AP Precalculus Quiz: Logarithmic Functions

What this quiz covers

This quiz focuses on Logarithmic Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $t=6\log_{2}\!\left(\tfrac{200}{30}\right)$
2. $t=6\log_{10}\!\left(\tfrac{200}{30}\right)$
3. $t=6\log_{2}\!\left(\tfrac{30}{200}\right)$
4. $t=\dfrac{\log\!\left(\tfrac{30}{200}\right)}{\log\!\left(\tfrac12\right)}$ (correct answer)

Explanation: This question tests AP Precalculus skills: understanding and applying logarithmic functions to solve exponential decay problems. The decay formula A(t) = 200(1/2)^(t/6) represents exponential decay where the substance halves every 6 years. To find when A(t) = 30, we set up the equation 30 = 200(1/2)^(t/6) and solve for t using logarithms. Choice D is correct because dividing both sides by 200 gives 30/200 = (1/2)^(t/6), then taking logarithms of both sides yields log(30/200) = (t/6)log(1/2), which rearranges to t = 6·log(30/200)/log(1/2). Choice A is incorrect because it uses log base 2 instead of the change of base formula, and has 200/30 instead of 30/200. To help students: Practice converting between exponential and logarithmic forms, and emphasize that when solving a^x = b, we get x = log(b)/log(a) using any consistent base.

Question 2

1. $t=8\dfrac{\log(0.20)}{\log(1/2)}$ (correct answer)
2. $t=8\dfrac{\log(1/2)}{\log(0.20)}$
3. $t=8\log\!\left(\dfrac{1/2}{0.20}\right)$
4. $t=8\dfrac{\log(0.20)}{\log 2}$

Explanation: This question tests AP Precalculus skills: using logarithms to solve half-life problems in radioactive decay. The decay formula A(t) = A₀(1/2)^(t/8) shows the amount halves every 8 days. To find when A(t) = 0.20A₀, we solve 0.20A₀ = A₀(1/2)^(t/8). Choice A is correct because dividing by A₀ gives 0.20 = (1/2)^(t/8), taking logarithms yields log(0.20) = (t/8)·log(1/2), and solving for t gives t = 8·log(0.20)/log(1/2). Choice D incorrectly uses log(2) in the denominator instead of log(1/2), which would give the negative of the correct answer. To help students: Remember that log(1/2) = -log(2), so both forms can work if signs are handled correctly. Practice recognizing the half-life formula structure with the time constant in the exponent.

Question 3

1. $t=\dfrac{\log(200{,}000/150{,}000)}{\log(1.03)}$ (correct answer)
2. $t=\dfrac{\log(1.03)}{\log(200{,}000/150{,}000)}$
3. $t=\dfrac{\log(200{,}000/150{,}000)}{\log(0.03)}$
4. $t=\log\!\left(\dfrac{200{,}000}{150{,}000}\right)+\log(1.03)$

Explanation: This question tests AP Precalculus skills: applying logarithms to solve exponential growth equations in demographic modeling. The population model P(t) = 150,000(1.03)^t represents 3% annual growth from 150,000. To find when P(t) = 200,000, we solve 200,000 = 150,000(1.03)^t. Choice A is correct because dividing by 150,000 gives 200,000/150,000 = (1.03)^t, then taking logarithms yields log(200,000/150,000) = t·log(1.03), so t = log(200,000/150,000)/log(1.03). Choice C incorrectly uses log(0.03) instead of log(1.03), confusing the growth rate 0.03 with the growth factor 1.03. To help students: Emphasize that for r% growth, we use the base (1 + r/100). Practice setting up ratios of final value to initial value before applying logarithms.

Question 4

1. $t=\dfrac{\log(2)}{\log(1.03)}$ (correct answer)
2. $t=\dfrac{\log(1.03)}{\log(2)}$
3. $t=\dfrac{\log(240{,}000)}{\log(120{,}000)}$
4. $t=\log\!\left(\dfrac{2}{1.03}\right)$

Explanation: This question tests AP Precalculus skills: understanding and applying logarithmic functions to solve exponential growth problems. The population model P(t) = 120,000(1.03)^t represents exponential growth with a 3% annual increase. To find when the population doubles to 240,000, we set up 240,000 = 120,000(1.03)^t and solve for t. Dividing both sides by 120,000 gives 2 = (1.03)^t, then taking logarithms of both sides yields log(2) = t·log(1.03), so t = log(2)/log(1.03). Choice A is correct because it properly applies the power rule of logarithms and isolates t. Choice B incorrectly inverts the fraction, while Choice C attempts to use the ratio of populations directly without proper logarithmic manipulation. To help students: Emphasize the power rule log(a^x) = x·log(a) and practice solving exponential equations step by step.

