Algebra 2 Quiz: Modeling With Exponential Logarithmic Equations

20 questions · exam conditions

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Modeling With Exponential Logarithmic EquationsQuestion 1 of 20

A town's population is 18,000 and increases by $2.4\%$ each year.

Write an exponential model for the population after $t$ years and estimate how many years it will take for the population to reach 25,000. Round to the nearest tenth of a year.

Model:

P(t)=18000(1.024)^t

. It reaches 25,000 after about

t\approx 13.8

years.

Model:

P(t)=18000(0.976)^t

. It reaches 25,000 after about

t\approx 13.8

years.

Model:

P(t)=18000(1.24)^t

. It reaches 25,000 after about

t\approx 1.4

years.

Model:

P(t)=25000(1.024)^t

. It reaches 25,000 after about

t\approx 0.0

years.

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Algebra 2 Quiz

Algebra 2 Quiz: Modeling With Exponential Logarithmic Equations

Practice Modeling With Exponential Logarithmic Equations in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Modeling With Exponential Logarithmic Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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

A town's population is 18,000 and increases by $2.4\%$ each year.

Write an exponential model for the population after $t$ years and estimate how many years it will take for the population to reach 25,000. Round to the nearest tenth of a year.

Model: $P(t)=18000(1.024)^t$ . It reaches 25,000 after about $t\approx 13.8$ years. (correct answer)
Model: $P(t)=18000(0.976)^t$ . It reaches 25,000 after about $t\approx 13.8$ years.
Model: $P(t)=18000(1.24)^t$ . It reaches 25,000 after about $t\approx 1.4$ years.
Model: $P(t)=25000(1.024)^t$ . It reaches 25,000 after about $t\approx 0.0$ years.

Explanation: 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 population growing 2.4% annually from 18,000: (1) Initial population P₀ = 18,000, growth rate r = 0.024. (2) Growth factor: 1 + 0.024 = 1.024. (3) Model: P(t) = 18000(1.024)^t. (4) To find when P = 25,000: set 25000 = 18000(1.024)^t. (5) Isolate: (1.024)^t = 25000/18000 = 1.389. (6) Take log: t·log(1.024) = log(1.389). (7) Solve: t = log(1.389)/log(1.024) ≈ 13.8 years. Choice A correctly creates the exponential growth model P(t) = 18000(1.024)^t for 2.4% growth and solves to get t ≈ 13.8 years when population reaches 25,000. Choice B uses 0.976 which would represent 2.4% decay—but the population is increasing! Choice C uses 1.24 which would mean 24% growth, not 2.4%—be careful with percents as decimals. Context language "increases by 2.4%" means multiply by 1.024 each year, and solving exponential equations requires logarithms!

Question 2

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. If a sound has intensity $2.5\times 10^4\,I_0$ , find its decibel level. Round to the nearest tenth and interpret your result in context.

$\text{dB}\approx 24.0$ dB, meaning the sound is 24.0 dB above the reference level.
$\text{dB}\approx 44.0$ dB, meaning the intensity is $44.0I_0$ .
$\text{dB}\approx 4.4$ dB, meaning the sound is 4.4 dB above the reference level.
$\text{dB}\approx 44.0$ dB, meaning the sound is 44.0 dB above the reference level. (correct answer)

Explanation: This question tests applying logarithmic models for sound intensity in decibels and interpreting the result. Decibels use dB = 10 log10(I/I0), measuring intensity ratios logarithmically. For I = 2.5 × 10^4 I0, dB = 10 log10(2.5 ×

10^4

) = 10 (log10(2.5) + log10(

10^4

)) ≈ 10 (0.398 + 4) ≈ 10 × 4.398 = 43.98 ≈ 44.0 dB, meaning 44.0 dB above reference—fantastic! Choice A accurately computes and interprets the level. Choice D misinterprets as intensity 44.0 I0, but dB is logarithmic, not linear—recall it represents ratio, not direct multiple! Strategy: plug into formula, split logs, calculate carefully; context explains the scale—you're nailing this!

