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Chapter 3: Problem 81

Students in a mathematics class took an exam and then took a retest monthly with an equivalent exam. The average scores for the class are given by the human memory model $$f(t)=80-17 \log (t+1), \quad 0 \leq t \leq 12$$ where \(t\) is the time in months. (a) Use a graphing utility to graph the model over the specified domain. (b) What was the average score on the original exam \((t=0) ?\) (c) What was the average score after 4 months? (d) What was the average score after 10 months?

Short Answer

Expert verified
The average score on the original exam (t=0) is 80. The average score after 4 months and 10 months can be calculated by substituting \(t=4\) and \(t=10\) into the function, resulting in \(80 - 17 \log(5)\) and \(80 - 17 \log(11)\) respectively.

Step by step solution

01

Graph the function

Firstly, plot the function \( f(t) = 80 - 17 \log(t + 1) \) over the domain \(0 \leq t \leq 12\). This can be done using any scientific or graphing calculator or online graphing tools.
02

Find the average score at t=0

Plug in the value \(t=0\) into the function to find the average score on the original exam: \(f(0) = 80 - 17 \log(0 + 1) = 80 - 17 \log(1)\). Since the logarithm of 1 is 0, the average score on the original exam is \(80 - 17*0 = 80\)
03

Find the average score at t=4

Plug in the value \(t=4\) into the function to find the average score after 4 months: \(f(4) = 80 - 17 \log(4 + 1) = 80 - 17 \log(5)\). Using a calculator, find the value of the logarithm and subtract the result from 80.
04

Find the average score at t=10

Plug in the value \(t=10\) to the function to find the average score after 10 months: \(f(10) = 80 - 17 \log(10 + 1) = 80 - 17 \log(11)\). Again, use a calculator to find the value of the logarithm and subtract the result from 80.

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Most popular questions from this chapter

Using Properties of Logarithms In Exercises \(15-20\) , use the properties of logarithms to rewrite and simplify the logarithmic expression. $$\log _{4} 8$$

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