Modeling with Exponential/Logarithmic Equations - Algebra 2

An investment grows according to $A(t) = 5000(1.06)^t$ dollars after $t$ years. How long will it take for the investment to triple in value?
Answer: $\frac{\log(3)}{\log(1.06)}$ years To triple: $5000(1.06)^t = 15000$ , so $(1.06)^t = 3$ . Taking log: $t\log(1.06) = \log(3)$ , therefore $t = \frac{\log(3)}{\log(1.06)}$ . Choice B incorrectly assumes continuous compounding formula. Choice C uses linear rather than exponential reasoning. Choice D fails to simplify after setting up the equation correctly.
A town has an initial population of 12,000 and grows by 3% each year. Create an exponential model for the population after $t$ years, then determine how many years it will take for the population to reach 15,000. Round to the nearest tenth of a year and interpret your result in context.
Answer: Model: $P(t)=12000(1.03)^t$ . It reaches 15,000 after $t\approx 7.4$ years. This question tests your ability to create exponential models from real-world contexts (like compound interest, population growth, radioactive decay, or logarithmic scales) and solve them to answer practical questions. Exponential models arise when quantities change by a constant percent per time period: if something grows r percent per period starting from initial value a, the model is y = $a(1 + r)^t$ for growth. For this population problem: (1) Initial population P₀ = 12,000, growth rate r = 0.03 (3% as decimal). (2) Model: P(t) = 12000 $(1.03)^t$ . (3) To find when P = 15,000, solve: 15,000 = 12,000 $(1.03)^t$ . (4) Divide by 12,000: 1.25 = $(1.03)^t$ . (5) Take log: log(1.25) = t·log(1.03). (6) Solve: t = log(1.25)/log(1.03) ≈ 7.4 years. Choice A correctly creates the growth model P(t) = 12000 $(1.03)^t$ and accurately calculates that the population reaches 15,000 after approximately 7.4 years. Choice B incorrectly uses $(0.97)^t$ which would represent 3% decay (population decreasing), not 3% growth. The exponential growth recipe: (1) Identify initial value and growth rate as decimal, (2) Write model: y = $a(1 + r)^t$ , (3) Set equal to target and solve using logarithms. Context language is key: "grows by 3%" means multiply by 1.03, while "decreases by 3%" would mean multiply by 0.97!
A bacteria culture starts with 200 bacteria and doubles every 3 hours. Create an exponential model for the population $P(t)$ after $t$ hours, then solve for when the population first reaches 10,000 bacteria. Round to the nearest tenth of an hour and interpret your result.
Answer: Model: $P(t)=200\cdot 2^{t/3}$ . $t\approx 16.9$ hours; the culture reaches 10,000 bacteria after about 16.9 hours. This question tests your ability to create exponential or logarithmic models from real-world contexts like compound interest, population growth, radioactive decay, or logarithmic scales and solve them to answer practical questions. Exponential models arise when quantities change by a constant percent per time period: if something grows r percent per period starting from initial value a, the model is y = $a(1 + r)^t$ for growth or y = $a(1 - r)^t$ for decay; for doubling times, it's often y = a * $2^{t/d}$ where d is the doubling period. For this bacteria culture starting at 200 and doubling every 3 hours, the model is P(t) = 200 * $2^{t/3}$ ; to reach 10,000, set 10,000 = 200 * $2^{t/3}$ , simplify to $2^{t/3}$ = 50, take log base 2: t/3 = log2(50) ≈ 5.64, so t ≈ 16.9 hours—excellent work on modeling growth! Choice B correctly creates the exponential model with the proper exponent for doubling time, solves using logarithms, and interprets with units—keep up the precision! A tempting distractor like Choice A uses $2^{3t}$ which would triple the exponent incorrectly, leading to rapid growth that's too fast—remember to set the exponent as t divided by the period! For transferable strategy, identify if it's doubling (base 2) or percent growth (1 + r), write the model accordingly, and solve by isolating the base^t = k, then t = log(k)/log(base)—you're getting great at this! Practice with tripling or other multiples to reinforce the concept.
