Problem 26 - A Graphical Approach to College Algebra

Question

Suppose that a sample of bacteria has a concentration of 2 million bacteria per milliliter and it doubles in concentration every 12 hours. Then the time \(T\) it takes for the sample to reach a concentration of \(C\) can be approximated by the following logarithmic function. \(T(C)=\frac{500}{29} \ln \frac{C}{2}\) (a) Find the domain of \(T .\) Interpret your answer. (b) How long does it take for the concentration of bacteria to increase by \(50 \% ?\) (c) Determine the concentration \(C\) after 15 hours by solving the equation \(T(C)=15\)

Step-by-Step Solution

Answer

(a) Domain: \(C > 0\). (b) 6.98 hours. (c) \(C \approx 4.768\) million bacteria/mL.

Step 1: Determine the Domain of T

The domain of a function is the set of input values (here, concentrations \( C \)) for which the function is defined. In \( T(C) = \frac{500}{29} \ln \frac{C}{2} \), the logarithm \( \ln \frac{C}{2} \) requires that \( \frac{C}{2} > 0 \). This implies \( C > 0 \). Thus, the domain of \( T \) is \( C > 0 \). This means the function is defined for any positive concentration of bacteria.

Step 2: Interpret the Domain

The domain \( C > 0 \) indicates that the function \( T(C) \) can only be used for positive concentrations because a non-positive concentration doesn't make sense in this biological context. It implies we are dealing with samples that contain some amount of bacteria.

Step 3: Calculate Time for 50% Increase

A 50% increase means the new concentration \( C \) is \( 1.5 \times 2 \) million bacteria/mL, starting from an initial 2 million bacteria/mL. Thus, \( C = 3 \). Substitute this into the function: \( T(3) = \frac{500}{29} \ln \frac{3}{2} \). Compute this to find \( T(3) \approx \frac{500}{29} \times 0.405 \approx 6.98 \) hours.

Step 4: Solve T(C) = 15 for C

Given \( T(C) = 15 \), substitute into the function: \( 15 = \frac{500}{29} \ln \frac{C}{2} \). Solve for \( \ln \frac{C}{2} \) as follows: \( \ln \frac{C}{2} = \frac{15 \times 29}{500} = 0.87 \). Exponentiate both sides to solve for \( C \): \( \frac{C}{2} = e^{0.87} \approx 2.384 \). Thus, \( C = 2 \times 2.384 \approx 4.768 \) million bacteria/mL.

Key Concepts

Domain of a Function

The domain of a function refers to all the possible inputs for which the function is defined. In the context of the logarithmic function for predicting bacterial concentration growth, represented as \( T(C) = \frac{500}{29} \ln \frac{C}{2} \), the domain is derived from the constraints of the logarithm. For logarithmic functions, the expression inside the logarithm should be greater than zero. So, the condition \( \frac{C}{2} > 0 \) must hold true, implying \( C > 0 \).
This means that the function \( T(C) \) is defined solely for positive values of \( C \).
In practical terms, it suggests that only positive bacterial concentrations can be used as input, which makes sense since a negative concentration isn't feasible in real-world scenarios. Therefore, the domain \( C > 0 \) indicates that we are considering samples with some non-zero quantity of bacteria.

Exponential Growth

Exponential growth occurs when a quantity increases by the same proportion in each equal time period. The problem describes such a scenario, where bacteria doubles every 12 hours. This is characterized by an exponential function, a standard model for rapid increase in populations.
The growth rate is determined by the factor by which the concentration increases. In this instance, doubling means the concentration at any time doubles compared to the previous period.
Mathematically, exponential growth can be expressed using the formula \( C(t) = C_0 \times 2^{t/T} \), where:

• \( C_0 \) is the initial concentration.
• \( T \) is the doubling time (12 hours here).
• \( t \) represents time elapsed.

This formula reflects the behavioral pattern described: as time progresses, the concentration increases exponentially, reinforcing why understanding these functions is crucial in biological studies.

Solving Equations

Solving logarithmic equations involves manipulating the equation to find unknown variable values. In the exercise, students are required to solve for \( C \) when \( T(C) = 15 \), demonstrating an applicative use of equations.
The given equation, \( T(C) = \frac{500}{29} \ln \frac{C}{2} = 15 \), needs to be rearranged to isolate \( C \). First, solve for \( \ln \frac{C}{2} \) by multiplying both sides by \( \frac{29}{500} \), resulting in \( \ln \frac{C}{2} = 0.87 \).
Exponentiating both sides helps remove the logarithm, resulting in \( \frac{C}{2} = e^{0.87} \). Using the approximation, \( e^{0.87} \approx 2.384 \), we solve for \( C \) as follows:

• Multiply both sides by 2: \( C = 2 \times 2.384 \).
• Conclude that \( C \approx 4.768 \) million bacteria/mL.

This process highlights how understanding the properties of logarithms and exponentiation is instrumental in finding solutions, turning abstract math into significant real-world applications.