Question 3

A culture of bacteria starts with 300 bacteria and increases by $18\%$ every hour. Create an exponential model for the population $P(t)$ after $t$ hours, then find how many hours it takes for the population to reach 2000. Round to the nearest tenth of an hour and interpret your result in context.

Model: $P(t)=300(1.18)^t$ . It reaches 2000 after $t\approx 11.3$ hours. (correct answer)
Model: $P(t)=300(0.82)^t$ . It reaches 2000 after $t\approx 11.3$ hours.
Model: $P(t)=300(1.18t)$ . It reaches 2000 after $t\approx 5.6$ hours.
Model: $P(t)=300(1.18)^{2000}$ . It reaches 2000 after $t\approx 11.3$ hours.

Explanation: 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 with 18% increase per hour: (1) Initial population P₀ = 300, (2) Growth rate r = 0.18, (3) Model: P(t) = 300(1.18)^t, (4) To find when P = 2000, solve 2000 = 300(1.18)^t, (5) Divide: (1.18)^t = 6.667, (6) Take log: t·log(1.18) = log(6.667), (7) Solve: t = log(6.667)/log(1.18) ≈ 11.3 hours. Choice A correctly creates the exponential model P(t) = 300(1.18)^t and solves to get t ≈ 11.3 hours, meaning the bacteria population reaches 2000 after approximately 11.3 hours. Choice C incorrectly writes P(t) = 300(1.18t) which is linear, not exponential—the exponent must be t alone, not 1.18t! Exponential growth modeling steps: (1) Identify initial value and percent growth rate, (2) Convert percent to decimal and add 1 for growth factor, (3) Write P(t) = P₀(growth factor)^t, (4) To solve for time: set equal to target, isolate exponential, take logs. Remember: exponential means the variable is in the exponent, not multiplied by a constant!

Question 4

The decibel level of a sound is modeled by $\text{dB}=10\log_{10}\left(\dfrac{I}{I_0}\right)$ , where $I$ is intensity and $I_0$ is a reference intensity.

A sound has intensity $2500I_0$ . What is its decibel level? Round to the nearest tenth of a decibel.

About $34.0$ dB, meaning the sound is $2500$ times the reference intensity. (correct answer)
About $3.4$ dB, meaning the sound is $2500$ times the reference intensity.
About $34.0$ dB, meaning the sound is $34$ times the reference intensity.
About $25000.0$ dB, meaning the sound is $2500$ times the reference intensity.

Explanation: 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 uses logarithms to compress wide ranges of sound intensities into manageable numbers. For sound with intensity 2500I₀: (1) Use formula dB = 10log₁₀(I/I₀). (2) Substitute: dB = 10log₁₀(2500I₀/I₀). (3) Simplify: dB = 10log₁₀(2500). (4) Calculate: log₁₀(2500) = log₁₀(2.5 × 10³) = log₁₀(2.5) + 3 ≈ 0.398 + 3 = 3.398. (5) Multiply: dB = 10(3.398) = 33.98. (6) Round: dB ≈ 34.0. (7) Interpret: 34 dB means intensity is 2500 times reference (not 34 times). Choice A correctly calculates 34.0 dB and explains that the sound is 2500 times the reference intensity. Choice B gets 3.4 dB by forgetting to multiply by 10—the formula has 10 in front! Choice C has correct decibel value but wrong interpretation—34 dB means 10^(34/10) ≈ 2512 times reference, not 34 times. Decibel formula: dB = 10log₁₀(I/I₀) where the 10 converts bels to decibels, and the log compresses the intensity ratio!

Question 5

A certain chemical concentration starts at 80 units and increases by $6\%$ each day. Create an exponential model for the concentration $C(t)$ after $t$ days, then solve for when it reaches 120 units. Round to the nearest tenth of a day and interpret your result.

Model: $C(t)=80(1.06)^t$ . It reaches 120 units after about $t\approx 7.0$ days. (correct answer)
Model: $C(t)=80(0.94)^t$ . It reaches 120 units after about $t\approx 7.0$ days.
Model: $C(t)=120(1.06)^t$ . It reaches 120 units after about $t\approx 7.0$ days.
Model: $C(t)=80(1.6)^t$ . It reaches 120 units after about $t\approx 0.5$ day.

