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AP Precalculus Quiz

AP Precalculus Quiz: Exponential And Logarithmic Equations And Inequalities

Practice Exponential And Logarithmic Equations And Inequalities in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 11

0 of 11 answered

Sound level is L=10log⁡10(I/I0)L=10\log_{10}(I/I_0)L=10log10​(I/I0​), with I0=10−12 W/m2I_0=10^{-12}\,\text{W/m}^2I0​=10−12W/m2. If L=70L=70L=70 dB, determine the sound intensity III in W/m2\text{W/m}^2W/m2.

Select an answer to continue

What this quiz covers

This quiz focuses on Exponential And Logarithmic Equations And Inequalities, 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.

All questions

Question 1

Sound level is L=10log⁡10(I/I0)L=10\log_{10}(I/I_0)L=10log10​(I/I0​), with I0=10−12 W/m2I_0=10^{-12}\,\text{W/m}^2I0​=10−12W/m2. If L=70L=70L=70 dB, determine the sound intensity III in W/m2\text{W/m}^2W/m2.

  1. I=10−5 W/m2I=10^{-5}\,\text{W/m}^2I=10−5W/m2 (correct answer)
  2. I=10−19 W/m2I=10^{-19}\,\text{W/m}^2I=10−19W/m2
  3. I=7⋅10−12 W/m2I=7\cdot 10^{-12}\,\text{W/m}^2I=7⋅10−12W/m2
  4. I=10−7 W/m2I=10^{-7}\,\text{W/m}^2I=10−7W/m2

Explanation: This question tests AP Precalculus skills in solving exponential and logarithmic equations and inequalities. Exponential equations involve variables in the exponent, while logarithmic equations involve finding the power to which a number must be raised to obtain another number. Understanding properties of logs and exponentials is crucial. In this problem, the given scenario involves the decibel formula L=10log₁₀(I/I₀), requiring solving for I when L=70 dB and I₀=10^(-12) W/m². Choice A is correct because it accurately applies the inverse relationship: 70=10log₁₀(I/10^(-12)), so 7=log₁₀(I/10^(-12)), giving I/10^(-12)=10^7, thus I=10^(-5) W/m². Choice D is incorrect because it results from misapplying the formula, perhaps confusing the base-10 logarithm with natural logarithm or making an arithmetic error with exponents. To help students: Practice converting between logarithmic and exponential forms. Reinforce that decibels use base-10 logarithms and involve a reference intensity.

Question 2

A bacteria culture is modeled by P(t)=800e0.15tP(t)=800e^{0.15t}P(t)=800e0.15t, where ttt is hours. Determine the population when t=ln⁡(2)0.15t=\frac{\ln(2)}{0.15}t=0.15ln(2)​, and interpret the result.

  1. P=1600P=1600P=1600, the population doubles (correct answer)
  2. P=800ln⁡(2)P=800\ln(2)P=800ln(2), the population doubles
  3. P=1200P=1200P=1200, the population increases by 50%50\%50%
  4. P=400P=400P=400, the population halves

Explanation: This question tests AP Precalculus skills in solving exponential and logarithmic equations and inequalities. Exponential equations involve variables in the exponent, while logarithmic equations involve finding the power to which a number must be raised to obtain another number. Understanding properties of logs and exponentials is crucial. In this problem, the given scenario involves bacterial growth modeled by P(t)=800e^(0.15t), requiring evaluation when t=ln(2)/0.15. Choice A is correct because substituting gives P=800e^(0.15×ln(2)/0.15)=800e^(ln(2))=800×2=1600, and this represents doubling of the initial population. Choice B is incorrect because it results from not evaluating the exponential expression, leaving it in terms of ln(2). To help students: Practice recognizing that e^(ln(a))=a for positive a. Encourage understanding the relationship between exponential growth rate and doubling time.

