Business Calculus Quiz: Exponential Business Models
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Exponential Business ModelsQuestion 1 of 14

An investor deposits $1,000 into Account A, which earns 5% annual interest compounded continuously. A second investor deposits $1,200 into Account B, which earns 4% annual interest compounded annually. Approximately how many years will it take for the balance in Account A to surpass the balance in Account B?

12.7 years
14.5 years
17.0 years
19.1 years
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Business Calculus Quiz

Business Calculus Quiz: Exponential Business Models

Practice Exponential Business Models in Business Calculus 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 Exponential Business Models, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

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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.

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Question 1

An investor deposits $1,000 into Account A, which earns 5% annual interest compounded continuously. A second investor deposits $1,200 into Account B, which earns 4% annual interest compounded annually. Approximately how many years will it take for the balance in Account A to surpass the balance in Account B?

  1. 12.7 years
  2. 14.5 years
  3. 17.0 years (correct answer)
  4. 19.1 years
Explanation: Let A(t)A(t) be the value of Account A after tt years, and B(t)B(t) be the value of Account B. The formulas are A(t)=1000e0.05tA(t) = 1000e^{0.05t} and B(t)=1200(1.04)tB(t) = 1200(1.04)^t. We need to find tt when A(t)>B(t)A(t) > B(t). We solve for A(t)=B(t)A(t) = B(t) to find the crossover point: 1000e0.05t=1200(1.04)t1000e^{0.05t} = 1200(1.04)^t. Divide by 1000: e0.05t=1.2(1.04)te^{0.05t} = 1.2(1.04)^t. Take the natural logarithm of both sides: ln⁡(e0.05t)=ln⁡(1.2(1.04)t)\ln(e^{0.05t}) = \ln(1.2(1.04)^t). This simplifies to 0.05t=ln⁡(1.2)+ln⁡((1.04)t)0.05t = \ln(1.2) + \ln((1.04)^t), which is 0.05t=ln⁡(1.2)+tln⁡(1.04)0.05t = \ln(1.2) + t \ln(1.04). Rearranging to solve for tt: t(0.05−ln⁡(1.04))=ln⁡(1.2)t(0.05 - \ln(1.04)) = \ln(1.2). Using a calculator, ln⁡(1.2)≈0.1823\ln(1.2) \approx 0.1823 and ln⁡(1.04)≈0.0392\ln(1.04) \approx 0.0392. So, t(0.05−0.0392)=0.1823t(0.05 - 0.0392) = 0.1823, which gives t(0.0108)=0.1823t(0.0108) = 0.1823. Thus, t≈0.1823/0.0108≈16.88t \approx 0.1823 / 0.0108 \approx 16.88 years. The closest answer is 17.0 years.

Question 2

The market value of a brand is growing exponentially. In 2020, its value was $1.8 million. In 2022, its value reached $2.7 million. Assuming the growth model is of the form $V(t) = V_0 e^{kt}wherewheret=0$ corresponds to the year 2018, what was the brand's value in 2018?

  1. $0.9 million
  2. $1.2 million (correct answer)
  3. $1.35 million
  4. $1.5 million
Explanation: Let t=0t=0 represent the year 2018. Then 2020 is t=2t=2 and 2022 is t=4t=4. We are given V(2)=1.8V(2) = 1.8 and V(4)=2.7V(4) = 2.7. The model is V(t)=V0ektV(t) = V_0 e^{kt}. We have two equations: 1.8=V0e2k1.8 = V_0 e^{2k} and 2.7=V0e4k2.7 = V_0 e^{4k}. To find kk, we can divide the second equation by the first: 2.71.8=V0e4kV0e2k\frac{2.7}{1.8} = \frac{V_0 e^{4k}}{V_0 e^{2k}}, which simplifies to 1.5=e2k1.5 = e^{2k}. Now we can find V0V_0 using the first equation: 1.8=V0e2k1.8 = V_0 e^{2k}. Since we know e2k=1.5e^{2k}=1.5, we substitute this in: 1.8=V0(1.5)1.8 = V_0(1.5). Solving for V0V_0 gives V0=1.8/1.5=1.2V_0 = 1.8 / 1.5 = 1.2. So the initial value in 2018 was $1.2 million.

Question 3

The value VV (in thousands of dollars) of a company's stock is modeled by V(t)=PektV(t) = P e^{kt}, where tt is the number of years since 2020. An analyst plots the natural logarithm of the stock's value, ln⁡(V(t))\ln(V(t)), against time tt. The resulting graph is a straight line with the equation y=0.075t+5.8y = 0.075t + 5.8. What do the slope and y-intercept of this line represent?

