91Ó°ÊÓ

Exponential functions are a crucial concept in mathematics, especially when dealing with growth and decay models. They are usually expressed in the form \( y = a \times b^x \), where \( a \) represents the initial value, \( b \) is the base or growth factor, and \( x \) is the exponent. In the context of medication entering the bloodstream, the situation is modeled by an exponential decay function. Here, the equation is given by \( R = 5 \times (0.95)^t \). This means the rate \( R \) at which the medication enters decreases over time \( t \). The base, in this case, \( 0.95 \), indicates a decay factor since it is less than 1. The function shows how the medication’s entry rate diminishes exponentially, gradually decreasing as time passes. An understanding of why exponential functions behave this way is tied to their property of consistent proportional changes; that is, each time unit results in a constant percentage change.

Mathematical modeling involves using mathematical expressions to represent real-world phenomena. In this scenario, the model describes how medication enters the bloodstream at a rate affected by time. Logarithmic and exponential functions are employed here to model the changing rate. Mathematical models are powerful because they allow predictions and analysis of natural phenomena, like how fast a medication affects the body. By solving equations such as \( \log_{10} \left( \frac{R}{5} \right) = t \log_{10} 0.95 \), we can express the variable \( R \) (rate) explicitly in terms of another variable \( t \) (time). This clarity helps medical professionals adjust dosage timings to achieve desired bloodstream concentrations. Simply put, mathematical modeling provides a quantifiable approach to understanding and managing situations scientifically and pragmatically.

Solving problems involving logarithms and exponents requires systematic approaches. In this exercise, following problem-solving steps helps in simplifying and understanding the solution process.

• Understand the concept: Grasp that logarithms convert multiplicative relationships into additive ones, simplifying complex calculations.
• Apply properties: Use logarithmic properties like \( \log_b a - \log_b c = \log_b \left( \frac{a}{c} \right) \), which combines logarithms into one.
• Simplify the equation: Reorganize the given equation to form a single logarithm. Here, transforming \( \log_{10} R - \log_{10} 5 = t \log_{10} 0.95 \) into \( \log_{10} \left( \frac{R}{5} \right) = t \log_{10} 0.95 \).
• Eliminate the logarithm: Use exponentiation to transform log equations into more manageable exponential ones, which in this case becomes \( \frac{R}{5} = (0.95)^t \).
• Solve for the variable: Finally, isolate the desired variable by straightforward algebraic operations, leading to \( R = 5 \times (0.95)^t \).

These steps not only help solve this problem effectively but are fundamental techniques in approaching many similar problems in mathematics.