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AP Precalculus -2.5 Exponential Function Modeling- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -2.5 Exponential Function Modeling- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -2.5 Exponential Function Modeling- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question 

Water hyacinth is an invasive plant species found in many lakes that typically grows at a rate of 7% per day. As part of a study, a scientist introduces a 150-gram sample of water hyacinth into a testing pool. Which of the following functions gives the amount of water hyacinth in the testing pool \( t \) weeks after the sample is introduced? (Note: 1 week is 7 days.)
(A) \( f(t) = 150 \left( 1 + 0.07^{(1/7)} \right)^t \)
(B) \( g(t) = 150 \left( 1.07^{(1/7)} \right)^t \)
(C) \( h(t) = 150 \left( 1 + 0.07^{(7)} \right)^t \)
(D) \( k(t) = 150 \left( 1.07^{(7)} \right)^t \)
▶️ Answer/Explanation
Detailed solution

Daily growth factor: \( 1 + 0.07 = 1.07 \).
In one week (7 days), growth factor = \( (1.07)^7 \).
If \( t \) is in weeks, amount after \( t \) weeks: \( 150 \cdot (1.07^7)^t = 150 \cdot (1.07)^{7t} \).
The given options:
(D) \( k(t) = 150 (1.07^{(7)})^t = 150 \cdot (1.07^7)^t \), matches.
✅ Answer: (D)

Question 

The sales of a new product, in items per month, is modeled by the expression \( 225 + 500 \log_{10}(15t + 10) \), where \( t \) represents the time since the product became available for purchase, in months. What is the number of items sold per month for time \( t = 6 \)?
(A) 725
(B) 1225
(C) 1700
(D) 5225
▶️ Answer/Explanation
Detailed solution

Plug \( t = 6 \):
\( 15t + 10 = 15(6) + 10 = 90 + 10 = 100 \)
\( \log_{10}(100) = 2 \)
Sales = \( 225 + 500 \times 2 = 225 + 1000 = 1225 \)
✅ Answer: (B)

Question 

At time \(t=0\) years, the population of a certain city was \(23,144\). During each of the next \(10\) years, the population decreased by \(4\%\) per year. Based on this information, which of the following models the population as a function of time \(t\), in years, for \(0\le t\le10\)?
(A) \(23,144-0.04t\)
(B) \(23,144-0.96t\)
(C) \(23,144(0.96)^{t}\)
(D) \(23,144(1.04)^{t}\)
▶️ Answer/Explanation
Detailed solution

1. Identify Model Type:
Constant percentage decrease implies exponential decay.

2. Determine Growth Factor:
Decay of \(4\%\) means the multiplier is \(1 – 0.04 = 0.96\).

3. Construct Function:
\(P(t) = \text{Initial Value} \cdot (\text{Factor})^t\)
\(P(t) = 23,144(0.96)^t\)

✅ Answer: (C)

Question 

The function \( S \) is given by \( S(t) = \frac{500,000}{1 + 0.4e^{kt}} \), where \( k \) is a constant. If \( S(4) = 300,000 \), what is the value of \( S(12) \)?
(A) 175,325
(B) 214,772
(C) 343,764
(D) 357,143
▶️ Answer/Explanation
Detailed solution

Given \( S(4) = 300,000 \):
\( 300000 = \frac{500000}{1 + 0.4 e^{4k}} \)
\( 1 + 0.4 e^{4k} = \frac{500000}{300000} = \frac{5}{3} \)
\( 0.4 e^{4k} = \frac{5}{3} – 1 = \frac{2}{3} \)
\( e^{4k} = \frac{2/3}{0.4} = \frac{2}{3} \cdot \frac{5}{2} = \frac{5}{3} \)
So \( e^{4k} = \frac{5}{3} \).
Now \( S(12) = \frac{500000}{1 + 0.4 e^{12k}} \).
Note \( e^{12k} = (e^{4k})^3 = \left(\frac{5}{3}\right)^3 = \frac{125}{27} \).
\( S(12) = \frac{500000}{1 + 0.4 \cdot \frac{125}{27}} \)
\( 0.4 \cdot \frac{125}{27} = \frac{2}{5} \cdot \frac{125}{27} = \frac{50}{27} \)
\( 1 + \frac{50}{27} = \frac{77}{27} \)
\( S(12) = 500000 \cdot \frac{27}{77} \approx 175324.675 \approx 175325 \)
✅ Answer: (A)

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