Question 5

1. $t\ge \dfrac{\log 2}{\log(1.06)}$ (correct answer)
2. $t\le \dfrac{\log 2}{\log(1.06)}$
3. $t\ge \dfrac{\log(1.06)}{\log 2}$
4. $t\ge \dfrac{\log 2}{\log(0.94)}$

Explanation: This question tests AP Precalculus skills: solving exponential inequalities for investment growth using logarithms. The problem requires finding when an investment doubles, which is a common financial application. We need to solve 3000(1.06)^t ≥ 6000, which simplifies to (1.06)^t ≥ 2. Choice A is correct because taking logarithms gives t·log(1.06) ≥ log(2), and since log(1.06) > 0, dividing preserves the inequality: t ≥ log(2)/log(1.06). Choice B has the wrong inequality direction, while Choice C inverts the fraction incorrectly. To help students: Emphasize that when the base is greater than 1, logarithms preserve inequality direction. Practice identifying doubling scenarios in exponential growth problems.

Question 6

1. $t=5730\log_2\!\left(\tfrac{1}{0.30}\right)$ years (correct answer)
2. $t=5730\log_{10}(0.30)$ years
3. $t=5730\log_2(0.30)$ years
4. $t=5730\left(\tfrac{1}{0.30}\right)$ years

Explanation: This question tests AP Precalculus skills: understanding and applying logarithmic functions to radioactive decay problems with half-life. The carbon-14 decay formula C(t) = C₀(1/2)^(t/5730) shows that the sample halves every 5730 years. To find when C(t) = 0.30C₀, we set up 0.30C₀ = C₀(1/2)^(t/5730) and solve for t. Dividing by C₀ gives 0.30 = (1/2)^(t/5730), then taking log base 2 of both sides yields log₂(0.30) = t/5730, so t = 5730·log₂(0.30) = 5730·log₂(1/0.30) years. Choice A is correct because log₂(0.30) = -log₂(1/0.30), and the negative sign makes the time positive since we're looking at decay. Choice C incorrectly keeps log₂(0.30) which would give a negative time, while Choice B uses the wrong logarithm base. To help students: Practice with half-life problems and emphasize that log₂(a) = -log₂(1/a) for positive a.

Question 7

1. $t=\dfrac{\log(5000/800)}{\log(1.25)}$ (correct answer)
2. $t=\dfrac{\log(1.25)}{\log(5000/800)}$
3. $t=\dfrac{\log(5000-800)}{\log(1.25)}$
4. $t=\dfrac{\log(5000/800)}{\log(0.25)}$

Explanation: This question tests AP Precalculus skills: solving exponential growth equations using logarithms in biological contexts. Logarithms transform the exponential equation into a form where we can isolate the time variable. We need to solve 800(1.25)^t = 5000, which simplifies to (1.25)^t = 5000/800 = 6.25. Choice A is correct because taking logarithms gives t·log(1.25) = log(6.25), so t = log(5000/800)/log(1.25). Choice B incorrectly inverts the fraction, while Choice C uses subtraction instead of division for the bacteria counts. To help students: Emphasize the importance of simplifying the exponential equation first. Practice recognizing that bacterial growth problems typically involve finding ratios, not differences.

Question 8

1. $t=\dfrac{\ln(4)}{0.18}$ (correct answer)
2. $t=\dfrac{\ln(1200)}{\ln(300)}$
3. $t=\dfrac{0.18}{\ln(4)}$
4. $t=\dfrac{\ln(0.18)}{\ln(4)}$

Explanation: This question tests AP Precalculus skills: understanding and applying natural logarithms to solve exponential growth problems with base e. The culture growth formula N(t) = 300e^(0.18t) represents continuous exponential growth. To find when N(t) = 1200, we set up 1200 = 300e^(0.18t) and solve for t. Dividing by 300 gives 4 = e^(0.18t), then taking natural logarithm of both sides yields ln(4) = 0.18t, so t = ln(4)/0.18. Choice A is correct because it properly applies the natural logarithm to isolate t from the exponential expression. Choice C incorrectly inverts the fraction, while Choice B attempts to use logarithms of the population values directly without proper manipulation. To help students: Emphasize that when dealing with base e, using natural logarithm (ln) simplifies calculations since ln(e^x) = x.

