F-LE Doubling Your Money

On January 1, 2010 Janice put $450 in a savings account that pays 4.2% interest once at the end of each year.

How much money will she have in January 2014?
Write an equation that can be used to calculate the amount of money, $B$ ( in dollars), that Janice has in her account in terms of $n$, the number of times interest has been paid on the account.
How long will it take for Janice to double her money?
Given that for small values of $r$, $\ln \left(1+\frac{r}{100} \right) \approx \frac{r}{100}$, show that if the interest rate is $\frac{r}{100}$, the amount of money in Janice’s account will double in approximately $\frac{70}{r}$ years. [The Rule of 70]

The Rule of 70 is commonly used in financial calculations to estimate the doubling time of investments.

Solution: Doubling Your Money

Students may answer this question by making a table of balances for the beginning of each year.

Generalizing from the table in (a), \)B = 450 \cdot 1.042^n = 450 \left ( 1+ \frac{4.2}{100} \right )^n\(
We want to find the first \)n$ for which $450 \cdot 1.042^n \ge 900$. First we solve the equation $$900 = 450 \left ( 1+ \frac{4.2}{100} \right )^x$$ for $x$: \begin{align} \frac{900}{450} &= (1 + .042)^x\\ 2 &= 1.042^x\\ \ln 2 &= x \ln(1.042)\\ \frac{\ln 2}{\ln(1.042)} &=x\\ x &\approx 16.847. \end{align} So after interest is paid 17 times, Janice will have slightly more than double her money: $453(1.042)^{17} \approx 905.66$. This will be in January of the year 2027.
Generalizing from part (c), we have: $$t = \frac{\ln 2}{\ln\left(1+\frac{r}{100}\right) }. $$ So using the approximation given in part d), we have $$t \approx \frac{\ln 2}{\frac{r}{100}} = \frac{100(\ln 2)}{r} \approx \frac{0.693 \cdot 100}{r} \approx \frac{70}{r}.$$