A town's population is 48,000 and decreases by $1.6\%$ each year. Create an exponential model for the population $P(t)$ after $t$ years, then solve for when the population will fall to 40,000. Round to the nearest tenth of a year and interpret your result.
Answer: Model: $P(t)=48000(0.984)^t$ . It falls to 40,000 after about $t\approx 11.4$ years. This question tests your ability to create exponential decay models for population decline and solve for when the population reaches a specific value. A 1.6% annual decrease means 98.4% of the population remains each year, giving decay factor 0.984: starting from 48,000, the model is P(t) = 48000 $(0.984)^t$ . To find when P = 40,000: set 40000 = 48000 $(0.984)^t$ , divide by 48000 to get 0.833 = $(0.984)^t$ , take log: log(0.833) = t·log(0.984), solve: t = log(0.833)/log(0.984) ≈ 11.4 years. Choice A correctly models the decay with base 0.984 (keeping 98.4% when losing 1.6%) and interprets that the population falls to 40,000 after about 11.4 years. Choice B incorrectly uses 1.016 as the base, which would represent 1.6% growth not decay—if population decreases by 1.6%, multiply by 0.984 not 1.016! Remember: r% decrease means multiply by (1 - r/100), while r% increase means multiply by (1 + r/100).
An account is opened with $2500 and earns $5\%$ interest compounded annually. You do not add or withdraw money. Which equation models the balance after $t$ years, and about how long will it take for the account to reach $4000? Round time to the nearest tenth of a year.
Answer: Model: $A(t)=2500(1.05)^t$ . It reaches $4000 after about $t\approx 9.6$ years. This question tests your ability to create exponential models from real-world contexts (like compound interest, population growth, radioactive decay, or logarithmic scales) and solve them to answer practical questions. Exponential models arise when quantities change by a constant percent per time period: if something grows r percent per period starting from initial value a, the model is y = $a(1 + r)^t$ for growth or y = $a(1 - r)^t$ for decay. For compound interest specifically: A = $P(1 + r)^t$ where P is principal, r is annual rate (as decimal), t is years, A is final amount. For modeling $2500 at 5% annual interest, find when it reaches $4000: (1) Identify model type: compound interest (exponential growth). (2) Write equation: A = 2500 $(1.05)^t$ . (3) Set equal to target: 4000 = 2500 $(1.05)^t$ . (4) Isolate exponential: $(1.05)^t$ = 1.6. (5) Take logarithm: log( $(1.05)^t$ ) = log(1.6). (6) Use power property: t·log(1.05) = log(1.6). (7) Solve: t = log(1.6)/log(1.05) ≈ 9.6 years. Choice A correctly creates the exponential model A(t) = 2500 $(1.05)^t$ for 5% growth and solves to get t ≈ 9.6 years when the account reaches $4000. Choice B uses 0.95 which would represent 5% decay, not growth—compound interest means the account grows! Choice C uses linear growth A = 2500 + 0.05t, but compound interest multiplies by (1.05) each year, not adds 0.05. Exponential modeling recipe: identify initial value (2500), growth rate as decimal (0.05), create growth factor (1 + 0.05 = 1.05), write model A = 2500 $(1.05)^t$ , then solve exponentially!
A medication dose starts at 100 mg in the bloodstream. Each hour, 15% of the medication is eliminated. Create an exponential decay model for the amount remaining after $t$ hours, then solve for when the amount drops to 25 mg. Round to the nearest tenth of an hour and interpret your result in context.