Explanation: This question tests your ability to create exponential growth models for chemical concentrations and solve for when a specific level is reached. A 6% daily increase means multiplying by 1.06 each day: starting at 80 units, the model is C(t) = 80(1.06)^t. To find when C = 120: set 120 = 80(1.06)^t, divide by 80 to get 1.5 = (1.06)^t, take log: log(1.5) = t·log(1.06), solve: t = log(1.5)/log(1.06) ≈ 7.0 days. Choice A correctly models 6% growth with base 1.06 and finds the concentration reaches 120 units after about 7.0 days. Choice D incorrectly uses base 1.6, perhaps misunderstanding 6% as 60%—remember that 6% growth means multiply by 1.06, not 1.6! The exponential model captures how the concentration compounds: each day's 6% increase applies to the previous day's higher amount, creating accelerating growth.

Question 6

A lab sample contains 1500 bacteria. After 4 hours, it contains 2400 bacteria. Assume the bacteria grow exponentially at a constant percent rate per hour. Create an exponential model $P(t)=P_0\,b^t$ (with $t$ in hours), then use it to predict when the population will reach 5000 bacteria. Round to the nearest tenth of an hour and interpret your result in context.

Model: $P(t)=1500\left(\frac{2400}{1500}\right)^{4t}$ . It reaches 5000 at $t\approx 0.6$ hours after the start.
Model: $P(t)=1500+\left(\frac{2400-1500}{4}\right)t$ . It reaches 5000 at $t\approx 15.6$ hours after the start.
Model: $P(t)=1500\left(\frac{1500}{2400}\right)^{t/4}$ . It reaches 5000 at $t\approx 8.9$ hours after the start.
Model: $P(t)=1500\left(\frac{2400}{1500}\right)^{t/4}$ . It reaches 5000 at $t\approx 8.9$ hours after the start. (correct answer)

Explanation: 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. When given two data points for exponential growth, find the growth factor b by using the ratio of populations over the time interval. For bacteria growing from 1500 to 2400 in 4 hours: (1) After 4 hours: 2400 = 1500·b^4, so b^4 = 2400/1500 = 1.6, (2) Take fourth root: b = (1.6)^(1/4) ≈ 1.1247, (3) Model: P(t) = 1500(1.1247)^t or equivalently P(t) = 1500(1.6)^(t/4), (4) To find when P = 5000: 5000 = 1500(1.6)^(t/4), (5) Divide: (1.6)^(t/4) = 3.333, (6) Take log: (t/4)log(1.6) = log(3.333), (7) Solve: t = 4·log(3.333)/log(1.6) ≈ 8.9 hours. Choice A correctly creates the model P(t) = 1500(2400/1500)^(t/4) = 1500(1.6)^(t/4) and solves to get t ≈ 8.9 hours, meaning the bacteria population reaches 5000 after approximately 8.9 hours from the start. Choice D incorrectly assumes linear growth, calculating average increase per hour—bacterial growth is exponential, not linear! Two-point exponential modeling: (1) From P(0) = P₀ and P(t₁) = P₁, find b from P₁ = P₀·b^(t₁), (2) Growth factor over t₁ hours: b^(t₁) = P₁/P₀, (3) Model: P(t) = P₀(P₁/P₀)^(t/t₁), (4) This form clearly shows the growth ratio. Understanding that exponential growth compounds makes the difference between linear and exponential models clear!

Question 7

An investment account starts with $6000 and earns $5\%$ interest compounded annually. Create an exponential model for the balance $A(t)$ after $t$ years. Then find the balance after 8 years, rounding to the nearest cent, and interpret your result in context.

