Question

Students in a seventh-grade class were given an exam. During the next 2 years, the same students were retested several times. The average score \(g\) can be approximated by the model \(g(t)=87-16 \log _{10}(t+1), \quad 0 \leq t \leq 24\) where \(t\) is the time (in months). (a) What was the average score on the original exam? (b) What was the average score after 6 months? (c) When did the average score drop below \(70 ?\)

Short answer

The average score on the original exam was 87. After six months, the average score was 73. The average score dropped below 70 around the 12th month.

Step-by-step solution

Compute the score on the original exam

To find the original average score, replace \(t\) with \(0\) in the function \(g(t)\) to get \(g(0)=87-16 \log _{10}(0+1) = 87 - 16 \log _{10}(1) = 87.\)

Calculate the average score after 6 months

Replace \(t\) with \(6\) in the equation \(g(t)\) as follows: \(g(6)=87-16 \log _{10}(6+1) = 87 - 16 * \log _{10}(7) = 73.\)

Find when the score drops below 70

Place \(g(t)<70\) in the equation and resolve for \(t\). This results in \(87 - 16 \log _{10} (t+1) < 70\), which simplifies to \(-16 \log _{10} (t+1) < -17\), thereby dividing by -16, remembering to revert the inequality as dividing or multiplying by a negative number. This means \(\log _{10} (t+1) > 1.0625\). Inverting the logarithm, yields \(t+1 > 10^{1.0625}\) or \(t > 10^{1.0625} - 1\). Evaluating the decimal yields \(t \approx 11.8\). Therefore, the average score dropped below 70 around the 12th month, as we round up in this case since the time cannot be fractional.

Key concepts

Algebraic Modeling

Algebraic modeling is an essential part of mathematics, allowing us to represent real-world situations through mathematical expressions and equations. It translates scenarios into an algebraic language that we can systematically solve. In our example related to the seventh-grade class exam scores, the model

\(g(t) = 87 - 16 \log_{10}(t+1)\)

uses logarithmic functions to describe how average scores change over time. To solve part (a) of the exercise, we see algebraic modeling in action as we substitute the initial time \(t=0\) into the equation, giving us the initial average score. This substitution simplifies the expression since \(\log_{10}(1)\) equals zero, leaving us with the average initial score of 87. Similarly, for part (b), we input \(t=6\) months into our model to find the average score at that time. This score accounts for the decay in the exam scores as time passes, which is perfectly captured by the slow reduction caused by the logarithmic term in the model.

Logarithms and Exponential Functions

Logarithms and exponential functions are two sides of the same coin in algebra. They are inverse operations, meaning they can undo each other. An exponential function like \(y = b^x\), where \(b\) is a positive number other than 1, can be rewritten in logarithmic form as \(x = \log_b(y)\). When dealing with logarithms, one of the most common bases is 10, also known as the common logarithm, symbolized as \(\log _{10}\).

Our textbook exercise offers a vivid demonstration of how logarithmic functions are used in algebraic models, represented by the average score function \(g(t)\). This is important because logarithms describe situations where growth or decay accelerates or decelerates in correlation to the current value. For instance, finding the average score after 6 months requires understanding of logarithms to determine the score's decline over time, illustrated by the calculation \(87 - 16 * \log _{10}(7) = 73\), which indicates that the scores decay in a non-linear fashion, as is typical for processes represented by logarithmic relations.

Problem Solving with Logarithms

Problem solving with logarithms often involves rewriting the equation to expose the variable of interest and using rules of logarithms to simplify expressions. When the average score was set to drop below 70, the equation to solve was:

\(87 - 16 \log_{10}(t+1) < 70\)

This inequality was manipulated using logarithmic properties and the understanding that dividing by a negative number changes the direction of the inequality, leading us to \(t > 10^{1.0625} - 1\). This solution process emphasizes the non-linear nature of logarithmic functions and can be nuanced as multiple steps are often required to isolate the variable. Understanding these steps is crucial when applying logarithms to solve problems, whether they are concerned with declining scores or other real-world phenomena that follow similar patterns. By conquering these problem-solving skills with logarithms, students can tackle a vast array of mathematical and practical situations effectively.