Problem 19 - A Graphical Approach to College Algebra

Question

The concentration of bacteria \(B\) in millions per milliliter after \(x\) hours is given by $$B(x)=3.5 e^{0.02 x}$$ (a) How many bacteria are there after 1 hour? (b) How many bacteria are there after 6.5 hours? (c) After how many hours will there be 6 million bacteria per milliliter?

Step-by-Step Solution

Answer

(a) 3.57 million after 1 hour; (b) 3.98 million after 6.5 hours; (c) 27 hours to reach 6 million bacteria.

Step 1: Calculate Bacteria After 1 Hour

We need to find the number of bacteria after 1 hour using the formula \(B(x) = 3.5 e^{0.02 x}\).Substitute \(x = 1\) into the formula: \[ B(1) = 3.5 e^{0.02 imes 1} = 3.5 e^{0.02} \]Use a calculator to find \(e^{0.02} \approx 1.0202\).Then, \(B(1) \approx 3.5 imes 1.0202 = 3.5707\).Therefore, after 1 hour, the concentration is approximately 3.57 million bacteria per milliliter.

Step 2: Calculate Bacteria After 6.5 Hours

We need to find the number of bacteria after 6.5 hours using the formula \(B(x) = 3.5 e^{0.02 x}\).Substitute \(x = 6.5\) into the formula: \[ B(6.5) = 3.5 e^{0.02 imes 6.5} = 3.5 e^{0.13} \]Use a calculator to find \(e^{0.13} \approx 1.138\).Then, \(B(6.5) \approx 3.5 imes 1.138 = 3.983\).Therefore, after 6.5 hours, the concentration is approximately 3.98 million bacteria per milliliter.

Step 3: Solve for Hours to Reach 6 Million Bacteria

To find out after how many hours there will be 6 million bacteria, set \(B(x) = 6\):\[ 6 = 3.5 e^{0.02x} \]Divide both sides by 3.5:\[ e^{0.02x} = \frac{6}{3.5} \approx 1.7143 \]Take the natural logarithm of both sides:\[ \ln(e^{0.02x}) = \ln(1.7143) \]This simplifies to:\[ 0.02x = \ln(1.7143) \]Calculate \(\ln(1.7143) \approx 0.539\).Solve for \(x\):\[ x = \frac{0.539}{0.02} = 26.95 \]Thus, it will take approximately 27 hours for the concentration to reach 6 million bacteria per milliliter.

Key Concepts

Bacteria Growth

Bacteria growth is a fascinating process, especially when modeled mathematically. In many natural scenarios, bacteria reproduce at an exponential rate, meaning they multiply quickly over a short period of time. This rapid multiplication can be described using exponential functions, which show how the concentration of bacteria increases over time.
In the given problem, the concentration of bacteria is expressed as a function of time:

• The function is denoted by \( B(x) = 3.5 e^{0.02x} \), where \( B(x) \) is the concentration in millions per milliliter, \( e \) is the base of the natural logarithm, and \( x \) is time in hours.

This mathematical model helps predict how fast the bacteria population grows in a controlled environment, which is useful in various scientific and medical fields.

Natural Exponential Function

The natural exponential function plays a crucial role in modeling continuous growth processes like bacteria growth. The function is represented as \( e^{x} \), where \( e \) is an irrational number approximately equal to 2.71828.
This function is unique because its derivative is itself, meaning that it models growth or decay processes with a constant relative rate of change. It's this property that makes it ideal for modeling bacteria growth.
In the bacteria growth model, the formula \( B(x) = 3.5 e^{0.02x} \) can be broken down as follows:

• The constant 3.5 represents the initial concentration of bacteria.
• \( e^{0.02x} \) models the continuous growth of the bacteria, where 0.02 is the growth rate.

This type of function appears frequently in fields like biology, finance, and physics due to its natural growth representation.

Problem Solving in Algebra

Problem solving in algebra, especially with exponential functions, often involves evaluating the function at specific values of the variable. The given problem requires finding the concentration of bacteria at particular times and determining when a certain concentration is achieved.
Let's walk through the steps.
First, calculate the bacteria count at given times:

• Substitute the time value into the function to find the corresponding bacteria concentration.
• Use a calculator to compute the exponential term, such as \( e^{0.02} \) or \( e^{0.13} \).

The next step is to solve for the time when the concentration reaches a specific value. This involves isolating the exponential term and using logarithms:

• Set the function equal to the target concentration.
• Divide by initial concentration.
• Apply the logarithm to both sides to solve for the exponent.
• Finally, solve for the time variable \( x \).

This exercise shows the power of algebra in solving real-world problems involving exponential growth processes.