Question

the function ( p(x) = 90 - 25 log_{2} x ) models the percentage, ( p(x) ), of students who could recall the important features of a classroom lecture as a function of time, where ( x ) represents the number of days that have elapsed since the lecture was given. the figure shows the graph of the function. after how many days do only a quarter of the students recall the important features of the classroom lecture? (let ( p(x) = 25 ) and solve for ( x ).) locate the point on the graph that conveys this information.
after how many days do only a quarter of the students recall the important features of the classroom lecture?
( square ) days
(round to one decimal place as needed.)

Explanation:

Step1: Set up the equation

We know that a quarter of the students recalling means \( P(x) = 25 \) (since the initial percentage is related to recall, and a quarter of the full recall context here is 25 as per the problem's instruction to set \( P(x)=25 \)). The function is \( P(x)=90 - 25\log_{2}x \). So we set up the equation:

$$ 25=90 - 25\log_{2}x $$

Step2: Isolate the logarithmic term

Subtract 90 from both sides:

$$ 25-90=- 25\log_{2}x $$

$$ - 65=- 25\log_{2}x $$

Divide both sides by - 25:

$$ \frac{-65}{-25}=\log_{2}x $$

$$ \frac{13}{5}=\log_{2}x $$

Step3: Convert logarithmic equation to exponential form

Recall that if \( \log_{a}b = c \), then \( b=a^{c} \). So for \( \log_{2}x=\frac{13}{5} \), we have:

$$ x = 2^{\frac{13}{5}} $$

Calculate \( 2^{\frac{13}{5}}=2^{2.6}\approx6.964467 \)

Step4: Round to one decimal place

Rounding 6.964467 to one decimal place gives 7.0

Answer:

\( 7.0 \)