Question 3

Sound level is L=10log⁡10 ⁣(I10−12)L=10\log_{10}\!\left(\tfrac{I}{10^{-12}}\right)L=10log10​(10−12I​) dB. If two machines differ by 151515 dB, what is the intensity ratio I2I1\tfrac{I_2}{I_1}I1​I2​​?​

  1. I2I1=101.5≈31.6\tfrac{I_2}{I_1}=10^{1.5}\approx 31.6I1​I2​​=101.5≈31.6 (correct answer)
  2. I2I1=1015≈1.0×1015\tfrac{I_2}{I_1}=10^{15}\approx 1.0\times 10^{15}I1​I2​​=1015≈1.0×1015
  3. I2I1=1.5\tfrac{I_2}{I_1}=1.5I1​I2​​=1.5
  4. I2I1=10−1.5≈0.0316\tfrac{I_2}{I_1}=10^{-1.5}\approx 0.0316I1​I2​​=10−1.5≈0.0316

Explanation: This question tests AP Precalculus skills in solving exponential and logarithmic equations and inequalities. Exponential equations involve variables in the exponent, while logarithmic equations involve finding the power to which a number must be raised to obtain another number. Understanding properties of logs and exponentials is crucial. In this problem, the given scenario involves comparing sound intensities using the decibel formula L = 10log₁₀(I/10^(-12)), with a 15 dB difference. Choice A is correct because the difference in decibels relates to the ratio of intensities: L₂ - L₁ = 10log₁₀(I₂/I₁) = 15, so log₁₀(I₂/I₁) = 1.5, therefore I₂/I₁ = 10^1.5 ≈ 31.6. Choice B is incorrect because it misinterprets the relationship, using 10^15 instead of 10^1.5, failing to divide the decibel difference by 10. To help students: Emphasize that decibel differences correspond to logarithms of intensity ratios. Practice working with properties of logarithms, especially log(a/b) = log(a) - log(b).

Question 4

Sound level is L=10log⁡10(I/I0)L=10\log_{10}(I/I_0)L=10log10​(I/I0​) with I0=10−12 W/m2I_0=10^{-12}\,\text{W/m}^2I0​=10−12W/m2. If one sound is 202020 dB louder, what is I2/I1I_2/I_1I2​/I1​?

  1. I2/I1=2I_2/I_1=2I2​/I1​=2
  2. I2/I1=10I_2/I_1=10I2​/I1​=10
  3. I2/I1=100I_2/I_1=100I2​/I1​=100 (correct answer)
  4. I2/I1=0.01I_2/I_1=0.01I2​/I1​=0.01

Explanation: This question tests AP Precalculus skills in solving exponential and logarithmic equations and inequalities. Exponential equations involve variables in the exponent, while logarithmic equations involve finding the power to which a number must be raised to obtain another number. Understanding properties of logs and exponentials is crucial. In this problem, the given scenario involves comparing sound intensities when one is 20 dB louder, using L=10log₁₀(I/I₀). Choice C is correct because a 20 dB difference means L₂-L₁=20=10log₁₀(I₂/I₀)-10log₁₀(I₁/I₀)=10log₁₀(I₂/I₁), so log₁₀(I₂/I₁)=2, giving I₂/I₁=10²=100. Choice B is incorrect because it results from thinking 20 dB means 20 times louder rather than understanding the logarithmic relationship. To help students: Emphasize that decibels are logarithmic, so equal dB differences represent equal ratios of intensities. Practice working with logarithmic scales in various contexts.

Question 5

A radioactive sample is modeled by A(t)=80(12)t/6A(t)=80\left(\frac12\right)^{t/6}A(t)=80(21​)t/6, where ttt is hours. How long will it take until only 202020 milligrams remain?

  1. t=3t=3t=3 hours
  2. t=6t=6t=6 hours
  3. t=12t=12t=12 hours (correct answer)
  4. t=24t=24t=24 hours

Explanation: This question tests AP Precalculus skills in solving exponential and logarithmic equations and inequalities. Exponential equations involve variables in the exponent, while logarithmic equations involve finding the power to which a number must be raised to obtain another number. Understanding properties of logs and exponentials is crucial. In this problem, the given scenario involves radioactive decay modeled by A(t)=80(1/2)^(t/6), requiring solving the equation 80(1/2)^(t/6) = 20. Choice C is correct because it accurately applies logarithms: dividing by 80 gives (1/2)^(t/6) = 1/4, recognizing that 1/4 = (1/2)^2, so t/6 = 2, therefore t = 12 hours. Choice B is incorrect because it results from solving (1/2)^t = 1/4 without accounting for the t/6 in the exponent. To help students: Practice recognizing when bases are related (like 1/4 = (1/2)^2) to avoid using logarithms unnecessarily. Reinforce the concept of half-life and how the exponent t/6 means the half-life is 6 hours.

