AP Precalculus: Logarithmic Functions (Drill 1)

Question 1. What is the domain of the function \( f(x) = \log_3 (2x - 6) \)?

• A) x > 6
• B) x > 3 ✓
• C) x ≥ 3
• D) x > 0

Explanation: Choice B is correct. A logarithm is defined only when its argument is strictly positive. Set the argument greater than zero: 2x − 6 > 0, so 2x > 6, giving x > 3. Choice A is incorrect because x > 6 results from solving 2x > 6 by subtracting 2 rather than dividing by 2; the student made an arithmetic error in isolating x. Choice C is incorrect because x ≥ 3 includes x = 3, which gives the argument 2(3) − 6 = 0; logarithms are undefined at zero, so the boundary must be strict (x > 3, not x ≥ 3). Choice D is incorrect because x > 0 is the domain of log₃(x) with argument x alone; the student ignored the transformation 2x − 6 and applied the base domain restriction directly to x.

Question 2. Which of the following correctly identifies a feature of the graph of \( f(x) = \log_2 (x + 4) \)?

• A) The graph has a vertical asymptote at x = 4.
• B) The graph has a vertical asymptote at x = −4 and passes through (−3, 0). ✓
• C) The graph has a vertical asymptote at x = −4 and passes through (0, 4) for the logarithmic function given.
• D) The graph has a horizontal asymptote at y = −4.

Explanation: Choice B is correct. The argument x + 4 equals zero when x = −4, so the vertical asymptote is at x = −4. The x-intercept occurs when f(x) = 0: log₂(x + 4) = 0 means x + 4 = 2⁰ = 1, so x = −3. The graph passes through (−3, 0). Both facts in Choice B are correct. Choice A is incorrect because the vertical asymptote is at x = −4, not x = 4; the student applied the shift in the wrong direction, confusing f(x + 4) with f(x − 4). Choice C is incorrect because the y-intercept is f(0) = log₂(0 + 4) = log₂(4) = 2, not 4; the student used the argument value 4 directly rather than evaluating the logarithm log₂(4) = 2. Choice D is incorrect because logarithmic functions have vertical asymptotes, not horizontal ones; the function grows without bound as x → ∞ and has no horizontal asymptote.

Question 3. The function g is defined by \( g(x) = -\log_5 (x - 2) + 1 \). Compared to the graph of \( f(x) = \log_5 x \), the graph of g is:

• A) reflected over the x-axis, shifted right 2 units, and shifted up 1 unit ✓
• B) reflected over the y-axis, shifted right 2 units, and shifted up 1 unit
• C) reflected over the x-axis, shifted left 2 units, and shifted down 1 unit
• D) reflected over the x-axis, shifted right 2 units, and shifted down 1 unit

Explanation: Choice A is correct. Compare g(x) = −log₅(x − 2) + 1 to f(x) = log₅(x) term by term. The negative sign in front reflects the graph over the x-axis (output values are negated). The (x − 2) inside the argument shifts the graph right 2 units (horizontal translation). The +1 outside shifts the graph up 1 unit (vertical translation). Choice B is incorrect because a reflection over the y-axis would require f(−x) = log₅(−x); the negative sign is applied to the entire output, not to x, so it reflects over the x-axis, not the y-axis. Choice C is incorrect because (x − 2) shifts the graph right, not left; a left shift of 2 would require (x + 2) inside the argument. Choice D is incorrect because the +1 outside the logarithm shifts the graph up, not down; a downward shift of 1 would require −1 outside the logarithm.

Question 4. A population of bacteria is modeled by \( P(t) = 500 \cdot 2^t \), where t is time in hours. A researcher uses the inverse function to determine the time at which the population reaches a given size. Which of the following expresses t as a function of P?

• A) \( t = \log_2 (500P) \)
• B) \( t = \log_2 \left(\frac{P}{500}\right) \) ✓
• C) \( t = \frac{\log_2 P}{500} \)
• D) \( t = 500 \cdot \log_2 P \)

Explanation: Choice B is correct. Starting from P = 500 · 2^t, isolate the exponential: divide both sides by 500 to get P/500 = 2^t. Apply log base 2 to both sides: log₂(P/500) = t. As a check: when P = 500 (t = 0), log₂(500/500) = log₂(1) = 0 ✓. When P = 1000 (t = 1), log₂(1000/500) = log₂(2) = 1 ✓. Choice A is incorrect because log₂(500P) multiplies rather than divides by 500; this would correspond to P = 2^t/500 rather than the original model. Choice C is incorrect because (log₂ P)/500 divides the logarithm by 500 after taking it, rather than dividing P by 500 before applying the logarithm; these are not equivalent operations. Choice D is incorrect because 500 · log₂ P multiplies the logarithm by 500 rather than dividing the argument by 500 first; the student moved the coefficient inside the log incorrectly.

Question 5. Which of the following statements about the function \( f(x) = 2\log_{10}(x - 1) \) is correct?

• A) The range of f is y ≥ 0, and f is increasing on its entire domain.
• B) The domain of f is x > 1, the range is all real numbers, and f is increasing on its entire domain. ✓
• C) The domain of f is x > 0, and f has a vertical asymptote at x = 0.
• D) The range of f is y > 0, and f has a horizontal asymptote at y = 0.

Explanation: Choice B is correct. The argument of the logarithm is (x − 1), which must be positive: x − 1 > 0 gives domain x > 1. The vertical asymptote is at x = 1. Since log₁₀(x − 1) takes all real values as x ranges over (1, ∞), multiplying by 2 does not restrict the range, f has range (−∞, ∞), all real numbers. Because log base 10 is an increasing function and x − 1 is an increasing function of x, f is increasing on its entire domain. All three statements in Choice B are correct. Choice A is incorrect because the range is not y ≥ 0; f(x) = 2log₁₀(x − 1) takes all real values, including negative outputs when 0 < x − 1 < 1 (i.e., 1 < x 1 (not x > 0) and the vertical asymptote is at x = 1 (not x = 0); the student ignored the horizontal shift of 1 and treated the function as if it were 2log₁₀(x). Choice D is incorrect because logarithmic functions do not have horizontal asymptotes; the range is all real numbers, not y > 0, and the function grows without bound.