AP Precalculus: Exponential and Logarithmic Equations (Drill 19)

Question 1. Which value of \( x \) satisfies the equation \( 2^{3x-1} = 32 \)?

• A) \( x = 1 \)
• B) \( x = \dfrac{5}{3} \)
• C) \( x = \dfrac{7}{3} \)
• D) \( x = 2 \) ✓

Explanation: Choice D is correct. Since \( 32 = 2^5 \), the equation becomes \( 2^{{3x-1}} = 2^5 \). Setting exponents equal: \( 3x - 1 = 5 \), so \( 3x = 6 \) and \( x = 2 \). Choice A is incorrect because the student confuses 32 with \( 2^3 = 8 \), setting \( 3x - 1 = 3 \), giving \( x = \frac{4}{3} \), not 1. Choice B is incorrect because the student correctly identifies \( 32 = 2^5 \) but skips adding 1 to both sides, solving \( 3x = 5 \) and getting \( x = \frac{5}{3} \) instead of 2. Choice C is incorrect because the student adds 2 to the right-side exponent rather than isolating: treating \( 3x = 5 + 2 = 7 \), giving \( x = \frac{7}{3} \).

Question 2. Which value of \( x \) satisfies the equation \( \log_3(2x + 3) = 3 \)?

• A) \( x = 12 \) ✓
• B) \( x = 13 \)
• C) \( x = 15 \)
• D) \( x = 3 \)

Explanation: Choice A is correct. Converting from logarithmic to exponential form: \( \log_3(2x+3) = 3 \) means \( 2x + 3 = 3^3 = 27 \). Solving: \( 2x = 24 \), so \( x = 12 \). Choice B is incorrect because the student subtracts 1 instead of 3: \( 2x = 27 - 1 = 26 \), giving \( x = 13 \). Choice C is incorrect because the student adds rather than subtracts the constant: \( 2x = 27 + 3 = 30 \), giving \( x = 15 \). Choice D is incorrect because the student sets the argument equal to the exponent directly: \( 2x + 3 = 3^1 = 3 \), giving \( x = 0 \), or misreads the equation as \( 2x + 3 = 3 \), yielding values near 0.

Question 3. A population of bacteria is modeled by \( P(t) = 500 \cdot 2^{t/4} \), where \( t \) is time in hours. After how many hours will the population equal 8,000?

• A) \( t = 4 \)
• B) \( t = 8 \)
• C) \( t = 16 \) ✓
• D) \( t = 32 \)

Explanation: Choice C is correct. Setting \( P(t) = 8000 \): \( 500 \cdot 2^{t/4} = 8000 \). Dividing by 500: \( 2^{{t/4}} = 16 = 2^4 \). Setting exponents equal: \( \frac{t}{4} = 4 \), so \( t = 16 \). Choice A is incorrect because the student drops the \( \frac{t}{4} \) structure, solving \( 2^t = 16 \) instead, giving \( t = 4 \). Choice B is incorrect because the student correctly finds \( 2^{{t/4}} = 16 \) but then halves the result without justification, getting \( t = 8 \) instead of 16. Choice D is incorrect because the student multiplies 4 by 8 (as if \( t = 4 \times 8 = 32 \)) rather than multiplying both sides of \( \frac{t}{4} = 4 \) by 4.

Question 4. Which value of \( x \) satisfies \( \log_2 x + \log_2(x - 6) = 4 \)?

• A) \( x = 2 \)
• B) \( x = 8 \) ✓
• C) \( x = -2 \)
• D) \( x = 10 \)

Explanation: Choice B is correct. Using the product rule: \( \log_2[x(x-6)] = 4 \). Converting: \( x(x-6) = 2^4 = 16 \). Expanding: \( x^2 - 6x - 16 = 0 \). Factoring: \( (x-8)(x+2) = 0 \), so \( x = 8 \) or \( x = -2 \). Since \( x \) must exceed 6 for both logs to be defined, \( x = 8 \) is the only valid solution. Choice A is incorrect because the value 2 does not satisfy the original equation (domain requires \( x > 6 \)) and likely arises from solving \( x(x-6) = 4 \) (using 4 instead of \( 2^4 = 16 \) on the right side) and making an arithmetic error. Choice C is incorrect because \( x = -2 \) is the rejected solution -- it is algebraically valid but fails the domain requirement since \( \log_2(-2) \) is undefined. Choice D is incorrect because the student applies the quadratic formula incorrectly, using \( b = -6 \) and \( c = -16 \) but computing the discriminant as \( 36 + 16 = 52 \) -- adding instead of subtracting the product term -- and then misreading the result as approximately 10.

Question 5. The concentration of a substance decays according to \( C(t) = C_0 \cdot e^{-kt} \). Measurements show that \( C(2) = 80 \) and \( C(5) = 20 \). Which of the following is the value of the decay constant \( k \)?

• A) \( k = \ln 2 \)
• B) \( k = \dfrac{\ln 4}{5} \)
• C) \( k = \dfrac{\ln 2}{5} \)
• D) \( k = \dfrac{\ln 4}{3} \) ✓

Explanation: Choice D is correct. Divide the two equations to eliminate \( C_0 \): \( \frac{C(2)}{C(5)} = \frac{C_0 e^{-2k}}{C_0 e^{-5k}} = e^{3k} \). So \( e^{3k} = \frac{80}{20} = 4 \). Taking the natural log: \( 3k = \ln 4 \), giving \( k = \frac{\ln 4}{3} \). Choice A is incorrect because the student uses \( t = 2 \) as the denominator instead of the time difference \( 5 - 2 = 3 \): solving \( e^{{2k}} = 4 \) gives \( k = \frac{\ln 4}{2} = \ln 2 \). Choice B is incorrect because the student correctly finds \( e^{{3k}} = 4 \) but then divides by \( t = 5 \) rather than by 3 (the time difference): \( k = \frac{\ln 4}{5} \). Choice C is incorrect because the student takes the square root of the ratio (\( \sqrt{4} = 2 \)), then divides by 5: \( k = \frac{\ln 2}{5} \).