Chapter 4.3, Problem 89E

To determine

The time required to double an investment at $6\%,7\%\text{ and }8\%$.

Answer to Problem 89E

The time required to double an investment at $6\%$ is $\underset{¯}{t\approx 11.552}$, the time required to double an investment at $7\%$ is $\underset{¯}{t\approx 9.902}$ and the time required to double an investment at $8\%$ is $\underset{¯}{t\approx 8.664}$.

Explanation of Solution

Given that, the time t required to double the amount of an investment at an interest rate r compounded continuously is $t=\frac{\ln 2}{r}$.

Substitute the value of $r=0.06$ in the formula $t=\frac{\ln 2}{r}$ and calculate the time required to double an investment at $6\%$ can be calculated as follows.

$\begin{array}{c}t=\frac{\ln 2}{0.06}\\ \approx \frac{0.693147}{0.06}\\ \approx 11.552\end{array}$

Hence, the time required to double an investment at $6\%$ is $\underset{¯}{t\approx 11.552}$.

Now, substitute the value of $r=0.07$ in the formula $t=\frac{\ln 2}{r}$ and calculate the time required to double an investment at $7\%$ can be calculated as follows.

$\begin{array}{c}t=\frac{\ln 2}{0.07}\\ \approx \frac{0.693147}{0.07}\\ \approx 9.902\end{array}$

Hence, the time required to double an investment at $7\%$ is $\underset{¯}{t\approx 9.902}$.

Substitute the value of $r=0.08$ in the formula $t=\frac{\ln 2}{r}$ and calculate the time required to double an investment at $8\%$ can be calculated as follows.

$\begin{array}{c}t=\frac{\ln 2}{0.08}\\ \approx \frac{0.693147}{0.08}\\ \approx 8.664\end{array}$

Hence, the time required to double an investment at $8\%$ is $\underset{¯}{t\approx 8.664}$.