Exponential and Logarithmic Equations and Inequalities - AP Precalculus

1. Sound level is $L=10\log_{10}\!\left(\tfrac{I}{10^{-12}}\right)$ dB. If $L=70$ dB, determine the intensity $I$ in $\text{W}/\text{m}^2$.​
   Answer: $I=10^{-5}\,\text{W}/\text{m}^2$ 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.
2. A radioactive isotope has half-life $8$ days and initial mass $50$ mg, modeled by $A(t)=50\left(\tfrac12\right)^{t/8}$. Determine the time when $A(t)=12.5$ mg.
   Answer: $t=16$ days 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.
3. A radioactive sample has half-life $6$ hours and initial mass $80$ mg, modeled by $M(t)=80\left(\tfrac12\right)^{t/6}$. Determine $t$ when $M(t)=10$ mg.​
   Answer: $t=18\text{ hours}$ 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.
4. A town's population follows $P(t)=12000(1.04)^t$, where $t$ is years since 2026. How long will it take for the population to reach $18000$?
   Answer: $t=10.5$ years 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.
5. Sound level is $L=10\log_{10}(I/I_0)$ with $I_0=10^{-12}\,\text{W/m}^2$. If one sound is $20$ dB louder, what is $I_2/I_1$?
   Answer: $I_2/I_1=100$ 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.
6. A bacteria culture is modeled by $P(t)=800e^{0.15t}$, where $t$ is hours. Determine the population when $t=\frac{\ln(2)}{0.15}$, and interpret the result.
   Answer: $P=1600$, the population doubles 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.
7. A radioactive sample is modeled by $A(t)=80\left(\frac12\right)^{t/6}$, where $t$ is hours. How long will it take until only $20$ milligrams remain?
   Answer: $t=12$ hours 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.
8. Sound level is $L=10\log_{10}\!\left(\tfrac{I}{10^{-12}}\right)$ dB. If two machines differ by $15$ dB, what is the intensity ratio $\tfrac{I_2}{I_1}$?​
   Answer: $\tfrac{I_2}{I_1}=10^{1.5}\approx 31.6$ 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).
9. A carbon sample follows $A(t)=A_0\left(\tfrac12\right)^{t/5730}$ with half-life $5730$ years. If $A(t)/A_0=0.25$, determine the sample's age $t$.
   Answer: $t=11460$ years 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.
10. A medicine decays as $A(t)=200e^{-0.08t}$, where $t$ is hours after dosing. How long will it take for the amount to drop below $60$ mg?​
    Answer: $t>15.1$ hours 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.
11. Sound level is $L=10\log_{10}(I/I_0)$, with $I_0=10^{-12}\,\text{W/m}^2$. If $L=70$ dB, determine the sound intensity $I$ in $\text{W/m}^2$.
    Answer: $I=10^{-5}\,\text{W/m}^2$ 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.

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