Chapter 8.8, Problem 5PPS

(a)

To determine

To find: The value of $k$ .

Answer to Problem 5PPS

The value of $k$ for the bacteria $k=0.071$ .

Explanation of Solution

Given information:

The given function is $y=80e^{kt}$ .

Calculation:

Calculate the equation.

  $\begin{array}{l}675=80e^{k\left(30\right)}\\ 8.44=e^{k\left(30\right)}\\ \ln 8.44=\ln e^{k\left(30\right)}\\ 2.13=30k\\ \frac{2.13}{30}=k\\ k=0.071\end{array}$

Therefore the value of $k$ for the bacteria $k=0.071$

(b)

To determine

To find: The bacteria reach a population of $6000$ cells.

Answer to Problem 5PPS

The population of bacteria will be $6000$ call after $60.8$ seconds.

Explanation of Solution

Given information:

The given function is $y=80e^{kt}$ .

Calculation:

Calculate the equation.

  $\begin{array}{c}y=80e^{\left(0.071\right)k}\\ 6000=80e^{0.071t}\\ \ln 25=\ln e^{0.071t}\\ 4.32=0.071t\\ t=\frac{4.32}{0.071}\\ =60.8\end{array}$ .

Therefore the population of bacteria will be $6000$ call after $60.8$ seconds.

(c)

To determine

To find: The long it will be before the number of cell of this bacteria.

Answer to Problem 5PPS

The number of cell bacteria in $30.85$ minutes .

Explanation of Solution

Given information:

The given function is $30.85$ .

Calculation:

Calculate the equation,

  $\begin{array}{c}35e^{0.0978t}>80e^{0.071t}\\ \ln 35e^{0.0978t}>\ln 80e^{0.071t}\\ \ln 35+\ln e^{0.0978t}>\ln 80+\ln e^{0.071t}\\ 0.0978t-0.071t>4.3820-3.555\\ 0.0268t>0.82\\ t>\frac{0.82668}{0.0268}\\ t>30.85\end{array}$

Therefore the number of cell bacteria in $30.85$ minutes