  1. The slope is the continuous annual growth rate of the stock's value, and the y-intercept is the natural logarithm of its initial value in thousands of dollars. (correct answer)
  2. The slope is the initial value of the stock in thousands of dollars, and the y-intercept is the continuous annual growth rate.
  3. The slope is the annual growth factor, and the y-intercept is the initial value in thousands of dollars.
  4. The slope is the percentage growth rate (7.5%), and the y-intercept is the initial value of the stock in thousands of dollars.
Explanation: Given the model V(t)=PektV(t) = P e^{kt}, we can take the natural logarithm of both sides: ln⁡(V(t))=ln⁡(Pekt)=ln⁡(P)+ln⁡(ekt)=ln⁡(P)+kt\ln(V(t)) = \ln(P e^{kt}) = \ln(P) + \ln(e^{kt}) = \ln(P) + kt. This can be written as ln⁡(V(t))=kt+ln⁡(P)\ln(V(t)) = kt + \ln(P). This equation is in the form of a line, y=mt+by = mt + b, where y=ln⁡(V(t))y = \ln(V(t)), the slope mm is the continuous growth rate kk, and the y-intercept bb is ln⁡(P)\ln(P), the natural logarithm of the initial value. Comparing this to the given equation y=0.075t+5.8y = 0.075t + 5.8, we see that the slope k=0.075k=0.075 is the continuous annual growth rate, and the y-intercept ln⁡(P)=5.8\ln(P)=5.8 is the natural logarithm of the initial value PP.

Question 4

The market relevance of a particular product is found to decay exponentially after its launch, with a half-life of 9 months. If the product's initial relevance score is 100, approximately how many months will it take for the score to fall to 20?

  1. 2.9 months
  2. 14.4 months
  3. 18.0 months
  4. 20.9 months (correct answer)
Explanation: The model for exponential decay using half-life is S(t)=S0(1/2)t/TS(t) = S_0 (1/2)^{t/T}, where S0S_0 is the initial score, TT is the half-life, and tt is time. Here, S0=100S_0 = 100 and T=9T = 9 months. The model is S(t)=100(0.5)t/9S(t) = 100(0.5)^{t/9}. We want to find tt when S(t)=20S(t) = 20. So, 20=100(0.5)t/920 = 100(0.5)^{t/9}. Dividing by 100 gives 0.2=(0.5)t/90.2 = (0.5)^{t/9}. To solve for tt, we take the natural logarithm of both sides: ln⁡(0.2)=ln⁡((0.5)t/9)\ln(0.2) = \ln((0.5)^{t/9}), which simplifies to ln⁡(0.2)=(t/9)ln⁡(0.5)\ln(0.2) = (t/9)\ln(0.5). Solving for tt: t=9ln⁡(0.2)ln⁡(0.5)≈9−1.6094−0.6931≈9(2.3219)≈20.897t = 9 \frac{\ln(0.2)}{\ln(0.5)} \approx 9 \frac{-1.6094}{-0.6931} \approx 9(2.3219) \approx 20.897 months.

Question 5

The number of households in a target demographic is currently 2 million and is projected to grow exponentially to 3 million over the next 5 years. A company currently generates $4.5 million in annual revenue from this demographic and maintains a constant revenue per household. Assuming the demographic growth trend continues, what is the company's projected annual revenue in 10 years?

  1. $6.75 million
  2. $9.00 million
  3. $10.125 million (correct answer)
  4. $13.50 million
Explanation: First, model the population growth D(t)=D0(b)tD(t) = D_0(b)^t. Let D0=2D_0=2 million. D(5)=3D(5) = 3 million. So 3=2(b)53 = 2(b)^5, which means the 5-year growth factor is b5=1.5b^5 = 1.5. In 10 years, the population will be D(10)=D0(b)10=D0(b5)2=2(1.5)2=2(2.25)=4.5D(10) = D_0(b)^{10} = D_0(b^5)^2 = 2(1.5)^2 = 2(2.25) = 4.5 million households. The current revenue per household is $4.5 million / 2 million households = $2.25 per household. Assuming this rate is constant, the projected revenue in 10 years will be 4.54.5 million households ×\times $2.25/household = $10.125 million.

Question 6

Two investment accounts start with the same principal. Account A grows at 6% annually compounded continuously, while Account B grows at 6.2% annually compounded quarterly. After how many complete years will Account B first exceed Account A by at least 2%?

  1. 8 years
  2. 12 years (correct answer)
  3. 15 years
  4. 18 years
Explanation: Account A: A(t) = Pe^(0.06t). Account B: B(t) = P(1.0155)^(4t). We need B(t) ≥ 1.02A(t), so P(1.0155)^(4t) ≥ 1.02Pe^(0.06t). This gives (1.0155)^(4t) ≥ 1.02e^(0.06t). Taking natural log: 4t ln(1.0155) ≥ ln(1.02) + 0.06t. Solving: 4t(0.0154) ≥ 0.0198 + 0.06t, so 0.0616t ≥ 0.0198 + 0.06t, giving 0.0016t ≥ 0.0198, so t ≥ 12.375. Therefore, after 12 complete years, Account B first exceeds Account A by at least 2%.

Question 7

An e-commerce platform's daily active users follow U(t)=25000(1.05)tU(t) = 25000(1.05)^t, where tt is days since launch. Competition analysis shows that without intervention, growth will plateau when the user base reaches 180,000. If the company implements growth strategies that extend the exponential phase until 250,000 users before plateauing begins, how many additional days of exponential growth are gained?

