

Thus 69.3/r is a better approximation than 72/r. You can verify this by plugging in your own numbers.

Answer 2

the rule of 72 is a general guide
of course it does not work for 72% interest for one year
where are you getting that high interest anyway?

it is more like for 7.2% for 10 years will double because of compounding

try it with different interest rates and the corresponding periods to see if 69.3 is better