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Chapter 1: Problem 35

A culture of bacteria originally numbers \(500 .\) After 2 hours there are 1500 bacteria in the culture. Assuming exponential growth, how many are there after 6 hours?

Short Answer

Expert verified
There are 13,500 bacteria after 6 hours.

Step by step solution

01

Understand the Formula for Exponential Growth

Exponential growth can be described by the formula \( N(t) = N_0 \times e^{kt} \), where \( N(t) \) is the population at time \( t \), \( N_0 \) is the initial population, \( k \) is the growth rate, and \( e \) is the base of the natural logarithm.
02

Identify Known Values

From the problem, we know the initial number of bacteria \( N_0 = 500 \), the number of bacteria after 2 hours \( N(2) = 1500 \), and we need to find \( N(6) \).
03

Determine the Growth Rate (k)

Use the formula \( N(t) = N_0 \times e^{kt} \) with the values at 2 hours. Set \( 1500 = 500 \times e^{2k} \). Solving for \( k \), we divide both sides by 500 to get \( 3 = e^{2k} \), then take the natural logarithm: \( \ln(3) = 2k \). Finally, \( k = \frac{\ln(3)}{2} \approx 0.5493 \).
04

Calculate the Bacteria Count at 6 Hours

Now, use \( k \) to find \( N(6) \). Plug \( k \) into \( N(6) = 500 \times e^{6k} \). Calculate \( e^{6k} = e^{3\ln(3)} = 3^3 = 27 \). Thus, \( N(6) = 500 \times 27 = 13500 \).
05

Review the Calculation

The calculation is complete. After 6 hours, the number of bacteria is \( 13500 \). Verify each step to ensure there are no errors in computation or formula application.

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Key Concepts

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

Bacteria Culture
Imagine you have a petri dish filled with bacteria, and initially, there are 500 bacteria. This setup is what we refer to as a 'bacteria culture'. In science, when we talk about bacteria growing, we often assume that they are reproducing at a constant rate, leading to what is known as exponential growth.

Bacteria are fascinating because they can double in a relatively short amount of time under the right conditions. Let's explore the concept of bacteria culture a bit more closely:
  • Bacteria are microorganisms that reproduce by cell division.
  • In a culture, you can think of it like a colony growing on a food source.
  • The amount of food (nutrients) and space can affect how quickly they grow.
When you set a fixed time and watch how many bacteria multiply, you a glimpse into their dynamic process of growth. This forms the basis of our problem involving exponential growth.
Natural Logarithm
When dealing with exponential growth, the idea of a natural logarithm comes in handy for simplifying calculations.No need to be scared of the term! The natural logarithm, represented by the symbol \( \ln \), is a way to undo the exponential base we often use in natural growth equations.

Let鈥檚 break it down further:
  • The base of the natural logarithm is the constant \( e \), which is approximately equal to 2.71828.
  • This logarithm helps us solve equations involving exponential functions by turning products into sums, which are easier to deal with mathematically.
  • For example, if you have \( e^{2k} = 3 \), taking the natural logarithm of both sides leads to a solution for \( k \).
The beauty of the natural logarithm lies in its ability to convert a rather complex-looking multiplication process into a simple addition problem, making calculations less prone to errors and easier to handle.
Growth Rate Calculation
To determine how fast the bacteria in our culture are growing, we need to calculate the 'growth rate', commonly denoted as \( k \) in mathematical formulas. This is essentially what makes your equation move from being abstract to something concrete and measurable.

Here's how growth rate calculation works:
  • The growth rate \( k \) is a constant in the exponential growth formula \( N(t) = N_0 \times e^{kt} \).
  • In our bacteria culture growing from 500 to 1500 in 2 hours, we first rearrange the formula to solve for \( k \).
  • Applying a natural logarithm, we solved \( \ln(3) = 2k \), leading to \( k = \frac{\ln(3)}{2} \), which calculates approximately to 0.5493.
Calculating \( k \) gives us an insight into how rapidly our culture grows, and it provides the necessary information to predict future population sizes accurately. This process of deriving \( k \) is foundational not only in bacterial growth but also in modeling many real-world phenomena that experience exponential growth.

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Most popular questions from this chapter

Which of the following functions has its domain identical with its range? (a) \(\quad f(x)=x^{2}\) (b) \(\quad g(x)=\sqrt{x}\) (c) \(h(x)=x^{3}\) (d) \(\quad i(x)=|x|\)

In the early 1920 s, Germany had tremendously high inflation, called hyperinflation. Photographs of the time show people going to the store with wheelbarrows full of money. If a loaf of bread cost \(1 / 4\) marks in 1919 and 2,400,000 marks in \(1922,\) what was the average yearly inflation rate between 1919 and \(1922 ?\)

On the graph of \(y=\sin x,\) points \(P\) and \(Q\) are at consecutive lowest and highest points. Find the slope of the line through \(P\) and \(Q\)

If \(g(x)=\ln (a x+2),\) where \(a \neq 0,\) what is the effect of increasing \(a\) on the vertical asymptote?

A population of animals oscillates sinusoidally between a low of 700 on January 1 and a high of 900 on July 1 (a) Graph the population against time. (b) Find a formula for the population as a function of time, \(t,\) in months since the start of the year.
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