Model: $A(t)=6000(1.5)^t$ . After 8 years, $A(8)\approx \$ 8{,}866.84$.
Model: $A(t)=6000+0.05t$ . After 8 years, $A(8)\approx \$ 6{,}000.40$.
Model: $A(t)=6000(0.95)^t$ . After 8 years, $A(8)\approx \$ 8{,}866.84$.
Model: $A(t)=6000(1.05)^t$ . After 8 years, $A(8)\approx \$ 8{,}866.84$. (correct answer)

Explanation: 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 compound interest at 5% annually: (1) Initial investment P = $6000, (2) Interest rate r = 0.05, (3) Model: A(t) = 6000(1.05)^t, (4) After 8 years: A(8) = 6000(1.05)^8, (5) Calculate: (1.05)^8 ≈ 1.477455, (6) Multiply: A(8) = 6000(1.477455) ≈ $8,866.84. Choice A correctly creates the exponential model A(t) = 6000(1.05)^t and evaluates it at t = 8 to get approximately $8,866.84, meaning the investment grows to $8,866.84 after 8 years of compound interest. Choice D incorrectly uses linear growth A(t) = 6000 + 0.05t which would only add $0.05 per year—compound interest multiplies by 1.05 each year, creating exponential growth! Compound interest evaluation: (1) Use A = P(1 + r)^t with P = principal, r = rate, t = time, (2) Calculate (1 + r)^t carefully with calculator, (3) Multiply by principal, (4) Round to nearest cent for money. The power of compounding: $6000 grows to $8,866.84 in 8 years, gaining $2,866.84 through exponential growth!

Question 8

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.

Model: $V(t)=18000(0.86)^t$ . It falls to $9000 after $t\approx 4.6$ years. (correct answer)
Model: $V(t)=18000(1.14)^t$ . It falls to $9000 after $t\approx 4.6$ years.
Model: $V(t)=18000-0.14t$ . It falls to $9000 after $t\approx 64{,}285.7$ years.
Model: $V(t)=18000(0.14)^t$ . It falls to $9000 after $t\approx 0.4$ years.

Explanation: 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!

Question 9

Compound interest is compounded annually. You deposit $2500 into an account earning $5\%$ interest per year. Create an exponential model for the balance $A(t)$ after $t$ years, then solve for how long it will take the balance to reach $4000. Round your time to the nearest tenth of a year and interpret your result in context.

Model: $A(t)=2500(1.05)^t$ . It reaches $4000 after about $t\approx 9.6$ years. (correct answer)
Model: $A(t)=2500(0.95)^t$ . It reaches $4000 after about $t\approx 9.6$ years.
Model: $A(t)=2500(1.5)^t$ . It reaches $4000 after about $t\approx 1.0$ year.
Model: $A(t)=2500+0.05t$ . It reaches $4000 after $t=30000$ years.

Explanation: This question tests your ability to create exponential models from compound interest contexts and solve them to find when an investment reaches a target value. Compound interest follows the exponential model A = P(1 + r)^t where P is principal (initial amount), r is annual interest rate as a decimal, t is time in years, and A is the final amount. For $2500 at 5% interest: A(t) = 2500(1 + 0.05)^t = 2500(1.05)^t. To find when it reaches $4000: set 4000 = 2500(1.05)^t, divide by 2500 to get 1.6 = (1.05)^t, take log of both sides: log(1.6) = t·log(1.05), solve: t = log(1.6)/log(1.05) ≈ 9.6 years. Choice A correctly creates the exponential growth model with growth factor 1.05 (representing 5% growth), solves using logarithms, and interprets that the investment reaches $4000 after approximately 9.6 years. Choice B incorrectly uses 0.95 as the base, which would represent 5% decay, not growth—compound interest grows money, it doesn't shrink it! The key insight is that r% growth means multiply by (1 + r/100) each period, so 5% growth gives base 1.05, while r% decay would give base (1 - r/100).

Question 10

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.

$\text{dB}\approx 33.0$ dB; the sound is about 33.0 decibels relative to the reference intensity. (correct answer)
$\text{dB}\approx 230.1$ dB; the sound is about 230.1 decibels relative to the reference intensity.
$\text{dB}\approx 3.3$ dB; the sound is about 3.3 decibels relative to the reference intensity.
$\text{dB}\approx -33.0$ dB; the sound is about -33.0 decibels relative to the reference intensity.

Explanation: 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.

