Chapter 3.4, Problem 130E

a.

To determine

To find: the time required for a $\$1000$ investment to double at interest rate $r$ , compounded continuously.

Answer to Problem 130E

After $18.48$ years the investment will be double.

Explanation of Solution

Given information :

Interest rate $r=3.75\%$ and total amount $P=\$1000$ .

Used formula:

The function of compound interest,

  $A=P{e}^{rt},\left(A=\text{balance after }t\text{ years},P=\text{initial investment},r=\text{yearly rates of interest}\right)$

The investment $P$ will double at,

  $\begin{array}{l}2P=P{e}^{rt}\\ \Rightarrow 2=e^{rt}\\ \Rightarrow \log (2)=\log (e^{rt})\\ \Rightarrow rt\log (e)=\log (2)\\ \Rightarrow t=\frac{\log (2)}{r\log (e)}\\ \Rightarrow t=\frac{\log (2)}{0.0375\log (e)}=18.48\text{ years}\text{.}\end{array}$

b.

To determine

To find: the time required for a $\$1000$ investment to triple at interest rate $r$ , compounded continuously.

Answer to Problem 130E

After $29.29$ years the investment will be double.

Explanation of Solution

Given information :

Interest rate $r=3.75\%$ and total amount $P=\$1000$ .

Used formula:

The function of compound interest,

  $A=P{e}^{rt},\left(A=\text{balance after }t\text{ years},P=\text{initial investment},r=\text{yearly rates of interest}\right)$

The investment $P$ will triple at,

  $\begin{array}{l}A=3P=P{e}^{rt}\\ \Rightarrow 3=e^{rt}\\ \Rightarrow \log (3)=\log (e^{rt})\\ \Rightarrow rt\log (e)=\log (3)\\ \Rightarrow t=\frac{\log (3)}{r\log (e)}\\ \Rightarrow t=\frac{\log (3)}{0.0375\log (e)}\approx 29.29\text{ years}\text{.}\end{array}$