Logarithmic Function Context and Data Modeling
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AP Precalculus › Logarithmic Function Context and Data Modeling
A chemistry class measures a radioactive sample that steadily loses mass. The instructor emphasizes that the process is exponential with a constant decay rate, so a log transformation linearizes the data. They model mass by $m(t)=m_0(1-d)^t$ and rewrite it as $\log_{10}(m)=\log_{10}(m_0)+t,\log_{10}(1-d)$. Here the logarithm uses base 10, and the slope equals $\log_{10}(1-d)$. The table gives measured masses in grams at integer hours. Students are asked to pick the log-linear equation that matches the measurements.
Based on the scenario, which logarithmic equation represents the scenario described?
$\log_{10}(m)=\log_{10}(50)+t,\log_{10}(0.90)$
$\log_{10}(m)=\log_{10}(50)+t,\log_{10}(0.95)$
$\log_{10}(m)=\log_{10}(45)+t,\log_{10}(0.95)$
$m=50,\log_{10}(0.95t)$
Explanation
This question tests AP Precalculus understanding of logarithmic functions and data modeling, specifically applying logarithms to radioactive decay with given initial conditions and decay rates. The exponential decay model m(t) = $m₀(1-d)^t$ becomes linear when log-transformed, allowing for easier analysis and prediction. In this scenario, the initial mass is m₀ = 50 grams, and the decay results in retaining 95% of mass each hour (decay rate d = 0.05, so 1-d = 0.95). Choice A is correct because it accurately represents the logarithmic form: log₁₀(m) = log₁₀(m₀) + t·log₁₀(1-d) = log₁₀(50) + t·log₁₀(0.95). Choice C is incorrect because it uses m₀ = 45 instead of 50, misreading the initial mass from the table. To help students: Emphasize careful reading of data tables to identify initial values (at t=0) versus later measurements. Practice setting up logarithmic models by first writing the exponential form, then applying logarithms to both sides systematically.