A common inhabitant of human intestines is the bacterium Escherichia coli, named after the German pediatrician Theodor Escherich

Question

19 days ago

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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

Mathematics

2 answers:

babunello [8.4K] · Answer 1 · 2026-03-08T18:58:52+03:00

A.) P(t) = P0e^(kt)
P(20/60) = 40 e^(20k/60)
80 = 40 e^(k/3)
e^(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 = e^(8k) = e^(8(3ln(2))) = e^(24ln(2)) = e^(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

PIT_PIT [9.1K] · Answer 2 · 2026-03-08T18:58:06+03:00

Consequently, the final results for the various sections are summarized as follows:

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

Part(b): After $8$ hours, the bacterial population totals $\fbox{\begin\\\ 671088640\\\end{minispce}}$ cells.

Part(d): After $8$ hours, the growth rate is $\fbox{\begin\\\ 16777216\\\end{minispace}}$ cells per hour.

Part(e): The duration needed for the bacterial population to hit $1$ million cells is $\fbox{\begin\\\ 4.87\\\end{minispace}}$ hours.

Additional Information:

As stated, a single cell of the Bacterium Escherichia coli splits into two approximately every $20$ minutes.

The initial population given in the prompt is $40$ cells.

The bacterial population growth function can be expressed as follows:

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

In the prior equation, $P_{0}$ denotes the initial count, $t$ stands for time, $P(t)$ indicates the population after $t$ hours, and $k$ describes the growth rate.

With an initial population of $40$ cells, the value of $P_{0}$ is $40$ .

Part(a): Find the relative growth or k.

The equation illustrating bacterial growth is expressed as:

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

Since each bacterium divides into two every $20$ minutes or $\dfrac{1}{3}$ hours.

With an initial count of $40$ cells, after $\dfrac{1}{3}$ hours, there will be $80$ cells.

To find $k$ , substitute the relevant values into equation (1).

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

Then apply the antilogarithm.

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

Thus, $k$ results in $3ln2$ .

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

Part(b): Calculate the bacterial population after $\bf8$ hours.

The equation for determining the population after $t$ hours is defined as:

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

Substituting the relevant figures gives us $40$ for $P_{0}$ , $3ln2$ for $k$ , and $8$ for $t$ .

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

Thus, after $8$ hours, the population is $671088640$ cells.

Part(d): Calculate the growth rate following $\bf8$ hours.

The growth rate is represented as the proportion of the bacterial population after $t$ hours to the initial population size.

In equation $P(t)=P_{0}e^{(kt)}$ , plug in the necessary values $8$ for $t$ .

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

Consequently, $\dfrac{P(8)}{P(0)}$ equals $\fbox{\begin\\\ 16777216\\\end{minispace}}$ .

Thus, the bacterial growth rate after $8$ hours is $16777216$ cells per hour.

Part(e): Calculate the time it takes to reach $1$ million cells.

Let $t$ hours denote the time it takes for the bacterial population to meet $1$ million cells.

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

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

Applying the antilogarithm yields.

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

Thus, it will take $4.87$ hours for the population to attain $1$ million cells.

For further information:

1. An inquiry regarding composite functions

2. A question involving the radius and center of a circle

3. A query about determining line intercepts

Answer specifications:

Level: High school

Discipline: Mathematics

Chapter: Exponential function

Key themes: Functions, exponential nature, growth rate, Bacterium Escherichia coli, relative growth, population dynamics, cells, cellular division, growth function, decay function,