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Chapter 3: Problem 1

Suppose that a culture of bacteria has an initial population of \(n=100\). If the population doubles every three days, determine the number of bacteria present after 30 days. How much time is required for the population to reach 4250 in number?

Short Answer

Expert verified
The number of bacteria after 30 days is 102400. It takes approximately 16.26 days for the population to reach 4250.

Step by step solution

01

- Determine the doubling factor

The population doubles every three days, so the doubling factor is 2.
02

- Calculate the number of doubling periods in 30 days

Divide the total number of days by the length of each doubling period: 30 days / 3 days per period = 10 periods
03

- Apply the exponential growth formula

The population after a given number of periods can be calculated using the formula: \( N = N_0 \times 2^{t/T} \) where \( N_0 \) is the initial population, \( t \) is the total time (30 days), and \( T \) is the doubling period (3 days). For 30 days: \( N = 100 \times 2^{30/3} = 100 \times 2^{10} = 100 \times 1024 = 102400 \)
04

- Calculate the time required to reach 4250 bacteria

Use the population growth formula in reverse to find the required time. Set \( N = 4250 \), \( N_0 = 100 \) and solve for \( t \): \( 4250 = 100 \times 2^{t/3} \) Divide both sides by 100: \( 42.5 = 2^{t/3} \) Take the logarithm base 2 of both sides: \( \text{log}_2(42.5) = t/3 \) Calculate the logarithm: \( \text{log}_2(42.5) \text{ is approximately 5.42} \) Thus, \( t/3 = 5.42 \), and solving for \( t \) we get: \( t = 5.42 \times 3 = 16.26 \text{ days} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Doubling Period
The concept of the doubling period is fundamental to understanding exponential growth in population. In this exercise, the bacteria culture doubles every three days. This doubling period tells us how frequently the population size will double.

For example, starting with an initial population of 100 bacteria, after three days, the population will double to 200 bacteria. After another three days, the population will double again to 400 bacteria, and so forth. This continuous doubling leads to rapid increases in population size over a relatively short period.

The doubling period, therefore, is a critical variable in the exponential growth formula, helping us predict population sizes at future timescales.
Exponential Growth Formula
Exponential growth describes a process where the quantity increases at a constant rate per time period. For the bacteria population in our exercise, we can use the exponential growth formula:

\[ N = N_0 \times 2^{t/T} \]
Where:
\ N \ - Future population
\ N_0 \ - Initial population (100 bacteria)
\ t \ - Total time (days)
\ T \ - Doubling period (3 days)

Applying this formula, for 30 days, we substitute to find:

\[ N = 100 \times 2^{30/3} = 100 \times 2^{10} = 100 \times 1024 = 102400 \]
This means after 30 days, the bacteria population will be 102,400.

By utilizing the exponential growth formula, we can conveniently determine how the population will evolve over time given the doubling period.
Logarithmic Calculation
Logarithmic calculations are handy when we need to solve for time or other variables in exponential growth scenarios. To determine how long it will take the bacteria population to reach a specific size, we invert the exponential formula and utilize logarithms.

For example, given a population goal of 4250 bacteria, we start with:
\[ N = 4250 = 100 \times 2^{t/3} \]
First, isolate the exponential term by dividing both sides by 100:
\[ 42.5 = 2^{t/3} \]
Next, apply the logarithm base 2 to both sides:
\[ \text{log}_2(42.5) = t/3 \]
Using a logarithm calculator, we find that \ \text{log}_2 (42.5) \ is approximately 5.42.
We then solve for t by multiplying both sides by 3:
\[ t = 5.42 \times 3 = 16.26 \text{ days} \]
Thus, it will take around 16.26 days for the bacteria population to reach 4250.

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Most popular questions from this chapter

Consider a radioactive substance with halflife 10 days. If there are initially \(5000 \mathrm{~g}\) of the substance, how much remains after 365 days?

(Tumor and Organism Growth) Ludwig von Bertalanffy (1901-1972) made valuable contributions to the study of organism growth, including the growth of tumors, based on the relationship between body size of the organism and the metabolic rate. He theorized that weight is directly proportional to volume and that the metabolic rate is proportional to surface area. This indicates that surface area is proportional to \(V^{2 / 3}\) because \(V\) is measured in cubic units and \(S\) in square units. Therefore, Bertalanffy studied the IVP \(d V / d t=a V^{2 / 3}-b V, V(0)=V_{0}\). (a) Solve this initial value problem to show that volume is given by \(V(t)=\) \(\left[\frac{a}{b}-\left(\frac{a}{b}-V_{0}^{1 / 3}\right) e^{-b t / 3}\right]^{3}\). (b) Find \(\lim _{t \rightarrow \infty} V(t)\) and explain what the limit represents. (c) This initial value problem is similar to that involving the Logistic equation \(d V / d t=a V^{2}-b V, V(0)=V_{0}\). Compare the limiting volume between the Logistic equation and the model proposed by Bertalanffy in the cases when \(a / b>1\) and when \(a / b<1\).

In a chemical reaction, chemical \(\mathrm{A}\) is converted to chemical \(B\) at a rate proportional to the amount of chemical A present. If half of chemical A remains after five hours, when does \(1 / 6\) of the initial amount of chemical A remain? How much of the initial amount remains after \(15 \mathrm{~h}\) ?

Five college students with the flu virus return to an isolated campus of 2500 students. If the rate at which this virus spreads is proportional to the number of infected students \(y\) and to the number not infected \(2500-\) \(y\), solve the initial value problem \(d y / d t=\) \(k y(2500-y), y(0)=5\) to find the number of infected students after \(t\) days if 25 students have the virus after one day. How many students have the flu after five days?

Solve the Logistic equation if \(r=1 / 100\) and \(a=10^{-8}\) given that \(y(0)=100,000\). Find \(y(25)\). What is the limiting population?
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