Question 86089: Students in a psychology class took a final examination. As a part of an experiment to see how much of the course they remembered over time, they took equivalent forms of the exam in monthly intervals thereafter. The average score, f(t), for the group t months was modeled by the function f(t) = 76 - 18 log (t+1), where 0< or equal to t < or equal to 12.
A) What was the average score when the exam was first given?
B) What was the average score after 4 months? 1 year?
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Since you did not specify the base of the logarithms, I'm presuming that this problem uses
logarithms to the base 10.
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The average score (S) the first time the exam was given occurs when t equals zero. This makes
the equation:
.
S = 76 - 18*log(0+1) = 76 - 18*log(1)
.
but log(1) = 0
.
so the equation reduces to:
.
S = 76 - 18*0 = 76
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So the first time the exam was given the average score was 76.
.
After 4 months the average score becomes:
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S = 76 - 18*log(4+1) = 76 - 18*log(5)
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Using a calculator you can find that log(5) = 0.69897. Substituting this value for log(5)
changes the equation to:
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S = 76 - 18*(0.69897) = 76 - 12.58146= 63.4185.
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So 4 months later when the students took the exam, they scored an average of 63.4 on it.
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A year (12 months) after they first took the exam they re-tested. The equation becomes:
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S = 76 - 18*log(12+1) = 76 - 18*log(13)
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and a calculator tells you that log(13) = 1.1139. Substituting this value for log(13) makes
the equation become:
.
S = 76 - 18*1.1139 = 76 - 20.0502 = 55.9498
.
So a year after they took the original exam, the students re-took it and scored an average
of 55.9 on it.
.
Hope this helps you to gain a little better understanding of working with logarithms.
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