Chapter 3.6, Problem 21E

Solution

To find: it will take until J’s account has a balance of $4150.

6 Years 9 Months

Given information:

J invests $2300 in a savings account with a 9% interest rate compounded quarterly,

Formula Used:

Compound Interest−

Value of an Investment Suppose a principal P is invested at a fixed annual interest rate r . The value of the investment after t years is:

• $A=P{\left(1+\frac{r}{k}\right)}^{kt}$ when interest compounds k times per year,
• $A=P{e}^{rt}$ when interest compounds continuously

Property Used:

Quotient rule:

  ${\log }_b\left(\frac{x}{y}\right)={\log }_{b}x-{\log }_{b}y$

Power rule:

  ${\log }_b{x}^m=m{\log }_{b}x$

Explanation:

From the given data, $A=\$4150,P=\$2300,r=0.09,k=4$

Substituting in the above formula for interest compounding 4 times a year,

  $\begin{array}{c}A=P{\left(1+\frac{r}{k}\right)}^{kt}\\ 4150=2300{\left(1+\frac{0.09}{4}\right)}^{4t}\\ \frac{4150}{2250}={1.0225}^{4t}\end{array}$

Take natural logarithm on both sides,

  $\ln \left(\frac{4150}{2250}\right)=\ln \left({1.0225}^{4t}\right)$

Using the quotient rule and the power rule,

  $\begin{array}{c}\ln 4150-\ln 2250=4t\ln 1.0225\\ t=\frac{\ln 4150-\ln 2250}{4\ln 1.0225}\\ t\approx 6.63\text{ years}\end{array}$

So, J’s account will have a balance of $4150 in about 6.63 years which is about 6 years and 9 months.