Answer: Model: $M(t)=100(0.85)^t$ . It drops to 25 mg after $t\approx 8.6$ hours. This question tests your ability to create exponential models from real-world contexts (like compound interest, population growth, radioactive decay, or logarithmic scales) and solve them to answer practical questions. Exponential models arise when quantities change by a constant percent per time period: for decay, if something decreases by r percent per period starting from initial value a, the model is y = $a(1 - r)^t$ . For this medication problem: (1) Initial amount M₀ = 100 mg, decay rate r = 0.15 (15% eliminated). (2) Remaining factor: 1 - 0.15 = 0.85 (85% remains each hour). (3) Model: M(t) = 100 $(0.85)^t$ . (4) To find when M = 25, solve: 25 = 100 $(0.85)^t$ . (5) Divide by 100: 0.25 = $(0.85)^t$ . (6) Take log: log(0.25) = t·log(0.85). (7) Solve: t = log(0.25)/log(0.85) ≈ 8.6 hours. Choice B correctly creates the decay model M(t) = 100 $(0.85)^t$ and accurately calculates that the medication drops to 25 mg after approximately 8.6 hours. Choice A incorrectly uses $0.15^t$ which would mean only 15% remains each hour (85% eliminated), decaying much too fast. The exponential decay recipe: (1) Identify decay rate r as decimal, (2) Calculate remaining factor: 1 - r, (3) Write model: y = $a(1 - r)^t$ , (4) Solve using logarithms. Context language guides you: "15% eliminated" means multiply by 0.85 each period!
An investment account starts with $1000 and earns $4\%$ interest compounded annually. Create an exponential model for the balance $A(t)$ after $t$ years, then solve for when the account balance doubles. Round to the nearest tenth of a year and interpret your result.
Answer: Model: $A(t)=1000(1.04)^t$ . The balance doubles after about $t\approx 17.7$ years. This question tests your ability to create compound interest models and solve for doubling time, a common financial calculation. With 4% annual interest, the growth factor is 1.04, giving model A(t) = 1000 $(1.04)^t$ for an initial $1000 investment. To find doubling time, set A = 2000: 2000 = 1000 $(1.04)^t$ , divide by 1000 to get 2 = $(1.04)^t$ , take log: log(2) = t·log(1.04), solve: t = log(2)/log(1.04) ≈ 17.7 years. Choice A correctly models 4% growth with base 1.04 and finds the balance doubles after about 17.7 years. Choice C incorrectly uses 1.4 as the base, confusing 4% with 40%—remember that 4% interest means multiply by 1.04, not 1.4! The "Rule of 72" provides a quick estimate: divide 72 by the interest rate to approximate doubling time, so 72/4 = 18 years, confirming our calculated 17.7 years.
A culture of bacteria starts with 300 bacteria and grows by 18% each hour. Create an exponential model for the population after $t$ hours, then find how many hours it takes for the population to exceed 2000. Round to the nearest tenth of an hour and interpret your result.
Answer: Model: $P(t)=300(1.18)^t$ . It exceeds 2000 after about $t\approx 11.2$ hours. This question tests your ability to create exponential models from real-world contexts (like compound interest, population growth, radioactive decay, or logarithmic scales) and solve them to answer practical questions. Exponential models arise when quantities change by a constant percent per time period: if something grows r percent per period starting from initial value a, the model is y = $a(1 + r)^t$ for growth or y = $a(1 - r)^t$ for decay. For bacterial growth: P = P₀ $(1 + r)^t$ where P₀ is initial population, r is growth rate (as decimal), t is time, P is population at time t. For this bacteria culture: (1) Identify initial population P₀ = 300 and growth rate r = 0.18 (18% as decimal). (2) Write model: P(t) = 300 $(1.18)^t$ . (3) To find when population exceeds 2000, set up inequality: 300 $(1.18)^t$ > 2000. (4) Divide by 300: $(1.18)^t$ > 6.667. (5) Take log of both sides: t·log(1.18) > log(6.667). (6) Solve for t: t > log(6.667)/log(1.18) ≈ 11.2 hours. (7) Interpret: The bacteria population will exceed 2000 after approximately 11.2 hours. Choice A correctly creates the exponential model P(t) = 300 $(1.18)^t$ and solves to get t ≈ 11.2 hours with proper interpretation. Choice B has the correct model but solves incorrectly—when you calculate log(6.667)/log(1.18), you should get approximately 11.2, not 6.9. Always double-check your logarithm calculations! Context language guides you: "grows by 8%" means b = 1.08, "decreases by 12%" means b = 0.88. Solving exponential equations for time: from a· $b^t$ = target, isolate $b^t$ = target/a, take log of both sides (any base, but use calculator's log or ln), apply power property: t·log(b) = log(target/a), solve: t = log(target/a)/log(b).
