Finite Mathematics - Compound Interest

If $P_{0}$ is deposited in a bank account with interest rate ( converted to a decimal) compounded times a year for years, then the final balance of the account is

$P = P_{0} \left (1+ \frac{r}{n} \right ) ^{nt}$

Dividing both sides of the formula by $P_{0}$ , this is

$\left (1+ \frac{r}{n} \right ) ^{nt} = \frac{P}{P_{0} }$

The money doubles, so substitute 2 for $\frac{P}{P_{0} }$ . Also, set and (bimonthly), and calculate as follows:

$\left (1+ \frac{0.07}{6} \right ) ^{6t} = 2$

$\ln \left [ \left (1+ \frac{0.07}{6} \right ) ^{6t} \right ]= \ln 2$

$6t \ln \left (1+ \frac{0.07}{6} \right ) = \ln 2$

$t = \frac{\ln 2}{ 6 \ln \left (1+ \frac{0.07}{6} \right ) }$

$\approx \frac{\ln 2}{ 6 \ln \left (1.011666667 \right ) }$

$\approx \frac{0.693147181 }{ 6 (0.011599136) }$

$\approx 9.960$

$0.960 \times 12 \approx 11.52$ , so this is 9 years and 11.52 months; rounding up to the next bimonth, this is 10 years even.

Compound Interest

Help Questions

Explanation