Chapter 0.3, Problem 38E

a.

To determine

The initial number of bacteria.

Answer to Problem 38E

The initial number of bacteria is 100.

Explanation of Solution

Given:

  $B=100e^{0.693t}$

Calculation:

For initial number of bacteria, put the time, $t=0$ :

  $\begin{array}{l}B=100e^{0.693\left(0\right)}\\ B=100e^0\\ B=100\left(1\right)\\ B=100\end{array}$

Hence, the initial number of bacteria is 100.

b.

To determine

The number of bacteria after 6 hours.

Answer to Problem 38E

The number of bacteria after 6 hours is 6394.

Explanation of Solution

Given:

  $t=6$

Calculation:

Substitute the values in the formula:

  $\begin{array}{l}B=100e^{0.693\left(6\right)}\\ B=100e^{4.158}\\ B\approx 6394\end{array}$

Hence, the number of bacteria after 6 hours is 6394.

c.

To determine

The time at which the bacteria grows to 200.

Answer to Problem 38E

The time at which the bacteria grows to 200 is 1 hours approximately.

Explanation of Solution

Given:

  $B=200$

Calculation:

Substitute the values in the formula:

  $\begin{array}{l}200=100e^{0.693t}\\ 2=e^{0.693t}\end{array}$

Take natural logarithm to both sides:

  $\begin{array}{l}\ln \left(2\right)=0.693t\ln \left(e\right)\\ \ln \left(2\right)=0.693t\left(1\right)\\ \ln \left(2\right)=0.693t\\ t=\frac{\ln \left(2\right)}{0.693}\\ t\approx 1\end{array}$

Hence, the time at which the bacteria grows to 200 is 1 hours approximately.