Question 9

1. $10^7$ (correct answer)
2. $7\cdot 10$
3. $10^{70}$
4. $10^{-7}$

Explanation: This question tests AP Precalculus skills: converting between logarithmic and exponential forms in real-world applications. The decibel scale uses logarithms to measure sound intensity on a more manageable scale. Given L = 10log₁₀(I/I₀) = 70, we need to solve for I/I₀ by reversing the logarithm. Choice A is correct because dividing by 10 gives log₁₀(I/I₀) = 7, which means I/I₀ = 10^7 by the definition of logarithm. Choice B incorrectly multiplies 7 by 10, while Choice C raises 10 to the 70th power instead of the 7th. To help students: Practice converting between log₁₀(x) = y and x = 10^y forms. Emphasize that in the decibel formula, we must first isolate the logarithm by dividing by 10.

Question 10

1. $t=\dfrac{\log 2}{\log(1.04)}$ (correct answer)
2. $t=\dfrac{\log(1.04)}{\log 2}$
3. $t=\dfrac{\log(160{,}000)-\log(80{,}000)}{\log(0.04)}$
4. $t=\log\!\big(2\cdot 1.04\big)$

Explanation: This question tests AP Precalculus skills: using logarithms to solve exponential growth equations in population modeling. The population model P(t) = 80,000(1.04)^t represents 4% annual growth from an initial population of 80,000. To find when the population doubles to 160,000, we solve 160,000 = 80,000(1.04)^t. Choice A is correct because dividing by 80,000 gives 2 = (1.04)^t, and taking logarithms yields log(2) = t·log(1.04), so t = log(2)/log(1.04). Choice B incorrectly inverts the fraction, which would give the reciprocal of the correct answer. To help students: Emphasize the pattern that when solving b^x = a, we get x = log(a)/log(b). Practice identifying what goes in the numerator (the result) versus denominator (the base).

Question 11

1. $t>\dfrac{\ln(10/50)}{-0.3}$ (correct answer)
2. $t<\dfrac{\ln(10/50)}{-0.3}$
3. $t>\dfrac{\ln(50/10)}{-0.3}$
4. $t>\dfrac{\log(10/50)}{0.3}$

Explanation: This question tests AP Precalculus skills: using natural logarithms to solve exponential decay inequalities. The decay formula M(t) = 50e^(-0.3t) uses the natural exponential base e with decay constant -0.3. To find when M(t) < 10, we solve 50e^(-0.3t) < 10. Choice A is correct because dividing by 50 gives e^(-0.3t) < 10/50, taking natural logarithms yields -0.3t < ln(10/50), and dividing by -0.3 (flipping the inequality) gives t > ln(10/50)/(-0.3). Choice C incorrectly uses ln(50/10) instead of ln(10/50), missing that we need the ratio of final to initial mass. To help students: Remember that dividing by a negative number reverses the inequality direction. Practice recognizing when to use natural logarithms (ln) versus common logarithms (log).

Question 12

1. $t\ge \dfrac{\log(1.5)}{\log(1.05)}$ (correct answer)
2. $t\ge \dfrac{\log(1.05)}{\log(1.5)}$
3. $t\le \dfrac{\log(1.5)}{\log(1.05)}$
4. $t\ge \dfrac{\log(1.5)}{\log(0.05)}$

Explanation: This question tests AP Precalculus skills: solving exponential inequalities to find minimum time periods. The account formula A(t) = 2000(1.05)^t models 5% annual compound interest from $2000. To find when A(t) ≥ 3000, we solve 2000(1.05)^t ≥ 3000. Choice A is correct because dividing by 2000 gives (1.05)^t ≥ 3000/2000 = 1.5, taking logarithms yields t·log(1.05) ≥ log(1.5), so t ≥ log(1.5)/log(1.05). Choice D incorrectly uses log(0.05) instead of log(1.05), confusing the interest rate with the growth factor. To help students: Remember that inequalities maintain direction when taking logarithms of positive numbers. Practice finding the smallest integer satisfying an inequality by calculating the exact value first.

