Question

module 12 (logarithms) assignment
score: 19/20 answered: 19/20
question 20
to measure a students retention of knowledge, the student is tested after a given amount of time. a students score on a math test t months after the school year is over can be approximated by this function:
r = 83 - 6\log_{2}(t + 1)
where r is the score as a percent.
what was the students score at the time the school year ended (t = 0)?
%
what was the students score after 5 months? round to one decimal place.
%
what was the students score after 14 months? round to one decimal place.
%
submit question

Explanation:

Part 1: Score when \( t = 0 \)

Step 1: Substitute \( t = 0 \) into the function

We have the function \( R = 83 - 6\log_{2}(t + 1) \). Substitute \( t = 0 \):
\( R = 83 - 6\log_{2}(0 + 1) \)

Step 2: Simplify the logarithm

Since \( \log_{2}(1)=0 \) (because \( 2^{0}=1 \)):
\( R = 83 - 6\times0 \)
\( R = 83 \)

Part 2: Score when \( t = 5 \)

Step 1: Substitute \( t = 5 \) into the function

Substitute \( t = 5 \) into \( R = 83 - 6\log_{2}(t + 1) \):
\( R = 83 - 6\log_{2}(5 + 1) \)
\( R = 83 - 6\log_{2}(6) \)

Step 2: Calculate the logarithm

We know that \( \log_{2}(6)=\frac{\ln(6)}{\ln(2)}\approx\frac{1.7918}{0.6931}\approx2.585 \)

Step 3: Calculate \( R \)

\( R = 83 - 6\times2.585 \)
\( R = 83 - 15.51 \)
\( R\approx67.5 \) (rounded to one decimal place)

Part 3: Score when \( t = 14 \)

Step 1: Substitute \( t = 14 \) into the function

Substitute \( t = 14 \) into \( R = 83 - 6\log_{2}(t + 1) \):
\( R = 83 - 6\log_{2}(14 + 1) \)
\( R = 83 - 6\log_{2}(15) \)

Step 2: Calculate the logarithm

\( \log_{2}(15)=\frac{\ln(15)}{\ln(2)}\approx\frac{2.7080}{0.6931}\approx3.907 \)

Step 3: Calculate \( R \)

\( R = 83 - 6\times3.907 \)
\( R = 83 - 23.442 \)
\( R\approx59.6 \) (rounded to one decimal place)

Answer:

s:

• When \( t = 0 \): \( \boldsymbol{83} \)
• When \( t = 5 \): \( \boldsymbol{67.5} \)
• When \( t = 14 \): \( \boldsymbol{59.6} \)