AP Precalculus: Logarithmic Expressions (Drill 1)

Question 1. Which of the following is equivalent to the equation \( 5^3 = 125 \)?

• A) \( \log_3 125 = 5 \)
• B) \( \log_5 3 = 125 \)
• C) \( \log_{125} 5 = 3 \)
• D) \( \log_5 125 = 3 \) ✓

Explanation: Choice D is correct. The exponential equation b^e = r converts to logarithmic form as log_b(r) = e. Here b = 5, e = 3, and r = 125, so the equivalent logarithmic form is log₅(125) = 3. Choice A is incorrect because log₃(125) = 5 has the base and exponent swapped; 3 is not the base of the original equation. Choice B is incorrect because log₅(3) = 125 places the exponent (3) as the argument and the result (125) as the output, reversing the roles of the exponent and the result. Choice C is incorrect because log₁₂₅(5) = 3 uses 125 as the base; in the original equation, 5 is the base, not 125.

Question 2. What is the value of \( \log_4 \frac{1}{64} \)?

• A) −16
• B) −3 ✓
• C) \( \frac{1}{3} \)
• D) 3

Explanation: Choice B is correct. Ask: 4 raised to what power equals 1/64? Since 4³ = 64, we have 4⁻³ = 1/64. Therefore log₄(1/64) = −3. Choice A is incorrect because −16 = −4², confusing squaring the base with the exponent; the student may have computed 4 × (−4) = −16 rather than thinking in terms of powers. Choice C is incorrect because 1/3 reverses the relationship; this would imply 4^1/3 = 1/64, but 4^1/3 ≈ 1.587, not 1/64. Choice D is incorrect because 3 is the exponent for 4³ = 64 (positive), not for 1/64; the student found the correct magnitude but dropped the negative sign.

Question 3. Which of the following is equivalent to \( \log_2 (8x^4) \)?

• A) \( 3 + 4\log_2 x \) ✓
• B) \( 3 \cdot 4\log_2 x \)
• C) \( 8 + 4\log_2 x \)
• D) \( 4\log_2 (8x) \)

Explanation: Choice A is correct. Apply the product rule first: log₂(8x⁴) = log₂(8) + log₂(x⁴). Since 2³ = 8, log₂(8) = 3. Then apply the power rule: log₂(x⁴) = 4log₂(x). Combined: 3 + 4log₂(x). Choice B is incorrect because it multiplies the two log terms rather than adding them; the product rule states log(ab) = log(a) + log(b), not log(a) · log(b). Choice C is incorrect because 8 replaces log₂(8) with the argument 8 itself; the student used the number 8 instead of evaluating log₂(8) = 3. Choice D is incorrect because the exponent 4 cannot be brought in front of the entire expression log₂(8x); the power rule applies only to the exponent on x, not to a factor inside the argument alongside a constant.

Question 4. The table below gives values of log_b(x) for selected values of x.

x	log_b(x)
2	0.43
3	0.68
5	1.00
6	1.11

Using the table, what is the value of log_b(12)?

• A) 0.68
• B) 1.11
• C) 1.54 ✓
• D) 1.79

Explanation: Choice C is correct. Write 12 as a product of values in the table: 12 = 4 × 3 = 2² × 3. Applying the product and power rules: log_b(12) = log_b(4) + log_b(3) = log_b(2²) + log_b(3) = 2·log_b(2) + log_b(3) = 2(0.43) + 0.68 = 0.86 + 0.68 = 1.54. Choice A is incorrect because 0.68 = log_b(3) alone; the student found one factor of 12 but ignored the other. Choice B is incorrect because 1.11 = log_b(6); 6 is a factor of 12 but the student did not account for the remaining factor of 2 (since 12 = 6 × 2, not just 6). Choice D is incorrect because 1.79 ≈ 0.68 + 1.11 = log_b(3) + log_b(6) = log_b(18), not log_b(12); the student used 3 × 6 = 18 rather than decomposing 12 correctly.

Question 5. A student claims that \( 2\ln x + \ln 3 - \ln x = \ln(3x) \). Which of the following best describes whether the student's claim is correct?

• A) The claim is correct because the power rule gives 2ln x = ln(2x), and applying the product rule then yields ln(3x).
• B) The claim is correct because 2ln x + ln 3 − ln x = ln(x²) + ln 3 − ln x = ln(3x²/x) = ln(3x). ✓
• C) The claim is incorrect because 2ln x = ln(x²), so the expression simplifies to ln(3x²) − ln x = ln(3x), which is only true when x = 1.
• D) The claim is incorrect because 2ln x − ln x = ln x, not ln(x²/x), so the expression equals ln(x) + ln 3 = ln(3x) only after applying the product rule, making the final answer correct but for a wrong intermediate reason.

Explanation: Choice B is correct. The student's final answer is correct and the reasoning in Choice B is valid. Step by step: 2ln x = ln(x²) by the power rule. Then ln(x²) + ln 3 = ln(3x²) by the product rule. Then ln(3x²) − ln x = ln(3x²/x) = ln(3x) by the quotient rule. The answer ln(3x) is correct for all x > 0. Choice A is incorrect because it states 2ln x = ln(2x), which applies the coefficient as a multiplier of the argument rather than as an exponent; the power rule gives 2ln x = ln(x²), not ln(2x). This is a named student error: confusing the power rule with scalar multiplication. Choice C is incorrect because it claims the result ln(3x) is only true when x = 1; in fact ln(3x²) − ln x = ln(3x²/x) = ln(3x) for all x > 0, not just x = 1. Choice D is incorrect in its characterization: 2ln x − ln x does equal ln x (which is ln(x²/x) simplified), but the reasoning path described is muddled and does not correctly represent the error; the student's original claim in the problem is actually fully correct.