Question 13

1. $t=6\log_2\!\left(\tfrac{50}{8}\right)$ hours (correct answer)
2. $t=6\log_{10}\!\left(\tfrac{50}{8}\right)$ hours
3. $t=6\log_2\!\left(\tfrac{8}{50}\right)$ hours
4. $t=6\left(\tfrac{50}{8}\right)$ hours

Explanation: This question tests AP Precalculus skills: understanding and applying logarithmic functions to solve exponential decay problems. The radioactive decay formula A(t) = 50(1/2)^(t/6) represents exponential decay where the sample halves every 6 hours. To find when A(t) = 8, we set up the equation 8 = 50(1/2)^(t/6) and solve for t using logarithms. Dividing both sides by 50 gives 8/50 = (1/2)^(t/6), then taking log base 2 of both sides yields log₂(8/50) = t/6, so t = 6log₂(8/50) = 6log₂(50/8) hours. Choice A is correct because it properly applies logarithm properties, noting that log₂(8/50) = -log₂(50/8). Choice B incorrectly uses log base 10 instead of log base 2, which doesn't match the base in the exponential function. To help students: Practice converting between exponential and logarithmic forms, and emphasize matching the logarithm base to the exponential base.

Question 14

1. $t=\dfrac{\log 2}{\log(1.03)}$ (correct answer)
2. $t=\dfrac{\log(1.03)}{\log 2}$
3. $t=\dfrac{\log 2}{\log(0.97)}$
4. $t=\log\big(2\cdot 1.03\big)$

Explanation: This question tests AP Precalculus skills: using logarithms to solve exponential growth problems involving population doubling. Logarithms transform exponential equations into linear ones, making it possible to isolate the variable in the exponent. The population doubles when P(t) = 240,000, so we need to solve 120,000(1.03)^t = 240,000, which simplifies to (1.03)^t = 2. Choice A is correct because taking logarithms gives t·log(1.03) = log(2), so t = log(2)/log(1.03). Choice B incorrectly inverts the fraction, while Choice C uses the wrong base (0.97 instead of 1.03). To help students: Emphasize setting up the equation correctly before applying logarithms. Practice recognizing when to use growth factor (1.03) versus decay factor in population models.

Question 15

1. $t=\dfrac{\log(0.10)}{\log(0.93)}$ (correct answer)
2. $t=\dfrac{\log(0.93)}{\log(0.10)}$
3. $t=\dfrac{\log(0.10)}{\log(1.07)}$
4. $t=\log\big(0.10\cdot 0.93\big)$

Explanation: This question tests AP Precalculus skills: solving exponential decay equations using logarithms in scientific contexts. Radioactive tracers follow predictable exponential decay patterns that can be analyzed with logarithms. We need to solve A₀(0.93)^t = 0.10A₀, which simplifies to (0.93)^t = 0.10. Choice A is correct because taking logarithms gives t·log(0.93) = log(0.10), so t = log(0.10)/log(0.93). Choice B incorrectly inverts the fraction, while Choice C uses 1.07 (which would represent growth) instead of the decay factor 0.93. To help students: Emphasize recognizing decay factors (less than 1) versus growth factors (greater than 1). Practice setting up equations by canceling common factors like A₀ before applying logarithms.

Question 16

1. $t\ge \dfrac{\log(20/60)}{\log(0.85)}$ (correct answer)
2. $t\le \dfrac{\log(20/60)}{\log(0.85)}$
3. $t\ge \dfrac{\log(60/20)}{\log(0.85)}$
4. $t\ge \dfrac{\log(20-60)}{\log(0.85)}$

Explanation: This question tests AP Precalculus skills: solving exponential inequalities using logarithms in pharmaceutical contexts. When dealing with decay (base < 1), the inequality direction reverses when we divide by the negative logarithm. We need to solve 60(0.85)^t ≤ 20, which gives (0.85)^t ≤ 1/3. Choice A is correct because taking logarithms and dividing by log(0.85) (which is negative) reverses the inequality: t ≥ log(20/60)/log(0.85) = log(1/3)/log(0.85). Choice B has the wrong inequality direction, while Choice C incorrectly uses 60/20 instead of 20/60. To help students: Emphasize that log(0.85) < 0, so dividing by it reverses inequalities. Practice recognizing when bases are less than 1 in decay problems.

