Chapter 6.6, Problem 56E

To determine

To calculate: The number of time periods elapsed.

Answer to Problem 56E

The required number of times period are $5\cdot 2$ and $36.2$ .

Explanation of Solution

Given Information:

Deposit amount $=\$1000$ .

Rate $=3.5\%$ .

Balance amounts are $\$1200$ and $\$3500$ .

Formula used:

Compound interest formula:

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

Where

  $\begin{array}{l}A=\text{final amount}\\ P=\text{initial principal balance}\\ n=\text{number of times interest applied per time period}\\ t=\text{time periods}\end{array}$

Calculation:

Consider deposit amount is. $\$1000$ and rate is $3.5\%$ .

Rate can be written as.

Rate $\begin{array}{l}=3.5\%\\ =\frac{3.5}{100}\\ =0.035\end{array}$

Now, for the balance $\$1200$ .

Use compound interest formula.

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

Substitute $1200$ for $A$ , $1000$ for $P$ , $0.035$ for $r$ and $12$ for $n$ in the above formula.

  $1200=1000{\left(1+\frac{0.035}{12}\right)}^{12t}$

Simplify the above equation.

  $\begin{array}{c}1200=1000{\left(1+\frac{0.035}{12}\right)}^{12t}\\ \frac{1200}{1000}={\left(\frac{12+0.035}{12}\right)}^{12t}\\ 1.2={\left(\frac{12.035}{12}\right)}^{12t}\\ 1.2={\left(1.0029\right)}^{12t}\end{array}$

Take logarithm each side of the above equation.

  $\log \left(1.2\right)=\log {\left(1.0029\right)}^{12t}$

Use power rule of logarithm $\log x^a=a\log x$ in the above equation.

  $\log \left(1.2\right)=\left(12t\right)\log \left(1.0029\right)$

Substitute $0.079$ for $\log \left(1.2\right)$ and $0.000125$ for $\log \left(1.0029\right)$ in the above equation.

  $0.079=\left(12t\right)\left(0.00125\right)$

Simplify the equation.

  $\begin{array}{l}0.079=\left(12t\right)\left(0.00125\right)\\ 0.079=t\left(0.015\right)\\ \frac{0.079}{0.015}=t\\ t=5.2\end{array}$

Thus, the number time periods $5.2$ .

For the balance $\$3500$ .

Use compound interest formula.

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

Substitute $3500$ for $A$ , $1000$ for $P$ , $0.035$ for $r$ and $12$ for $n$ in the above formula.

  $3500=1000{\left(1+\frac{0.035}{12}\right)}^{12t}$

Simplify the above equation.

  $\begin{array}{c}3500=1000{\left(1+\frac{0.035}{12}\right)}^{12t}\\ \frac{3500}{1000}={\left(\frac{12+0.035}{12}\right)}^{12t}\\ 3.5={\left(\frac{12.035}{12}\right)}^{12t}\\ 3.5={\left(1.0029\right)}^{12t}\end{array}$

Take logarithm each side of the above equation.

  $\log \left(3.5\right)=\log {\left(1.0029\right)}^{12t}$

Use power rule of logarithm $\log x^a=a\log x$ in the above equation.

  $\log \left(1.2\right)=\left(12t\right)\log \left(1.0029\right)$

Substitute $0.544$ for $\log \left(3.5\right)$ and $0.000125$ for $\log \left(1.0029\right)$ in the above equation.

  $0.544=\left(12t\right)\left(0.00125\right)$

Simplify the equation.

  $\begin{array}{l}0.544=\left(12t\right)\left(0.00125\right)\\ 0.544=t\left(0.015\right)\\ \frac{0.544}{0.015}=t\\ t=36.2\end{array}$

Hence, the require time period are $5.2$ and $36.2$ .