The intensity of sound is measured in decibels using $dB = 10\log\left(\frac{I}{I_0}\right)$ , where $I$ is the sound intensity and $I_0$ is a reference intensity. If one sound is 40 dB and another is 70 dB, how many times more intense is the louder sound?
Answer: 1000 times more intense than the quieter sound For 40 dB: $40 = 10\log\left(\frac{I_1}{I_0}\right)$ , so $I_1 = I_0 \cdot 10^4$ . For 70 dB: $70 = 10\log\left(\frac{I_2}{I_0}\right)$ , so $I_2 = I_0 \cdot 10^7$ . The ratio is $\frac{I_2}{I_1} = \frac{10^7}{10^4} = 10^3 = 1000$ . Choice A uses the difference in dB values directly. Choice C multiplies 70 × 100 incorrectly. Choice D divides 70 by 30 incorrectly.
The decibel level of a sound can be modeled by $\text{dB}=10\log_{10}(I/I_0)$ , where $I$ is the sound intensity and $I_0$ is a reference intensity. A sound has intensity $500$ times the reference (so $I/I_0=500$ ). Find the decibel level to the nearest tenth and interpret your result.
Answer: $\text{dB}\approx 27.0$ dB; the sound is about 27.0 decibels above the reference level. This question tests your ability to create exponential or logarithmic models from real-world contexts like compound interest, population growth, radioactive decay, or logarithmic scales and solve them to answer practical questions. Logarithmic models appear in scientific scales (pH, decibels, Richter) or as inverses when solving exponential equations: for decibels, dB = 10 log10(I/I0), compressing intensity ratios logarithmically. For I/I0 = 500, dB = 10 log10(500) ≈ 10 × 2.699 ≈ 26.99 or 27.0—this means the sound is 27.0 decibels above the reference level, corresponding to 500 times the intensity—fantastic application! Choice A correctly evaluates the log, rounds to the nearest tenth, and interprets as decibels above reference—spot on! A tempting distractor like Choice D misinterprets the log as a direct multiplier, saying 27 times instead of using the formula—remember, it's 10 times the log for decibels! Strategy: plug into the formula directly, use log properties like log(500) = log( $5×10^2$ ) = log5 + 2; calculators are key. You're rocking these concepts—keep practicing!
An investment account starts with $1000 and grows by $4\%$ per year, compounded annually. (1) Create an exponential model for the balance after $t$ years. (2) Use your model to determine when the account balance will double. Round to the nearest tenth of a year and interpret your result in context.
Answer: Model: $A(t)=1000(1.04)^t$ . The balance doubles after $t\approx 17.7$ years. This question tests exponential modeling for investments and finding doubling time. For annual growth r%, A(t)= $a(1+r)^t$ . Here, $1000 at 4%, A(t)=1000 $(1.04)^t$ . Doubling: 1000 $(1.04)^t$ =2000, $(1.04)^t$ =2, t=log(2)/log(1.04)≈0.3010/0.0170≈17.67≈17.7 years, meaning it doubles in about 17.7 years—outstanding! Choice A correctly models growth and solves with logs. Choice D uses linear, giving huge t—exponential for compounding! Use the recipe: set target=2a, solve t=log(2)/log(1+r)—you're mastering this!
A medication dose starts at 120 mg in the bloodstream. Each hour, $12\%$ of the amount is eliminated. Write an exponential decay model for the amount remaining after $t$ hours and find how long it takes for the amount to drop to 40 mg. Round to the nearest tenth of an hour.
