Advanced Functions Help!?

Students participating in a psychology experiment and took MHF4U were given an exam. Every month for a year after the exam, the students were retested to see how much of the material they remembered. The average scores for the group are given by the human memory model: f(t)=75 - (6log(t+1))/(log e) where e is the number in your calculator and t is time in months 0< t < 12.

a) What was the average score on the original exam (t=0)

b) When was the average score equal to 61.2%

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  • 3 months ago
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    There is a mistake in the question.  We are told 0< t < 12.  Later we are told to use t=0 for the ‘original exam’.  This makes no sense, but I’ll use t=0.

    I assume ‘log’ is base 10. I’ll use log₁₀ for clarity.

    f(t)=75 - (6log₁₀(t+1))/(log e)

    ‘e’ is called Euler’s number, or sometimes ‘the base of natural logarithms’.  It is 2.7183 approximately.  You can get it on your calculator using the ‘eˣ’ key (may require SHIFT or 2nd Function).  Then find e¹.

    log₁₀(e)= log₁₀(2.7183) = 0.4343 approx.

    __________________________

    a) What was the average score on the original exam (t=0)

    All we need to do is find f when t=0, i.e. evaluate f(0).

    f(0) = 75 – (6log₁₀(0+1)) / 0.4343

    You should know that log₁₀(1) = 0, but you can use your calculator if you don’t know it:

    f(0) = 75 – (6 * 0) / 0.4343

    . . . ,= 75 (answer)

    ______________

    b) Assuming the score are percentages, then we need to find the value of t which makes f(t) = 61.2

    61.2 = 75 – (6log₁₀(t+1)) / 0.4343

    6log₁₀(t+1) = 0.4343*(75 – 61.2)

    . . . . . . . . .  = 5.9934

    log₁₀(t+1) = 5.9934 / 6 = 0.9989

    You don’t really need a calculator as 0.9989 is near-enough 1.  Since log₁₀(10) = 1 we

    know t+1=10 approx.  But if you don’t want to do it that way:

    t+1 = 10^0.9989

    . . . = 9.98

    t = 8.98

    So, to the nearest month, the answer is 9 months.

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