Chapter 3, Problem 119RE

To determine

To find:

The time in which the money double.

Answer to Problem 119RE

The money to be double in $10.05$ years.

Explanation of Solution

Given information:

Deposit $\$7550$ in an account that pays $6.9\%$ interest, compounded continuously.

Concept used:

For continuous compounding apply the formula:

  $A=P{e}^{rt}$

For the money to be double Substitute these values in the above formula.

  $\begin{array}{c}A=2\left(7550\right)\\ =15100\end{array}$

  $P=7550$

  $\begin{array}{c}r=6.9\%\\ =0.069\end{array}$

  $\begin{array}{c}15100=7550e^{0.069t}\\ \frac{15100}{7550}=e^{0.069t}\\ 2=e^{0.069t}\end{array}$

Take logarithm on both sides:

  $\begin{array}{c}2=e^{0.069t}\\ \ln 2=\ln \left(e^{0.069t}\right)\\ 0.6931=0.069t\\ \frac{0.6931}{0.069}=t\\ 10.0449=t\\ t\approx 10.05\end{array}$

Hence, the money to be double in $10.05$ years.