Question 11

A lab culture starts with 50 cells and grows by 20% each day. Create an exponential model for the number of cells after $t$ days, then solve for how many days it takes to reach 500 cells. Round to the nearest tenth of a day and interpret your result in context.

Model: $N(t)=50(1.20)^t$ . It reaches 500 cells after $t\approx 12.6$ days. (correct answer)
Model: $N(t)=50(0.80)^t$ . It reaches 500 cells after $t\approx 12.6$ days.
Model: $N(t)=50(1.02)^t$ . It reaches 500 cells after $t\approx 117.0$ days.
Model: $N(t)=500(1.20)^t$ . It reaches 500 cells after $t\approx 12.6$ days.

Explanation: 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 cell culture problem: (1) Initial count N₀ = 50 cells, growth rate r = 0.20 (20% as decimal). (2) Growth factor: 1 + 0.20 = 1.20 (population multiplies by 1.20 each day). (3) Model: N(t) = 50(1.20)^t. (4) To find when N = 500, solve: 500 = 50(1.20)^t. (5) Divide by 50: 10 = (1.20)^t. (6) Take log: log(10) = t·log(1.20). (7) Solve: t = log(10)/log(1.20) = 1/log(1.20) ≈ 12.6 days. Choice A correctly creates the growth model N(t) = 50(1.20)^t and accurately calculates that the culture reaches 500 cells after approximately 12.6 days (a 10-fold increase). Choice C incorrectly uses growth factor 1.02 (only 2% growth), which would take much longer to reach 500 cells. The exponential growth strategy: (1) Convert percent growth to growth factor: 1 + r, (2) Write model: y = a(1 + r)^t, (3) For 10-fold growth with factor b: time = log(10)/log(b). Understanding growth factors makes calculations systematic!

Question 12

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.

Model: $P(t)=15000(1.03)^t$ . It reaches 15,000 after $t\approx 7.4$ years.
Model: $P(t)=12000(1.3)^t$ . It reaches 15,000 after $t\approx 0.9$ years.
Model: $P(t)=12000(1.03)^t$ . It reaches 15,000 after $t\approx 7.4$ years. (correct answer)
Model: $P(t)=12000(0.97)^t$ . It reaches 15,000 after $t\approx 7.4$ years.

Explanation: 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!

Question 13

A radioactive sample has a half-life of 12 years and initially has a mass of 160 grams. Create an exponential model for the remaining mass $N(t)$ after $t$ years, then solve for when 20 grams remain. Round your answer to the nearest tenth of a year and interpret your result.

Model: $N(t)=160\left(\tfrac12\right)^{t/12}$ . It reaches 20 g after about $t\approx 36.0$ years. (correct answer)
Model: $N(t)=160\left(\tfrac12\right)^{12t}$ . It reaches 20 g after about $t\approx 0.3$ years.
Model: $N(t)=20\left(\tfrac12\right)^{t/12}$ . It reaches 20 g after about $t\approx 36.0$ years.
Model: $N(t)=160(0.5)t/12$ . It reaches 20 g after about $t\approx 10.7$ years.

Explanation: This question tests your ability to create exponential models for radioactive decay using half-life and solve for when a specific amount remains. For half-life problems, use N(t) = N₀(1/2)^(t/h) where N₀ is initial amount, h is half-life, and t/h represents the number of half-lives elapsed. With N₀ = 160 g and h = 12 years: N(t) = 160(1/2)^(t/12). To find when N = 20: set 20 = 160(1/2)^(t/12), divide by 160 to get 1/8 = (1/2)^(t/12), recognize 1/8 = (1/2)³, so (1/2)^(t/12) = (1/2)³, therefore t/12 = 3, giving t = 36 years. Choice A correctly models decay with the half-life formula and finds that 20 g remains after 36 years (exactly 3 half-lives). Choice B incorrectly uses (1/2)^(12t) instead of (1/2)^(t/12), which would mean 12 half-lives per year instead of one half-life per 12 years—the exponent should be t divided by the half-life period! Understanding that t/h counts half-lives makes these problems systematic: from 160 to 20 is 160→80→40→20, which is 3 halvings.

