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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 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 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.