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I need some help understanding this. So as far as I can tell. The rule of 72 is used to determine when prices will double in years. This is done by 72 divided by the rate, or interest. So it would look like 72/6 if the percent was only 6. Which would equal 12 years. But here's my issues, there is the rule of 72 doubling time and actual doubling time. Can someone please explain this part. Like if the percent is one then the rule of 72 doubling time is 72 years but actual doubling time in only 70 years. How is this actually calculated? Can someone show me and actually equation. I will love you long time lol.

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The Rule of 72 isn't exact. It's a rule of thumb.

The actual doubling time for a periodic growth rate depends on the periodic rate $r$ and the number of periods $n$:

$$(1+r)^n = 2.$$

As $r$ varies from $1$ to $15% percent, the product that results in a doubling time varies a bit, but not much.

If $r=0.01$ then $$1 \times n = \ln 2 / \ln 1.01 \approx 69.7,$$ but if $r=0.15$ then $$15 \times n = 15 \times \ln 2 / \ln 1.15 \approx 74.4.$$

So, financial planners and the like use The Rule of $72$ because (a) it's not that far off from truth, and (b) it divides evenly a lot of the interest rates they're likely to talk to clients about, and (c) it rhymes.

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$72$ is an approximation, not an exact value. It is handy because it has so many factors of $2$ and $3$ it is easy to divide into. It comes about because it is close to $100 \cdot \log(2) \approx 69.3$. If the compounding were instant, the correct value would be $69.3$, but because the compounding is annual you want a number a little higher. As the interest rate rises the rule becomes more accurate because the compounding is farther from continuous. In your example of $6\%$ the correct doubling time is $11.89$ years, very close to $12$.

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The underlying mathematical reason is that when $|x|$ is small, we have $$\ln(1+x) \approx x$$

So suppose the compound annual interest rate is $\frac{r}{100}$, we want to find $n$ such that $$(1+\frac{r}{100})^n = 2$$ which gives $n = \dfrac{\ln 2}{\ln(1+\frac{r}{100})} \approx \dfrac{\ln 2}{r/100} = \dfrac{100 \ln 2}{r}$

Then we replace $100 \ln 2$ by 72, since it has a lot of factors as @Ross Millikan has said, and also since we want something slightly larger than $100 \ln 2$(because $\frac{r}{100}$ is slightly larger than $\ln(1+\frac{r}{100})$)

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