Answer: Model: $M(t)=120(0.88)^t$ . It drops to 40 mg after about $t\approx 8.8$ hours. This question tests your ability to create exponential models from real-world contexts (like compound interest, population growth, radioactive decay, or logarithmic scales) and solve them to answer practical questions. Exponential models arise when quantities change by a constant percent per time period: if something grows r percent per period starting from initial value a, the model is y = $a(1 + r)^t$ for growth or y = $a(1 - r)^t$ for decay. For medication decreasing by 12% each hour from 120 mg: (1) Identify model type: exponential decay. (2) Initial amount a = 120, decay rate r = 0.12. (3) Decay factor: 1 - 0.12 = 0.88. (4) Model: M(t) = 120 $(0.88)^t$ . (5) To find when M = 40: set 40 = 120 $(0.88)^t$ . (6) Isolate: $(0.88)^t$ = 40/120 = 1/3. (7) Take log: t·log(0.88) = log(1/3). (8) Solve: t = log(1/3)/log(0.88) ≈ 8.8 hours. Choice A correctly creates the exponential decay model M(t) = 120 $(0.88)^t$ for 12% elimination and solves to get t ≈ 8.8 hours when amount drops to 40 mg. Choice B uses 1.12 which would mean 12% growth—but medication is being eliminated, not increasing! Choice C uses 0.12 as the base, but "12% eliminated" means 88% remains, so base is 0.88. Understanding decay: if 12% is eliminated, then 100% - 12% = 88% remains, giving decay factor 0.88 for the exponential model!
A 500 mg dose of a medication decreases by 12% each hour. Create an exponential decay model for the amount remaining after $t$ hours, then determine how long it takes for the amount to drop to 200 mg. Round to the nearest tenth of an hour and interpret your answer.
Answer: Model: $M(t)=500(0.88)^t$ . It reaches 200 mg after about $t\approx 7.4$ hours. This question tests your ability to create exponential models from real-world contexts (like compound interest, population growth, radioactive decay, or logarithmic scales) and solve them to answer practical questions. Exponential models arise when quantities change by a constant percent per time period: if something grows r percent per period starting from initial value a, the model is y = $a(1 + r)^t$ for growth or y = $a(1 - r)^t$ for decay. For medication decay: M = M₀ $(1 - r)^t$ where M₀ is initial dose, r is decay rate (as decimal), t is time, M is amount remaining. For this medication problem: (1) Identify initial dose M₀ = 500 mg and decay rate r = 0.12 (12% as decimal). (2) Write model: M(t) = 500 $(0.88)^t$ since 1 - 0.12 = 0.88. (3) To find when amount drops to 200 mg, set up equation: 200 = 500 $(0.88)^t$ . (4) Divide by 500: 0.4 = $(0.88)^t$ . (5) Take log of both sides: log(0.4) = t·log(0.88). (6) Solve for t: t = log(0.4)/log(0.88) ≈ 7.4 hours. (7) Interpret: The medication amount will drop to 200 mg after approximately 7.4 hours. Choice A correctly creates the exponential decay model M(t) = 500 $(0.88)^t$ and solves to get t ≈ 7.4 hours with proper interpretation. Choice B incorrectly uses 1.12 as the factor, which would represent 12% growth, not decay—remember that decay means multiply by (1 - r), not (1 + r)! Exponential modeling recipe: (1) Identify initial value a (amount at t = 0), (2) Identify growth or decay: does it increase or decrease? What's the percent rate r per period?, (3) Determine growth factor: b = 1 + r for growth (example: 5% growth means multiply by 1.05), b = 1 - r for decay (example: 15% decay means multiply by 0.85), (4) Write model: y = a· $b^t$ where t is time in appropriate units, (5) For solving: set equal to target value and take logs. Understanding the model structure makes solving systematic!
A machine part is worth $900 when new and loses $12\%$ of its value each year. Create an exponential decay model for its value $V(t)$ after $t$ years, then solve for when its value will be $500. Round to the nearest tenth of a year and interpret your result.