Question 14

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.

$[H^+]\approx 2.5\times 10^{-4}\,\text{mol/L}$ , meaning the hydrogen ion concentration is about $2.5\times 10^{-4}$ mol per liter. (correct answer)
$[H^+]\approx 3.6\times 10^{-10}\,\text{mol/L}$ , meaning the hydrogen ion concentration is about $3.6\times 10^{-10}$ mol per liter.
$[H^+]\approx 4.0\times 10^{3}\,\text{mol/L}$ , meaning the hydrogen ion concentration is about $4.0\times 10^{3}$ mol per liter.
$[H^+]\approx 2.5\times 10^{4}\,\text{mol/L}$ , meaning the hydrogen ion concentration is about $2.5\times 10^{4}$ mol per liter.

Explanation: 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!

Question 15

A radioactive sample has a half-life of 12 years and starts with 160 grams. Create an exponential model for the mass remaining after $t$ years, then find how long it will take until only 20 grams remain. Round to the nearest tenth of a year and interpret your result in context.

Model: $N(t)=160\left(\tfrac12\right)^{t/12}$ . It reaches 20 g after $t\approx 36.0$ years. (correct answer)
Model: $N(t)=160\left(\tfrac12\right)^{12t}$ . It reaches 20 g after $t\approx 0.3$ years.
Model: $N(t)=160(2)^{t/12}$ . It reaches 20 g after $t\approx 36.0$ years.
Model: $N(t)=160-12t$ . It reaches 20 g after $t\approx 11.7$ years.

Explanation: This question tests modeling radioactive decay with half-life and solving for time to a remaining mass. Half-life models use y = a (1/2)^{t/h}, where h is half-life. For 160 g with h=12 years, N(t)=160 (1/2)^{t/12}. To reach 20 g: 160 (1/2)^{t/12} = 20, (1/2)^{t/12} = 20/160 = 1/8 = (1/2)^3, so t/12=3, t=36 years exactly—well done interpreting as three half-lives! Choice A correctly uses the half-life exponent t/12 and finds t=36.0. Choice B flips to 12t, leading to tiny t—exponent should be t divided by half-life, not multiplied! Strategy: recognize half-life means b=0.5, exponent t/h for number of half-lives, solve with logs or powers—you've got the hang of it!

Question 16

A medication dose starts at 120 mg in the bloodstream. Each hour, $25\%$ of the medication is eliminated. Create an exponential decay model for the amount $M(t)$ after $t$ hours, then solve for when the amount drops to 30 mg. Round to the nearest tenth of an hour and interpret your result.

Model: $M(t)=120(0.75)^t$ . It drops to 30 mg after about $t\approx 4.8$ hours. (correct answer)
Model: $M(t)=120(1.25)^t$ . It drops to 30 mg after about $t\approx 4.8$ hours.
Model: $M(t)=120(0.25)^t$ . It drops to 30 mg after about $t\approx 1.0$ hour.
Model: $M(t)=120-0.25t$ . It drops to 30 mg after $t=360$ hours.

Explanation: This question tests your ability to create exponential decay models from contexts where a quantity decreases by a constant percentage and solve for specific values. When 25% is eliminated each hour, 75% remains, so the decay factor is 0.75: if you start with 120 mg and keep 75% each hour, after t hours you have M(t) = 120(0.75)^t. To find when M = 30: set 30 = 120(0.75)^t, divide by 120 to get 0.25 = (0.75)^t, take log: log(0.25) = t·log(0.75), solve: t = log(0.25)/log(0.75) ≈ 4.8 hours. Choice A correctly models the decay with base 0.75 (keeping 75% means multiplying by 0.75) and interprets that the medication drops to 30 mg after about 4.8 hours. Choice C incorrectly uses 0.25 as the base, confusing the elimination rate with the retention rate—if 25% is eliminated, then 75% remains, so multiply by 0.75 not 0.25! The key insight: when r% is removed, (100-r)% remains, so the decay factor is (100-r)/100.

