Chapter 11.7, Problem 14E

(a)

To determine

To find:

An exponential function that models the data.

Answer to Problem 14E

  $y=\left(4.7871\right){\left(1.7687\right)}^x$

Explanation of Solution

Given:

The data of number of the bacteria, in millions.

Concept used:

Formula:

The equation is

  $y=a\ast b^x$

Calculation:

The table is

Time (hours)	$0$	$1$	$2$	$3$	$4$
Bacteria	$5$	$8$	$15$	$26$	$48$

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

  $\begin{array}{l}a=4.7871\\ b=1.7687\end{array}$

From equation (1)

  $\begin{array}{l}y=a\ast b^x\\ y=\left(4.7871\right){\left(1.7687\right)}^x\end{array}$

(b)

To determine

To find:

The equation in terms of base $e$ .

Answer to Problem 14E

  $N=4.7871\left(e^{.5702}\right)t$

Explanation of Solution

Given:

The data of number of the bacteria, in millions.

Concept used:

Formula:

The equation is

  $N=N_0{e}^{kt}$

Calculation:

The table is

Time (hours)	$0$	$1$	$2$	$3$	$4$
Bacteria	$5$	$8$	$15$	$26$	$48$

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

  $\begin{array}{l}a=4.7871\\ b=1.7687\end{array}$

From equation (1)

  $\begin{array}{l}y=a\ast b^x\\ y=\left(4.7871\right){\left(1.7687\right)}^x\\ N=N_0{e}^{kt}\\ N=4.7871{\left(e^{\ln 1.7687}\right)}^t\\ N=4.7871\left(e^{\ln 1.7687}\right)t\\ N=4.7871\left(e^{.5702}\right)t\end{array}$

(c)

To determine

To find:

The time required for a culture is double at the given model.

Answer to Problem 14E

  $t\approx 1.22$

Explanation of Solution

Given:

The data of number of the bacteria, in millions.

Concept used:

Formula:

The equation is

  $N=N_0{e}^{rt}$

Calculation:

The table is

Time (hours)	$0$	$1$	$2$	$3$	$4$
Bacteria	$5$	$8$	$15$	$26$	$48$

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

The time needed to quadruple an investment replace $N$ by $2N_0$ if

From equation (1)

  $\begin{array}{l}2N_0=N_0{e}^{rt}\\ 2=e^{rt}\\ \ln 2=\ln e^{rt}\\ \ln 2=rt\\ t=\frac{\ln 2}{0.5702}\\ t\approx 1.22\end{array}$