Question 6

A town’s population follows P(t)=12000(1.04)tP(t)=12000(1.04)^tP(t)=12000(1.04)t, where ttt is years since 2026. How long will it take for the population to reach 180001800018000?

  1. t=5.0t=5.0t=5.0 years
  2. t=10.5t=10.5t=10.5 years (correct answer)
  3. t=4.16t=4.16t=4.16 years
  4. t=20.8t=20.8t=20.8 years

Explanation: This question tests AP Precalculus skills in solving exponential and logarithmic equations and inequalities. Exponential equations involve variables in the exponent, while logarithmic equations involve finding the power to which a number must be raised to obtain another number. Understanding properties of logs and exponentials is crucial. In this problem, the given scenario involves a town's population growth modeled by P(t)=12000(1.04)^t, requiring solving the equation 18000=12000(1.04)^t. Choice B is correct because it accurately applies logarithms to solve: dividing by 12000 gives 1.5=(1.04)^t, then taking log of both sides yields t=log(1.5)/log(1.04)≈10.5 years. Choice C is incorrect because it results from using natural log instead of common log without proper conversion, or from arithmetic errors in the calculation. To help students: Practice isolating the exponential term before applying logarithms. Encourage checking answers by substituting back into the original equation to verify that P(10.5)≈18000.

Question 7

A carbon sample follows A(t)=A0(12)t/5730A(t)=A_0\left(\tfrac12\right)^{t/5730}A(t)=A0​(21​)t/5730 with half-life 573057305730 years. If A(t)/A0=0.25A(t)/A_0=0.25A(t)/A0​=0.25, determine the sample’s age ttt.

  1. t=2865t=2865t=2865 years
  2. t=5730t=5730t=5730 years
  3. t=11460t=11460t=11460 years (correct answer)
  4. t=14325t=14325t=14325 years

Explanation: This question tests AP Precalculus skills in solving exponential and logarithmic equations and inequalities. Exponential equations involve variables in the exponent, while logarithmic equations involve finding the power to which a number must be raised to obtain another number. Understanding properties of logs and exponentials is crucial. In this problem, the given scenario involves carbon dating with A(t)=A₀(1/2)^(t/5730), requiring solving for t when A(t)/A₀=0.25. Choice C is correct because 0.25=1/4=(1/2)², so (1/2)^(t/5730)=(1/2)², giving t/5730=2, thus t=11460 years, which represents two half-lives. Choice B is incorrect because it represents only one half-life, which would give A(t)/A₀=0.5, not 0.25. To help students: Practice recognizing when ratios are powers of 1/2 for quick solutions. Reinforce the concept that each half-life reduces the amount by half, so n half-lives reduce it to (1/2)ⁿ of the original.

Question 8

A radioactive isotope has half-life 888 days and initial mass 505050 mg, modeled by A(t)=50(12)t/8A(t)=50\left(\tfrac12\right)^{t/8}A(t)=50(21​)t/8. Determine the time when A(t)=12.5A(t)=12.5A(t)=12.5 mg.

  1. t=16t=16t=16 days (correct answer)
  2. t=8t=8t=8 days
  3. t=24t=24t=24 days
  4. t=4t=4t=4 days

Explanation: This question tests AP Precalculus skills in solving exponential and logarithmic equations and inequalities. Exponential equations involve variables in the exponent, while logarithmic equations involve finding the power to which a number must be raised to obtain another number. Understanding properties of logs and exponentials is crucial. In this problem, the given scenario involves radioactive decay modeled by A(t)=50(1/2)^(t/8), requiring solving the equation 12.5=50(1/2)^(t/8). Choice A is correct because it accurately applies the half-life concept: 12.5/50=0.25=(1/2)^2, so (1/2)^(t/8)=(1/2)^2, giving t/8=2, thus t=16 days. Choice B is incorrect because it results from confusing the half-life period with the answer, thinking one half-life gives the desired amount. To help students: Emphasize that half-life means the time for half the substance to remain. Encourage recognizing when amounts are powers of 1/2 of the initial value for simpler calculations.

