Chapter 11.7, Problem 16E

(a)

To determine

To find:

A function that models the amount in the account.

Answer to Problem 16E

  $y=\left(2137.52\right){\left(1.0534\right)}^x$

Explanation of Solution

Given:

The data of number of the amount.

Concept used:

Formula:

The equation is

  $y=a\times b^x$

Calculation:

The table is

Date	$\text{Jan}1,1955$	$\text{Jan}1,1956$	$\text{Jan}1,1957$	$\text{Jan}1,1958$	$\text{Jan}1,1959$
Balance	$2137.52$	$2251.61$	$2371.79$	$2498.39$	$2631.74$

  $y=a\ast b^x\mathrm{....................}\left(1\right)$

  $\begin{array}{l}a=2137.52\\ b=1.0534\end{array}$

From equation (1)

  $\begin{array}{l}y=a\times b^x\\ y=\left(2137.52\right){\left(1.0534\right)}^x\end{array}$

(b)

To determine

To find:

The equation in terms of base $e$ .

Answer to Problem 16E

  $N=2137.52\left(e^{\mathrm{.0.0520}}\right)x$.

Explanation of Solution

Given:

The data of number of the amount.

Concept used:

Formula:

The equation is

  $N=N_0{e}^{kt}$

Calculation:

The table is

Date	$\text{Jan}1,1955$	$\text{Jan}1,1956$	$\text{Jan}1,1957$	$\text{Jan}1,1958$	$\text{Jan}1,1959$
Balance	$2137.52$	$2251.61$	$2371.79$	$2498.39$	$2631.74$

  $y=a\times b^x\mathrm{....................}\left(1\right)$

  $\begin{array}{l}a=2137.52\\ b=1.0534\end{array}$

From equation (1)

  $\begin{array}{l}y=a\times b^x\\ y=\left(2137.52\right){\left(1.0534\right)}^x\end{array}$

  $\begin{array}{l}N=N_0{e}^{kt}\\ N=2137.52{\left(e^{\ln 1.0534}\right)}^x\\ N=2137.52\left(e^{\ln 1.0534}\right)x\\ N=2137.52\left(e^{\mathrm{.0.0520}}\right)x\end{array}$

(c)

To determine

To find:

The time required for an amount is half-life at the given model.

Answer to Problem 16E

  $r\approx 5.2\%$

Explanation of Solution

Given:

The data of number of the amount.

Concept used:

Formula:

The equation is

  $N=N_0{e}^{rt}$

Calculation:

The table is

Date	$\text{Jan}1,1955$	$\text{Jan}1,1956$	$\text{Jan}1,1957$	$\text{Jan}1,1958$	$\text{Jan}1,1959$
Balance	$2137.52$	$2251.61$	$2371.79$	$2498.39$	$2631.74$

  $N=N_0{e}^{rt}\mathrm{....................}\left(1\right)$

The time needed to quadruple an investment replace $N$ by $\frac{1}{2}{N}_0$ if

From equation (1)

  $\begin{array}{l}\frac{1}{2}{N}_0=N_0{e}^{rt}\\ 2631.74=2137.52e^{4r}\\ 4r=\ln \frac{2631.74}{2137.52}\\ r\approx 0.0520\\ r\approx 5.2\%\end{array}$