Question 17

A town has an initial population of 18,000 and grows by $2.2\%$ per year. Create an exponential model for the population after $t$ years, then determine the population after 15 years. Round to the nearest person and interpret your result in context.

Model: $P(t)=18000(1.022)^t$ . After 15 years, $P\approx 24{,}978$ people. (correct answer)
Model: $P(t)=18000(0.978)^t$ . After 15 years, $P\approx 24{,}978$ people.
Model: $P(t)=18000(1.22)^t$ . After 15 years, $P\approx 24{,}978$ people.
Model: $P(t)=18000+0.022t$ . After 15 years, $P\approx 18{,}000$ people.

Explanation: This question tests creating exponential population models and predicting future values. Growth by r% yearly uses P(t) = a (1 + r)^t. For 18,000 growing 2.2%, P(t)=18000(1.022)^t. After 15 years: 18000(1.022)^15 ≈ 18000 × 1.38765 ≈ 24,978 people, meaning the town grows to about 24,978 in 15 years—impressive! Choice A correctly models with 1.022 and computes the value. Choice D uses linear addition, but population growth is multiplicative—exponential fits percentage increase! Build skills by confirming rate as decimal, raising to t, rounding appropriately—keep it up, you're awesome!

Question 18

A lab culture starts with 500 cells and doubles every 4 hours.

Write an exponential model for the number of cells after $t$ hours and determine when the culture will reach 20,000 cells. Round to the nearest tenth of an hour.

Model: $N(t)=500\cdot 2^{t/4}$ . It reaches 20,000 cells after about $t\approx 21.3$ hours. (correct answer)
Model: $N(t)=500\cdot 2^{4t}$ . It reaches 20,000 cells after about $t\approx 1.3$ hours.
Model: $N(t)=500(1.25)^t$ . It reaches 20,000 cells after about $t\approx 21.3$ hours.
Model: $N(t)=20000\cdot 2^{t/4}$ . It reaches 20,000 cells after about $t\approx 0.0$ hours.

Explanation: 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. When something doubles every d time units, the model is N(t) = N₀ · 2^(t/d) where t/d counts how many doubling periods have passed. For 500 cells doubling every 4 hours: (1) Initial cells N₀ = 500, doubling time d = 4 hours. (2) Model: N(t) = 500 · 2^(t/4). (3) To find when N = 20,000: set 20000 = 500 · 2^(t/4). (4) Isolate: 2^(t/4) = 20000/500 = 40. (5) Take log: (t/4)·log(2) = log(40). (6) Solve for t/4: t/4 = log(40)/log(2) ≈ 5.322. (7) Multiply: t = 4(5.322) ≈ 21.3 hours. Choice A correctly creates the doubling model N(t) = 500 · 2^(t/4) and solves to get t ≈ 21.3 hours when culture reaches 20,000 cells. Choice B uses 2^(4t) which would mean 4 doublings per hour—but it doubles every 4 hours, so exponent should be t/4! Choice C tries to use growth factor 1.25, but "doubles" means multiply by 2, not by 1.25. Doubling formula key: use base 2 with exponent t/d where d is doubling time, giving N = N₀ · 2^(t/d)!

Question 19

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.

Model: $V(t)=900(0.88)^t$ . It will be worth $500 after about $t\approx 4.9$ years. (correct answer)
Model: $V(t)=900(1.12)^t$ . It will be worth $500 after about $t\approx 4.9$ years.
Model: $V(t)=900(0.12)^t$ . It will be worth $500 after about $t\approx 1.3$ years.
Model: $V(t)=900-0.12t$ . It will be worth $500 after about $t\approx 3333.3$ years.

Explanation: 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.

Question 20

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.

Model: $P(t)=48000(0.984)^t$ . It falls to 40,000 after about $t\approx 11.4$ years. (correct answer)
Model: $P(t)=48000(1.016)^t$ . It falls to 40,000 after about $t\approx 11.4$ years.
Model: $P(t)=40000(0.984)^t$ . It falls to 40,000 after about $t\approx 11.4$ years.
Model: $P(t)=48000-0.016t$ . It falls to 40,000 after $t=500,000$ years.

Explanation: 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).