Answer: Model: $V(t)=900(0.88)^t$ . It will be worth $500 after about $t\approx 4.9$ years. This question tests your ability to create exponential decay models for depreciation and solve for when an asset reaches a specific value. Losing 12% of value annually means retaining 88%, so the decay factor is 0.88: starting at $900, the model is V(t) = 900 $(0.88)^t$ . To find when V = 500: set 500 = 900 $(0.88)^t$ , divide by 900 to get 0.556 = $(0.88)^t$ , take log: log(0.556) = t·log(0.88), solve: t = log(0.556)/log(0.88) ≈ 4.9 years. Choice A correctly models the depreciation with base 0.88 (keeping 88% when losing 12%) and interprets that the machine part will be worth $500 after about 4.9 years. Choice C incorrectly uses 0.12 as the base, confusing the loss rate with the retention rate—if 12% is lost, then 88% remains, so multiply by 0.88 not 0.12! This models exponential decay where value decreases by a constant percentage, common in depreciation calculations.
The decibel level of a sound is given by $\text{dB}=10\log_{10}\left(\frac{I}{I_0}\right)$ , where $I$ is the sound intensity and $I_0$ is a reference intensity. A sound has intensity $I=2000I_0$ . Find its decibel level, rounded to the nearest tenth, and interpret the result in context.
Answer: $\text{dB}\approx 33.0$ dB; the sound is about 33.0 decibels relative to the reference intensity. This question tests your ability to apply logarithmic models for sound intensity using the decibel scale. The decibel formula is dB = 10log₁₀(I/I₀) where I is the sound intensity and I₀ is a reference intensity; this logarithmic scale compresses the huge range of audible intensities into manageable numbers. For I = 2000I₀: dB = 10log₁₀(2000I₀/I₀) = 10log₁₀(2000) = 10 × 3.301 ≈ 33.0 dB. Choice A correctly calculates 33.0 dB and interprets this as the sound level being about 33.0 decibels relative to the reference intensity. Choice B appears to calculate 10 × log₁₀(2000) × log₁₀(10) or some other error, getting an unrealistically high 230.1 dB—remember to simply evaluate 10 × log₁₀(2000)! The decibel scale is logarithmic because human hearing perceives intensity ratios rather than differences: a 10-fold increase in intensity adds 10 dB, so I = 2000I₀ means approximately 33 dB above reference.
A radioactive sample has a half-life of 50 years and starts with 200 g. Create an exponential model for the amount $N(t)$ after $t$ years, then solve for when 25 g remains. Round to the nearest year and interpret your result.
Answer: Model: $N(t)=200\left(\tfrac12\right)^{t/50}$ . $t\approx 150$ years; 25 g remains after about 150 years. This question tests your ability to create exponential or logarithmic models from real-world contexts like compound interest, population growth, radioactive decay, or logarithmic scales and solve them to answer practical questions. Exponential models arise when quantities change by a constant percent per time period: if something grows r percent per period starting from initial value a, the model is y = $a(1 + r)^t$ for growth or y = $a(1 - r)^t$ for decay; for half-life, it's y = a * $(0.5)^{t/h}$ where h is half-life. For this radioactive sample with 50-year half-life starting at 200 g, model N(t) = 200 * $(0.5)^{t/50}$ ; to reach 25 g, $(0.5)^{t/50}$ = 25/200 = 1/8 = $(0.5)^3$ , so t/50 = 3, t = 150 years—perfect exact solution! Choice B correctly uses the half-life in the exponent, equates powers for solving, and interprets accurately—great intuition! A tempting distractor like Choice A puts 50t in the exponent, making decay way too fast—remember, it's t divided by half-life for the number of halvings! Strategy tip: for half-life problems, set $(0.5)^{t/h}$ = fraction remaining, solve with logs or recognize powers of 1/2 like 1/8 = $(1/2)^3$ . You're building a strong foundation—keep going!
