Chapter 4.6, Problem 2E

Population Growth These exercises use the population growth model.

2. Bacteria Culture A certain culture of the bacterium Rhodobacter sphaeroides initially has 25 bacteria and is observed to double every 5 hours.

1. (a) Find an exponential model n(t) = n₀2^t/a for the number of bacteria in the culture after t hours.
2. (b) Estimate the number of bacteria after 18 hours.
3. (c) After how many hours will the bacteria count reach 1 million?

To determine

To find: An exponential model $n\left(t\right)=n_0{2}^{\frac{t}{a}}$ for the number of bacteria in the culture after $t$ hours.

Answer to Problem 2E

The exponential model for the number of bacteria in the culture after $t$ hours is $n\left(t\right)=25\left(2^{\frac{t}{5}}\right)$.

Explanation of Solution

Given:

The initial bacteria population is $25$.

The bacteria population is double in $5\text{hours}$.

Formula used: Exponential growth model (doubling time)

$n\left(t\right)=n_0{2}^{\frac{t}{a}}$ (I)

Where,

• $n_0$ is the initial size of the population.
• $a$ is the doubling time.
• $n\left(t\right)$ is size of the population at time $t$.

Calculation:

The given initial bacteria population is $25$. and the bacteria population is double in $5\text{hours}$.

Substitute $25$ for $n_0$ and $5$ for $a$ in equation (I) to find the model for the number of bacteria in the culture after $t$ hours,

$n\left(t\right)=25\left(2^{\frac{t}{5}}\right)$.

Thus, the exponential model for the number of bacteria in the culture after $t$ hours is $n\left(t\right)=25\left(2^{\frac{t}{5}}\right)$.

To determine

To find: The number of bacteria after $18\text{hours}$.

Answer to Problem 2E

The number of bacteria after $18\text{hours}$ is $303.14$.

Explanation of Solution

Given:

From part (a), the exponential model for the number of bacteria in the culture after $t$ hours is given below,

$n\left(t\right)=25\left(2^{\frac{t}{5}}\right)$ (II)

Calculation:

Substitute $18$ for $t$ in equation (II) to find the number of bacteria after $18\text{hours}$,

$\begin{array}{c}n\left(18\right)=25\left(2^{\frac{18}{5}}\right)\\ =25\left(12.125733\right)\\ =303.1433133\\ \approx 303.14\end{array}$

Thus, the number of bacteria in the culture after $18\text{hours}$ is $303.14$.

To determine

To find: The time in which bacteria count reach to $1\text{million}$.

Answer to Problem 2E

The bacteria count reach to $1\text{million}$ in $76.44\text{hours}$.

Explanation of Solution

Given:

From part (a), the exponential model for the number of bacteria in the culture after $t$ hours is given below,

$n\left(t\right)=25\left(2^{\frac{t}{5}}\right)$

Calculation:

Substitute $1\text{million}$ for $n\left(t\right)$ in equation (II) to find the time in which bacteria population reach to $1\text{million}$,

$\begin{array}{l}1000000=25{\left(2\right)}^{\frac{t}{5}}\\ {10}^6=5^2\left(2^{\frac{t}{5}}\right)\end{array}$

Simplify the above expression taking $\log$ to each side,

$\begin{array}{c}\log {10}^6=\log \left(5^2{2}^{\frac{t}{5}}\right)\\ \log {10}^6=\log 5^2+\log 2^{\frac{t}{5}}\left(\because {\log }_a(AB)={\log }_{a}A+{\log }_{b}B\right)\\ 6\log 10=2\log 5+\frac{t}{5}\log 2\left(\because \log a^b=b\log a\right)\\ 5\left(6-2\log 5\right)=t\log 2\left(\because \log 10=1\right)\end{array}$

Further simplify the above expression,

$\begin{array}{c}t=\frac{5\left(6-2\left(0.69897\right)\right)}{0.301030}\left(\because \log 5=0.69897,\log 2=0.301030\right)\\ =\frac{5\left(4.60206\right)}{0.301030}\\ =\frac{23.0103}{0.301030}\\ =76.438561\end{array}$

Thus, the bacteria count reach to $1\text{million}$ in $76.44\text{hours}$.