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TriKri · #5 November 27th, 2006, 12:18 PM

If we say $\begin{array}{cc}\text{lim}\\R \rightarrow 0\end{array}$ (I don't really knoh how to use limens, but I do so anyway), then the formula for the number of years until doubling the amount:

$N\ =\ \frac{log(2)}{\displaystyle{log\left(1+ \frac{R}{100\ \%}\right)}}\ =\ \frac{ln(2)}{\displaystyle{ln\left(1+ \frac{R}{100\ \%}\right)}}$
Can be as well written as
$N = \frac{ln(2)}{R/100\ \%} = \frac{100\cdot ln(2)\ \%}{R}$
since
$ln(x) = x - 1$ when $\begin{array}{cc}\text{lim}\\x \rightarrow 1\end{array}$ .

Now $100 \cdot ln(2)$ isn't 72, it is actually closer to 69.3. So we would get the formula
$N = \frac{69.3\ \%}{R}$
assuming that $\begin{array}{cc}\text{lim}\\R \rightarrow 0\end{array}$ . But that can't true, since the amount would not increase at all and the number of years until doubling would be infinite. Thus it's unfair to use the constant 69 or 69.3. I believe 72 is more of an average value for "different common R-values". To see what R-value they have used to get the value 72, we'd have to solve the equation

$\left(1 + \frac{R}{100}\right)^{\frac{72}{R}} = 2\ \Rightarrow\ ...\ \Rightarrow\ \frac{\displaystyle{ln\left(1+\frac{R}{100}\right)}}{R/100} = \frac{ln(2)}{0.72} \approx 0.963$