A car worth $\$18{,}000$ depreciates by $14\%$ each year. Create an exponential model for its value $V(t)$ after $t$ years, then find how long it takes for the value to fall to $9000. Round to the nearest tenth of a year and interpret your result in context.
Answer: Model: $V(t)=18000(0.86)^t$ . It falls to $9000 after $t\approx 4.6$ years. This question tests your ability to create exponential models from real-world contexts (like compound interest, population growth, radioactive decay, or logarithmic scales) and solve them to answer practical questions. Exponential models arise when quantities change by a constant percent per time period: if something grows r percent per period starting from initial value a, the model is y = $a(1 + r)^t$ for growth or y = $a(1 - r)^t$ for decay. For car depreciation at 14% per year: (1) Initial value V₀ = $18,000, (2) Depreciation rate r = 0.14, (3) Model: V(t) = 18000 $(1 - 0.14)^t$ = 18000 $(0.86)^t$ , (4) To find when V = $9000, solve 9000 = 18000 $(0.86)^t$ , (5) Divide: $(0.86)^t$ = 0.5, (6) Take log: t·log(0.86) = log(0.5), (7) Solve: t = log(0.5)/log(0.86) ≈ 4.6 years. Choice A correctly creates the exponential decay model V(t) = 18000 $(0.86)^t$ and solves to get t ≈ 4.6 years, meaning the car's value falls to $9000 after approximately 4.6 years. Choice C incorrectly uses 0.14 as the base—this would mean keeping only 14% each year, but depreciation of 14% means keeping 86% (losing 14%)! Depreciation modeling insight: (1) "Depreciates by r%" means value retains (100-r)% each year, (2) Decay factor = 1 - 0.14 = 0.86, (3) After t years: V(t) = V₀ $(0.86)^t$ , (4) Note that halving value (18000 to 9000) takes about 4.6 years at 14% depreciation. Understanding that depreciation removes value helps avoid the 0.14 vs 0.86 confusion!
Sound intensity level in decibels is modeled by $\text{dB}=10\log_{10}\left(\frac{I}{I_0}\right)$ , where $I_0$ is a reference intensity. A sound has intensity $5000I_0$ . Create an equation and find its decibel level. Round to the nearest tenth of a decibel and interpret the result.
Answer: About $37.0$ dB, meaning the sound level is approximately 37.0 decibels relative to the reference intensity. This question tests your ability to create exponential or logarithmic models from real-world contexts (like compound interest, population growth, radioactive decay, or logarithmic scales) and solve them to answer practical questions. The decibel scale is a logarithmic model where dB = 10log₁₀(I/I₀), measuring sound intensity relative to a reference level I₀. Logarithmic scales compress large ranges of values into manageable numbers. For this sound intensity problem: (1) Start with the decibel formula: dB = 10log₁₀(I/I₀). (2) Given that I = 5000I₀, substitute: dB = 10log₁₀(5000I₀/I₀). (3) Simplify the fraction: dB = 10log₁₀(5000). (4) Calculate the logarithm: log₁₀(5000) = log₁₀(5 × 10³) = log₁₀(5) + 3 ≈ 0.699 + 3 = 3.699. (5) Multiply by 10: dB = 10 × 3.699 = 36.99. (6) Round to nearest tenth: dB ≈ 37.0. (7) Interpret: The sound level is approximately 37.0 decibels relative to the reference intensity. Choice A correctly applies the logarithmic formula to get dB ≈ 37.0 with proper rounding and interpretation. Choice C appears to have forgotten to multiply by 10 in the formula—the decibel formula has a factor of 10 in front of the logarithm, so dB = 10log₁₀(I/I₀), not just log₁₀(I/I₀)! Logarithmic models appear in scientific scales (pH, decibels, Richter) or as inverses when solving exponential equations. For decibels: every 10-fold increase in intensity adds 10 dB (since log₁₀(10) = 1), so 10I₀ → 10 dB, 100I₀ → 20 dB, 1000I₀ → 30 dB, etc. Understanding logarithmic properties helps you check your work!
