AP Precalculus -2.15 Semi-log Plots- MCQ Exam Style Questions - Effective Fall 2023
AP Precalculus -2.15 Semi-log Plots- MCQ Exam Style Questions – Effective Fall 2023
AP Precalculus -2.15 Semi-log Plots- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.
Question
(B) \( f(x) = x^2 \) and \( g(x) = 3x^2 \)
(C) \( f(x) = 2^x \) and \( g(x) = 3 \cdot 2^x \)
(D) \( f(x) = \ln(2x) \) and \( g(x) = 3 \ln(2x) \)
▶️ Answer/Explanation
A semi-log plot has a logarithmic scale on the vertical axis and a linear scale on the horizontal axis. Exponential functions \( y = k \cdot a^x \) appear as straight lines because \( \log y = \log k + x \log a \), which is linear in \( x \).
For \( f(x) = 2^x \) and \( g(x) = 3 \cdot 2^x \):
In a semi-log plot, \( \log f(x) = x \log 2 \) and \( \log g(x) = \log 3 + x \log 2 \).
Both are lines with slope \( \log 2 \), but different intercepts, so they appear as parallel lines.
✅ Answer: (C)
Question
(B) \( \frac{1}{10}t + 5 \)
(C) \( 2.5 \cdot 2^{(t/10)} \)
(D) \( 3 + 2^{(t/10)} \)
▶️ Answer/Explanation
A semi-log plot appears linear when the data follow an exponential model. In such a plot, a straight line corresponds to exponential growth of the form \( N(t) = a \cdot b^{t} \). The slope of the line in the semi-log plot relates to the growth factor over time.
Given the answer choice (C): \( N(t) = 2.5 \cdot 2^{(t/10)} \)
This means the population doubles every 10 days (since when \( t \) increases by 10, \( 2^{(t/10)} \) doubles). At \( t = 0 \), \( N(0) = 2.5 \) thousand, and at \( t = 10 \), \( N(10) = 2.5 \cdot 2 = 5 \) thousand, matching a point from the semi-log data if the plot is linear.
✅ Answer: (C)
Question
(B) Exponential growth
(C) Exponential decay
(D) Logarithmic
▶️ Answer/Explanation
1. Understand Semi-Log Plots:
A semi-log plot typically scales the \(y\)-axis logarithmically while keeping the \(x\)-axis linear. This is used to test for exponential relationships.
If the graph of \((\log y)\) versus \(x\) is a straight line, the original data follows an exponential model.
2. Analyze the Pattern:
The problem states the points follow a decreasing linear pattern on this plot. A linear pattern confirms the function is exponential.
The fact that it is decreasing (negative slope) indicates that as \(x\) increases, \(\log y\) decreases, which means \(y\) decreases toward 0.
3. Conclusion:
An exponential function with a negative rate of change corresponds to Exponential Decay.
✅ Answer: (C)
Question
▶️ Answer/Explanation
A semi‑log plot has a logarithmic vertical axis and linear horizontal axis. For an exponential function, the semi‑log plot appears linear.
From the table: \( m(t) \) grows roughly by a constant factor (exponential). In a semi‑log plot, points \( (t, m(t)) \) are plotted with \( m(t) \) on a log scale.
Check values:
- log₁₀(30)≈1.48,
- log₁₀(50)≈1.70,
- log₁₀(83)≈1.92,
- log₁₀(139)≈2.14,
- log₁₀(231)≈2.36
increases roughly linearly in \( t \).
Graph C shows vertical axis labeled 1, 10, 100, 1000 (log scale) and likely shows points forming a straight line.
✅ Answer: (C)



