Drill 1 · Math · Semi-log Plots
AP Precalculus: Semi-log Plots (Drill 20) is a Math practice drill covering Semi-log Plots. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
This AP(r) Precalculus drill focuses on semi-log plots (Topic 2.15): interpreting graphs where one axis uses a logarithmic scale, recognizing that a straight line on a semi-log plot signals an exponential relationship, and converting between semi-log linear equations and explicit exponential functions.
Question 1. A semi-log plot shows a straight line with the vertical axis representing \( \log_{10} y \). The line passes through \( (0, 1) \) and \( (2, 3) \). Which of the following exponential functions best models the data?
Explanation: Choice A is correct. The slope of the line is \( m = \frac{3-1}{2-0} = 1 \) and the y-intercept is \( b = 1 \). The semi-log equation is \( \log_{10} y = x + 1 \). Converting: \( y = 10^{x+1} = 10^1 \cdot 10^x = 10 \cdot 10^x \). Choice B is incorrect because the student correctly finds slope \( m = 1 \) but ignores the y-intercept, writing \( \log_{10} y = x \) and thus \( y = 10^x \). Choice C is incorrect because the student misreads the y-intercept as 2 (reading (0, 2) rather than (0, 1)), giving \( \log_{10} y = x + 2 \) and \( y = 100 \cdot 10^x \). Choice D is incorrect because the student correctly identifies the initial value as \( y = 10^1 = 10 \) at \( x = 0 \) but assumes base 2 (from a doubling model) rather than base 10, writing \( y = 10 \cdot 2^x \).
Question 2. A researcher plots data on a semi-log graph with time \( t \) in years on the horizontal axis and \( \log_{10}(P) \) on the vertical axis, where \( P \) is a population. The plotted points appear to lie on a straight line. Which of the following conclusions is best supported?
Explanation: Choice B is correct. A straight line on a semi-log plot means \( \log_{10} P = mt + b \), which is equivalent to \( P = 10^b \cdot (10^m)^t \) -- an exponential function with constant growth factor \( 10^m \) per year. This means the population grows by a constant percentage each year. Choice A is incorrect because constant additive growth describes a linear model, in which \( P \) itself (not \( \log P \)) would be linear in \( t \). Choice C is incorrect because a straight line on a semi-log plot implies a constant exponential growth rate, not a decreasing one. A decreasing rate would appear as a concave-down curve on the semi-log graph. Choice D is incorrect because polynomial functions do not produce straight lines on semi-log plots. A polynomial \( P \) would yield a curved pattern on such a graph.
Question 3. The table below shows values of \( t \) and \( \log_{10}(N) \) for a bacterial culture.
| t | log10(N) |
|---|---|
| 0 | 2.0 |
| 1 | 2.3 |
| 2 | 2.6 |
| 3 | 2.9 |
| 4 | 3.2 |
Explanation: Choice D is correct. The values of \( \log_{10}(N) \) increase by a constant 0.3 for each unit increase in \( t \), so \( \log_{10} N \) is a linear function of \( t \), which means \( N \) is an exponential function of \( t \). The growth factor per unit time is \( 10^{0.3} \approx 1.995 \approx 2 \), so \( N \approx 100 \cdot 2^t \). Choice A is incorrect because the value 0.3 is the constant rate of change of \( \log_{10} N \), not of \( N \) itself. The population grows multiplicatively, not by a constant additive amount. Choice B is incorrect because it correctly identifies \( N \) as exponential but misidentifies the growth factor as 0.3. The slope 0.3 is the change in \( \log_{10} N \) per unit time; the actual growth factor is \( 10^{0.3} \approx 2 \), not 0.3. Choice C is incorrect because constant rate of change in \( \log_{10} N \) signals an exponential function for \( N \), not a quadratic. A quadratic \( N \) would produce a curved -- not linear -- pattern on the semi-log plot.
Question 4. A semi-log plot of experimental data produces the linear equation \( \log_{10} y = 0.5x + 2 \). Which of the following is an equivalent equation for \( y \) in terms of \( x \)?
Explanation: Choice B is correct. From \( \log_{10} y = 0.5x + 2 \), convert to exponential form: \( y = 10^{0.5x + 2} = 10^2 \cdot 10^{0.5x} = 100 \cdot (10^{0.5})^x = 100 \cdot (\sqrt{10})^x \). Choice A is incorrect because the student treats the y-intercept 2 as a coefficient rather than an exponent, writing \( y = 2 \cdot 10^{0.5x} \) instead of \( y = 10^2 \cdot 10^{0.5x} \). Choice C is incorrect because the student computes \( 10^2 = 10 \) (taking 10 to the first power rather than the second), giving a leading factor of 10 instead of 100. Choice D is incorrect because the student correctly evaluates \( 10^2 = 100 \) but ignores the coefficient 0.5 in the exponent, treating \( 10^{{0.5x}} \) as \( 10^x \), and writing \( y = 100 \cdot 10^x \).
Question 5. A semi-log plot shows \( \log_2(P) \) on the vertical axis and time \( t \) on the horizontal axis. The graph is a straight line passing through \( (0, 2) \) and \( (4, 4) \). What is the doubling time of \( P \) -- the length of time required for \( P \) to double?
Explanation: Choice C is correct. The line through \( (0, 2) \) and \( (4, 4) \) has slope \( m = \frac{4-2}{4-0} = \frac{1}{2} \). So \( \log_2 P = \frac{1}{2}t + 2 \), which means \( P = 2^{\frac{t}{2}+2} = 4 \cdot 2^{t/2} \). For \( P \) to double, \( \log_2 P \) must increase by 1 (since \( \log_2(2P) = \log_2 P + 1 \)). The time needed is \( \Delta t = \frac{1}{m} = \frac{1}{1/2} = 2 \). The doubling time is 2. Choice A is incorrect because the student uses the slope \( \frac{1}{2} \) directly as the doubling time rather than taking its reciprocal (misreading or rounding it to 1). Choice B is incorrect because the student reads the x-coordinate of the second point (4) as the doubling time, without computing the slope or recognizing that the doubling time equals 1 divided by the slope. Choice D is incorrect because the student multiplies the x-range (4) by the reciprocal of the slope (2) instead of simply computing the reciprocal: \( 4 {BS}times 2 = 8 \) rather than \( 1 \div (1/2) = 2 \).