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Chapter 7: Problem 55

Suppose a culture has 100 bacteria to begin with and the number of bacteria doubles every 2 hours. Then the number \(N\) of bacteria after \(t\) hours is given by \(N=100 \cdot 2^{\frac{t}{2}} .\) How many bacteria will be present after 3\(\frac{1}{2}\) hours?

Short Answer

Expert verified
336 bacteria

Step by step solution

01

Understand the Given Problem

We have a bacterial culture that starts with 100 bacteria. The bacteria double every 2 hours. We need to find the number of bacteria after 3.5 hours.
02

Use the Formula

The number of bacteria after \( t \) hours is given by the equation \( N = 100 \cdot 2^{\frac{t}{2}} \). Here, \( t = 3.5 \) hours.
03

Substitute Time into the Formula

Substitute \( t = 3.5 \) into the formula: \[ N = 100 \cdot 2^{\frac{3.5}{2}} \].
04

Calculate the Exponent Value

First, calculate the exponent by dividing 3.5 by 2: \[ \frac{3.5}{2} = 1.75 \]. The expression becomes \( N = 100 \cdot 2^{1.75} \).
05

Calculate the Power of 2

Calculate \( 2^{1.75} \). This can be approximated using a calculator: \( 2^{1.75} \approx 3.363 \).
06

Final Calculation

Substitute the value from Step 5 into the formula: \[ N = 100 \cdot 3.363 \]. Multiply to find \( N \approx 336.3 \). Round to the nearest whole number since we can't have a fraction of a bacterium: \( N \approx 336 \).

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Key Concepts

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

Bacterial Growth
Bacterial growth is a fascinating example of exponential growth, where the population size grows at a constant rate. In many environments, bacteria reproduce by dividing, which means that the population can double in numbers over a fixed period. This idealized growth model helps explain how a small initial quantity of bacteria can quickly become a large number. In this exercise, the culture starts with 100 bacteria and doubles every 2 hours. This doubling characteristic is a fundamental property of bacterial growth.
Doubling Time
Doubling time refers to the period it takes for a population to double in size. For the bacteria in this problem, the doubling time is 2 hours. Understanding doubling time is crucial in predicting how fast bacteria can grow. It's helpful to visualize that each doubling leads to twice the number of individuals present before. For example:
  • At 0 hours, there are 100 bacteria.
  • At 2 hours, they double to 200 bacteria.
  • At 4 hours, they double again to 400 bacteria.
Knowing doubling time allows predictions about future population sizes using exponential equations.
Exponentiation
Exponentiation is a mathematical operation that involves raising a base number to a power, which is the exponent. Here, the base is 2, and the exponent represents how many times the population doubles. In our exercise, we use the formula: \[N = 100 \cdot 2^{\frac{t}{2}}\]Each time increase in the exponent corresponds to bacteria doubling. When calculating for the 3.5-hour mark, we substituted 3.5 into the formula which required calculating \(2^{1.75}\), the result of raising 2 to the power of 1.75.
Problem Solving Steps
Following a structured problem solving approach is vital to solving exponential growth problems efficiently. Here are the steps used:
  • Step 1: Understand the Problem: Clearly identify initial conditions and what needs to be found. Here, we start with 100 bacteria doubling every 2 hours and need to find the count at 3.5 hours.
  • Step 2: Use the Right Formula: Identify or derive the formula that matches the problem, which is \(N = 100 \cdot 2^{\frac{t}{2}}\).
  • Step 3: Substitute Values: Insert known values into the formula to solve for the unknown. For this exercise, substituting \(t = 3.5\) hours.
  • Step 4: Calculate: Simplify the equation, calculate exponentiation, and perform any arithmetic to find the solution.
This structured approach ensures clarity and accuracy, crucial aspects of problem solving.

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Most popular questions from this chapter

Simplify. $$ \sqrt{4 x^{3} y^{2}} $$

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REVIEW Which of the following sentences is true about the graphs of \(y=2(x-3)^{2}+1\) and \(y=2(x+3)^{2}+1 ?\) F Their vertices are maximums G The graphs have the same shape with different vertices. H The graphs have different shapes with different vertices. J One graph has a vertex that is a maximum while the other graph has a vertex that is a minimum.
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