Question 9

A medicine decays as A(t)=200e−0.08tA(t)=200e^{-0.08t}A(t)=200e−0.08t, where ttt is hours after dosing. How long will it take for the amount to drop below 606060 mg?​

  1. t>9.6t>9.6t>9.6 hours
  2. t>15.1t>15.1t>15.1 hours (correct answer)
  3. t>2.4t>2.4t>2.4 hours
  4. t>42.0t>42.0t>42.0 hours

Explanation: This question tests AP Precalculus skills in solving exponential and logarithmic equations and inequalities. Exponential equations involve variables in the exponent, while logarithmic equations involve finding the power to which a number must be raised to obtain another number. Understanding properties of logs and exponentials is crucial. In this problem, the given scenario involves medicine decay modeled by A(t)=200e^(-0.08t), requiring solving the inequality 200e^(-0.08t)<60. Choice B is correct because it accurately applies logarithms: dividing by 200 gives e^(-0.08t)<0.3, taking ln yields -0.08t<ln(0.3)≈-1.204, so t>15.05 hours. Choice A is incorrect because it results from solving for when the amount equals 60 rather than drops below 60, or from calculation errors. To help students: Practice solving exponential inequalities and remembering to flip inequality signs when dividing by negative numbers. Encourage verification by checking boundary values.

Question 10

Sound level is L=10log⁡10 ⁣(I10−12)L=10\log_{10}\!\left(\tfrac{I}{10^{-12}}\right)L=10log10​(10−12I​) dB. If L=70L=70L=70 dB, determine the intensity III in W/m2\text{W}/\text{m}^2W/m2.​

  1. I=10−5 W/m2I=10^{-5}\,\text{W}/\text{m}^2I=10−5W/m2 (correct answer)
  2. I=7×10−12 W/m2I=7\times 10^{-12}\,\text{W}/\text{m}^2I=7×10−12W/m2
  3. I=107 W/m2I=10^{7}\,\text{W}/\text{m}^2I=107W/m2
  4. I=10−19 W/m2I=10^{-19}\,\text{W}/\text{m}^2I=10−19W/m2

Explanation: This question tests AP Precalculus skills in solving exponential and logarithmic equations and inequalities. Exponential equations involve variables in the exponent, while logarithmic equations involve finding the power to which a number must be raised to obtain another number. Understanding properties of logs and exponentials is crucial. In this problem, the given scenario involves the decibel formula L = 10log₁₀(I/10^(-12)), requiring solving for I when L = 70 dB. Choice A is correct because it accurately applies the inverse relationship: 70 = 10log₁₀(I/10^(-12)), so 7 = log₁₀(I/10^(-12)), which means 10^7 = I/10^(-12), therefore I = 10^7 × 10^(-12) = 10^(-5) W/m². Choice B is incorrect because it results from misapplying the logarithm, treating 70 as 7 × 10^(-12) instead of working with the logarithmic equation. To help students: Practice converting between logarithmic and exponential forms. Reinforce the meaning of decibels as a logarithmic scale and the reference intensity of 10^(-12) W/m².

Question 11

A radioactive sample has half-life 666 hours and initial mass 808080 mg, modeled by M(t)=80(12)t/6M(t)=80\left(\tfrac12\right)^{t/6}M(t)=80(21​)t/6. Determine ttt when M(t)=10M(t)=10M(t)=10 mg.​

  1. t=18 hourst=18\text{ hours}t=18 hours (correct answer)
  2. t=3 hourst=3\text{ hours}t=3 hours
  3. t=36 hourst=36\text{ hours}t=36 hours
  4. t=−18 hourst=-18\text{ hours}t=−18 hours

Explanation: This question tests AP Precalculus skills in solving exponential and logarithmic equations and inequalities. Exponential equations involve variables in the exponent, while logarithmic equations involve finding the power to which a number must be raised to obtain another number. Understanding properties of logs and exponentials is crucial. In this problem, the given scenario involves radioactive decay following M(t) = 80(1/2)^(t/6), requiring solving the equation 80(1/2)^(t/6) = 10. Choice A is correct because it accurately applies the properties of exponentials: dividing both sides by 80 gives (1/2)^(t/6) = 1/8, recognizing that 1/8 = (1/2)³, so t/6 = 3, therefore t = 18 hours. Choice B is incorrect because it results from solving t/6 = 1/2 instead of recognizing that 1/8 = (1/2)³. To help students: Practice recognizing powers of common bases like 1/2. Encourage using the half-life concept to check answers - after 3 half-lives (18 hours), the mass should be 80 × (1/2)³ = 10 mg.