Question

A common inhabitant of human intestines is the bacterium Escherichia coli, named after the German pediatrician Theodor Escherich, who identified it in 1885. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 40 cells. (a) Find the relative growth rate. k = 3ln(2) Correct: Your answer is correct. hr−1 (b)f cells after 8 hours. cells (d) Find the rate of growth after 8 hours. (Round your answer to the nearest integer.) cells/h (e) When will the population reach a million cells? h

Answer 1

A.) P(t) = P0exp(kt)
P(20/60) = 40 exp(20k/60)
80 = 40 exp(k/3)
exp(k/3) = 80/40 = 2
k/3 = ln(2)
k = 3ln(2)

b.) P(8) = 40(2)^24 = 40(16777216) = 671088640 cells

d.) Rate of change = exp(8k) = exp(8(3ln(2)) = exp(24ln(2)) = exp(16.6355) = 16777216 cells / hour

e.) P(t) = 40(2)^3t; t in hours
1,000,000 = 40(8)^t
25,000 = 8^t
ln(25,000) = t ln(8)
t = ln(25,000)/ln(8) = 4.87 hours

Answer 2

Thus, the final answer for the different parts is as follows:

Part(a): The relative growth or the value of $k$ is $\fbox{\begin\\\ \math k=3ln2\\\end{minispace}}$.

Part(b): The population of the bacteria after $8$ hours is $\fbox{\begin\\\ 671088640\\\end{minispce}}$ cells.

Part(d): The rate of growth after $8$ hours is $\fbox{\begin\\\ 16777216\\\end{minispace}}$ cells per hour.

Part(e): The time required for the population of the bacteria to reach a count of $1$ million is $\fbox{\begin\\\ 4.87\\\end{minispace}}$ hours.

Further explanation:

In the question it is given that a cell of the bacteria Bacterium Escherichia coli divides into two cells in every $20$ minutes.

According to the data given in the question the initial population of the bacteria is $40$ cells.

Consider the function for increase in the population of the bacteria as follows:

$\fbox{\begin\\\ \math P(t)=P_{0}e^{(kt)}\\\end{minispace}}$

In the above equation $P_{0}$ represents the initial population, $t$ represents the time, $P(t)$ is the population after $t$ hours and $k$ is the relative growth.

It is given that the initial population is $40$ cells so, the value of $P_{0}$ is $40$.

Part(a): Determine the relative growth or the value of k.

The function which represents the growth in the population of the bacteria is as follows:

$P(t)=P_{0}e^{(kt)}$                 (1)

Since, each cell of the bacteria divides into two cells in every $20$ minutes or $\dfrac{1}{3}$ hours.

Since, the initial population is $40$ cells so the population after $\dfrac{1}{3}$ hours is $80$ cells.

To obtain the value of $k$ substitute $\dfrac{1}{3}$ for $t$, $40$ for $P_{0}$ and $80$ for $P(t)$ in equation (1).

$\begin{aligned}80&=40\times e^{(k/3)}e^{(k/3)}\\ \dfrac{80}{40}e^{(k/3)}&=2\end{aligned}$

Take antilog in the above equation.

$\begin{aligned}\dfrac{k}{3}&=ln2\\k&=3ln2\end{aligned}$

Therefore, the value of $k$ is $3ln2$.

Thus, the relative growth of the bacteria is $\fbox{\begin\\\ \math k=3ln2\\\end{minispace}}$.

Part(b):Determine the population of the bacteria after $\bf8$ hours.

The equation to determine the population after $t$ hours is as follows:

$\fbox{\begin\\\ \math P(t)=P_{0}e^{(kt)}\\\end{minispace}}$

Substitute $40$ for $P_{0}$, $3ln2$ for $k$, $8$ for $t$ in the above equation.

$\begin{aligned}P(8)&=40e^{(8\eimes 3ln2)}\\&=40e^{(24ln2)}\\&=40\times 2^{24}\\&=671088640\end{alighned}$

Therefore, the population of the bacteria after $8$ hours is $671088640$ cells.

Part(d): Determine the rate of growth after $\bf8$ hours.

The rate of growth is defined as the ratio of the population of the bacteria after $t$ hours to the initial population of the bacteria.

Substitute $8$ for $t$ in the equation $P(t)=P_{0}e^{(kt)}$.

$\begin{aligned}P(8)&=P_{0}e^{(8\times 3ln2)}\\\dfrac{P(8)}{P(0)}&= e^{(24ln2)}\\\dfrac{P(8)}{P(0)}&=16777216\end{aligned}$

Therefore, the value of $\dfrac{P(8)}{P(0)}$ is $\fbox{\begin\\\ 16777216\\\end{minispace}}$.

This implies that the rate of growth of bacteria after $8$ hours is $16777216$ cells per hour.

Part(e):Determine the time in which the population of the bacteria becomes $1$ million cells.

Consider the time in which the population of the bacteria reaches a count of $1$ million cells as $t$ hours.

Substitute $1000000$ for $P(t)$, $40$ for $P_{0}$ and $3ln2$ for $k$ in the equation $P(t)=P_{0}e^{(kt)}$.

$\begin{aligned}1000000&=40e^{(3t\times ln2)}\\e^{(ln2^{(3t)})}&=25000\\2^{(3t)}&=25000\\8^t&=25000\end{aligned}$

Take antilog in the above equation.

$\begin{aligned}t&=\dfrac{ln25000}{ln2}\\t&=4.87\end{aligned}$

Therefore, the time required for the population of the bacteria to reach a count of $1$ million is $4.87$ hours.

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Answer details:

Grade: High school

Subject: Mathematics

Chapter: Exponential function

Keywords: Functions, exponential function, rate of growth, Bacterium Escherichia coli, relative growth, population, cells, relative growth, cell divides in two, growth function, decay function,