Question Statement

Students in a mathematics class took a final exam and were retested in monthly intervals thereafter. The average score $S(t)$, after $t$ months, was given by the function:

$S(t)=68-20\log (t+1),t\ge 0$

a. What was the average score when they initially took the test ($t=0$)? b. What was the average score: i. after 4 months? ii. after 24 months? c. Graph the function. d. After what time was the average score 50?

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Background and Explanation

This problem involves a logarithmic decay model, often used to represent the "forgetting curve" in psychology and education. To solve these parts, we use substitution for specific time values and algebraic manipulation to solve for $t$ when the score is known, specifically using the relationship between logarithms and exponents.

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Solution

Part (a)

To find the initial average score, we substitute $t=0$ into the function. Since $\log (1)=0$, the logarithmic term drops out.

$\begin{array}{rl}S(0)& =68-20\log (0+1)\\ & =68-20\log (1)\\ & =68-20\times 0\\ & =68\end{array}$

The initial average score was 68.

Part (b)

We substitute the given values of $t$ into the function and use a calculator to evaluate the common logarithm (${\log }_{10}$).

i. After 4 months ($t=4$): $\begin{array}{rl}S(4)& =68-20\log (4+1)\\ & =68-20\log (5)\\ & \approx 68-20(0.69897)\\ & \approx 68-13.98\\ & =54.02\end{array}$

ii. After 24 months ($t=24$): $\begin{array}{rl}S(24)& =68-20\log (24+1)\\ & =68-20\log (25)\\ & \approx 68-20(1.3979)\\ & \approx 68-27.96\\ & =40.04\end{array}$

Part (c)

To graph the function, we can calculate several points to see the trend of the curve.

The graph shows a steep initial decline that gradually levels off over time:

Part (d)

We are given the score $S(t)=50$ and need to solve for $t$. We isolate the logarithmic term and then convert the equation to exponential form.

$\begin{array}{rl}50& =68-20\log (t+1)\\ 20\log (t+1)& =68-50\\ 20\log (t+1)& =18\\ \log (t+1)& =\frac{18}{20}\\ \log (t+1)& =0.9\end{array}$

Now, convert from logarithmic form to exponential form (base 10): $\begin{array}{rl}t+1& =10^{0.9}\\ t+1& \approx 7.9433\\ t& =7.9433-1\\ t& =6.9433\end{array}$

The average score was 50 after approximately 6.94 months.

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Key Formulas or Methods Used

• Logarithmic Function: $S(t)=a-b{\log }_{10}(t+c)$
• Logarithmic Property: ${\log }_{10}(1)=0$
• Inverse Property: If ${\log }_{10}(x)=y$, then $x=10^y$
• Substitution: Evaluating functions by replacing variables with constants.

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Summary of Steps

1. Initial Value: Set $t=0$ to find the starting score.
2. Evaluation: Plug in specific time values ($t=4,t=24$) and use a calculator for the log values.
3. Graphing: Create a table of values and plot the points to visualize the logarithmic decay.
4. Solving for Time: Set the function equal to the target score, isolate the $\log$ term, and use the base-10 exponential to solve for $t$.