Question 17

1. $t=\dfrac{\log(40/90)}{\log(0.85)}$ (correct answer)
2. $t=\dfrac{\log(90/40)}{\log(0.85)}$
3. $t=\dfrac{\log(40/90)}{\log(1.15)}$
4. $t=\log\!\left(\dfrac{40}{90}\right)\log(0.85)$

Explanation: This question tests AP Precalculus skills: applying logarithms to solve exponential decay problems in pharmacology. The medication formula Q(t) = 90(0.85)^t shows 15% hourly decay from 90 mg. To find when Q(t) = 40, we solve 40 = 90(0.85)^t. Choice A is correct because dividing by 90 gives 40/90 = (0.85)^t, taking logarithms yields log(40/90) = t·log(0.85), so t = log(40/90)/log(0.85). Choice B incorrectly uses log(90/40) instead of log(40/90), inverting the ratio of final to initial amount. To help students: Always set up the ratio as (final value)/(initial value) on the left side. Remember that decay factors are less than 1, making their logarithms negative.

Question 18

1. $t>\dfrac{\log(0.25)}{\log(0.85)}$ (correct answer)
2. $t<\dfrac{\log(0.25)}{\log(0.85)}$
3. $t>\dfrac{\log(4)}{\log(0.85)}$
4. $t>\dfrac{\log(0.85)}{\log(0.25)}$

Explanation: This question tests AP Precalculus skills: understanding and applying logarithmic functions to solve exponential decay inequalities. The medication decay formula M(t) = 200(0.85)^t shows that 85% of the medication remains each hour. To find when M(t) < 50, we solve the inequality 200(0.85)^t < 50. Dividing by 200 gives (0.85)^t < 0.25, and taking logarithms yields t·log(0.85) < log(0.25). Since log(0.85) is negative (because 0.85 < 1), dividing by log(0.85) reverses the inequality sign, giving t > log(0.25)/log(0.85). Choice A is correct because it properly handles the inequality reversal when dividing by a negative logarithm. Choice B incorrectly maintains the less-than sign, while Choice C uses log(4) instead of log(0.25), though they are equivalent since 0.25 = 1/4. To help students: Emphasize that logarithms of numbers between 0 and 1 are negative, and practice solving exponential inequalities with careful attention to sign changes.

Question 19

1. $t=\dfrac{\log(8000/5000)}{\log(1.06)}$ (correct answer)
2. $t=\dfrac{\log(1.06)}{\log(8000/5000)}$
3. $t=\log\!\left(\dfrac{8000}{5000\cdot 1.06}\right)$
4. $t=\dfrac{\log(8000)-\log(5000)}{1.06}$

Explanation: This question tests AP Precalculus skills: understanding and applying logarithmic functions to solve compound interest problems. The account growth formula A(t) = 5000(1.06)^t represents 6% annual compound interest. To find when the account reaches $8000, we set up 8000 = 5000(1.06)^t and solve for t using logarithms. Dividing both sides by 5000 gives 8000/5000 = (1.06)^t, which simplifies to 1.6 = (1.06)^t. Taking logarithms of both sides yields log(1.6) = t·log(1.06), so t = log(1.6)/log(1.06) = log(8000/5000)/log(1.06). Choice A is correct because it properly applies logarithm properties and the change of base formula. Choice B incorrectly inverts the fraction, while Choice D incorrectly attempts to divide by 1.06 instead of log(1.06). To help students: Practice using logarithm properties like log(a/b) = log(a) - log(b) and emphasize the importance of applying logarithms to both sides of an equation.

Question 20

1. $t>\dfrac{\log(1.25)}{\log(1.025)}$ (correct answer)
2. $t<\dfrac{\log(1.25)}{\log(1.025)}$
3. $t>\dfrac{\log(1.025)}{\log(1.25)}$
4. $t>\log\!\left(\dfrac{1.25}{1.025}\right)$

Explanation: This question tests AP Precalculus skills: understanding and applying logarithmic functions to solve exponential growth inequalities. The population model P(t) = 80,000(1.025)^t represents 2.5% annual growth. To find when P(t) > 100,000, we solve the inequality 80,000(1.025)^t > 100,000. Dividing by 80,000 gives (1.025)^t > 1.25, and taking logarithms yields t·log(1.025) > log(1.25). Since log(1.025) is positive (because 1.025 > 1), dividing by log(1.025) maintains the inequality direction, giving t > log(1.25)/log(1.025). Choice A is correct because it properly applies logarithms and maintains the correct inequality sign. Choice B incorrectly reverses the inequality, while Choice C inverts the fraction. To help students: Practice solving exponential inequalities and emphasize that the sign of log(a) depends on whether a is greater than or less than 1.