The pH of a solution is given by $\text{pH}=-\log_{10}([H^+])$ , where $[H^+]$ is the hydrogen ion concentration in moles per liter. If a solution has pH 3.6, create an equation and solve for $[H^+]$ . Give your answer in scientific notation (to two significant figures) and interpret it in context.
Answer: $[H^+]\approx 2.5\times 10^{-4}\,\text{mol/L}$ , meaning the hydrogen ion concentration is about $2.5\times 10^{-4}$ mol per liter. This question tests your ability to create exponential or logarithmic models from real-world contexts (like compound interest, population growth, radioactive decay, or logarithmic scales) and solve them to answer practical questions. The pH scale is a logarithmic model where pH = -log₁₀([H⁺]), relating acidity to hydrogen ion concentration on a compressed scale. To find [H⁺] from pH = 3.6: (1) Start with pH = -log₁₀([H⁺]), (2) Substitute: 3.6 = -log₁₀([H⁺]), (3) Multiply by -1: -3.6 = log₁₀([H⁺]), (4) Convert from log to exponential form: [H⁺] = $10^{-3.6}$ , (5) Calculate: [H⁺] ≈ 2.5 × $10^{-4}$ mol/L. Choice A correctly solves the logarithmic equation to get [H⁺] ≈ 2.5 × $10^{-4}$ mol/L, meaning the hydrogen ion concentration is about 0.00025 moles per liter. Choice B appears to have computed $10^{-3.6}$ incorrectly or made a sign error—remember that pH = -log([H⁺]) means [H⁺] = $10^{-pH}$ , not $10^{pH}$ ! Logarithmic equation solving strategy: (1) To solve $\log_b$ (x) = y, convert to exponential: x = $b^y$ , (2) For pH problems: from pH = -log₁₀([H⁺]), get [H⁺] = $10^{-pH}$ , (3) Use calculator for $10^{-3.6}$ , (4) Express in scientific notation. The negative in the pH formula is crucial—higher pH means lower [H⁺] concentration!
You deposit $1200 into an account that earns 3.8% interest compounded annually. Create an exponential model for the account balance after $t$ years, then determine how long it will take for the balance to double. Round to the nearest tenth of a year and interpret your result.
Answer: Model: $A(t)=1200(1.038)^t$ . It doubles after about $t\approx 18.6$ years. This question tests your ability to create exponential models from real-world contexts (like compound interest, population growth, radioactive decay, or logarithmic scales) and solve them to answer practical questions. Exponential models arise when quantities change by a constant percent per time period: for compound interest specifically, A = $P(1 + r)^t$ where P is principal, r is annual rate (as decimal), t is years, A is final amount. To find doubling time, set A = 2P and solve for t. For this investment problem: (1) Identify principal P = 1200 and interest rate r = 0.038 (3.8% as decimal). (2) Write model: A(t) = 1200 $(1.038)^t$ . (3) To find doubling time, set A = 2400 (double of 1200): 2400 = 1200 $(1.038)^t$ . (4) Divide by 1200: 2 = $(1.038)^t$ . (5) Take log of both sides: log(2) = t·log(1.038). (6) Solve for t: t = log(2)/log(1.038) ≈ 18.6 years. (7) Interpret: The account balance will double after approximately 18.6 years. Choice A correctly creates the exponential model A(t) = 1200 $(1.038)^t$ and solves the doubling equation to get t ≈ 18.6 years with proper interpretation. Choice B has the correct model but makes a calculation error—when you compute log(2)/log(1.038), you should get approximately 18.6, not 26.3. This might be from using natural log (ln) on one part and log base 10 on another—be consistent with your logarithm base! Solving exponential equations for time: from a· $b^t$ = target, isolate $b^t$ = target/a, take log of both sides (any base, but use calculator's log or ln), apply power property: t·log(b) = log(target/a), solve: t = log(target/a)/log(b). For doubling time with rate r: t = log(2)/log(1 + r). This gives the "Rule of 72": doubling time ≈ 72/r% for small rates!

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