  1. 6.8 days (correct answer)
  2. 7.9 days
  3. 8.4 days
  4. 9.2 days
Explanation: Without intervention, exponential growth ends when U(t) = 180,000: 25000(1.05)^t = 180,000, so (1.05)^t = 7.2, giving t = ln(7.2)/ln(1.05) = 1.9741/0.0488 = 40.45 days. With intervention, exponential growth continues until U(t) = 250,000: 25000(1.05)^t = 250,000, so (1.05)^t = 10, giving t = ln(10)/ln(1.05) = 2.3026/0.0488 = 47.18 days. Additional days = 47.18 - 40.45 = 6.73 ≈ 6.8 days.

Question 8

A company aims to increase its capital assets from $2.5 million to $10 million in 6 years. If the growth is modeled by continuous compounding, what continuous annual interest rate is required to achieve this goal?

  1. 23.1% (correct answer)
  2. 26.0%
  3. 33.6%
  4. 50.0%
Explanation: The formula for continuous compounding is A=PertA = Pe^{rt}, where PP is the principal amount, AA is the final amount, rr is the annual interest rate, and tt is the time in years. We are given P=2.5P = 2.5 million, A=10A = 10 million, and t=6t = 6 years. We need to solve for rr. 10=2.5er(6)10 = 2.5 e^{r(6)}. First, divide both sides by 2.5: 4=e6r4 = e^{6r}. Next, take the natural logarithm of both sides: ln⁡(4)=ln⁡(e6r)\ln(4) = \ln(e^{6r}), which simplifies to ln⁡(4)=6r\ln(4) = 6r. Finally, solve for rr: r=ln⁡(4)6≈1.38636≈0.23105r = \frac{\ln(4)}{6} \approx \frac{1.3863}{6} \approx 0.23105. As a percentage, this is approximately 23.1%.

Question 9

A startup company has 50,000 users. For the first 4 months, its user base grows by 15% each month. After a change in marketing strategy, the user base grows by 5% each month for the next 8 months. Approximately how many users will the company have at the end of the full year?

  1. 89,800
  2. 129,200 (correct answer)
  3. 156,900
  4. 267,400
Explanation: This is a two-stage discrete growth problem. Let U(t)U(t) be the number of users after tt months. The initial number is U(0)=50,000U(0) = 50,000. For the first 4 months, the growth factor is 1.151.15. The number of users after 4 months is U(4)=50,000(1.15)4≈50,000(1.7490)≈87,450U(4) = 50,000(1.15)^4 \approx 50,000(1.7490) \approx 87,450. For the next 8 months, this new amount grows by a factor of 1.051.05 each month. The final number of users after 12 months is U(12)=U(4)(1.05)8≈87,450(1.4775)≈129,164U(12) = U(4)(1.05)^8 \approx 87,450(1.4775) \approx 129,164. The closest answer is 129,200.

Question 10

A bank offers two savings accounts. Account X offers a 4.80% nominal annual interest rate compounded monthly. Account Y offers a 4.88% nominal annual interest rate compounded quarterly. Which statement is correct?

  1. Account X should be chosen because its effective annual yield is approximately 0.08% higher.
  2. Account Y should be chosen because its effective annual yield is approximately 0.08% higher.
  3. Account X should be chosen because compounding monthly is always better than compounding quarterly.
  4. Account Y should be chosen because its effective annual yield is approximately 0.06% higher. (correct answer)
Explanation: To compare the accounts, we must calculate the effective annual yield (APY) for each. The formula is APY=(1+r/n)n−1APY = (1 + r/n)^n - 1. For Account X: r=0.048r = 0.048, n=12n = 12. APYX=(1+0.048/12)12−1=(1.004)12−1≈0.04907APY_X = (1 + 0.048/12)^{12} - 1 = (1.004)^{12} - 1 \approx 0.04907, or 4.907%. For Account Y: r=0.0488r = 0.0488, n=4n = 4. APYY=(1+0.0488/4)4−1=(1.0122)4−1≈0.04970APY_Y = (1 + 0.0488/4)^4 - 1 = (1.0122)^4 - 1 \approx 0.04970, or 4.970%. Since APYY>APYXAPY_Y > APY_X, Account Y is the better choice. The difference in yields is 4.9704.970% - 4.907% = 0.063%. Therefore, Account Y should be chosen because its APY is approximately 0.06% higher.

Question 11

An industrial machine purchased for $250,000 depreciates in value exponentially. Its value after 5 years is $100,000. How many additional years after year 5 will it take for the machine's value to fall to $25,000?

  1. 5.0 years
  2. 7.6 years (correct answer)
  3. 12.6 years
  4. 15.2 years
Explanation: The value model is V(t)=PbtV(t) = P b^t, where P=250,000P=250,000. We are given V(5)=100,000V(5)=100,000. So, 100,000=250,000⋅b5100,000 = 250,000 \cdot b^5, which gives b5=0.4b^5 = 0.4. The model is V(t)=250,000⋅(0.4)t/5V(t) = 250,000 \cdot (0.4)^{t/5}. We need to find the time tt when the value is 25,00025,000: 25,000=250,000⋅(0.4)t/525,000 = 250,000 \cdot (0.4)^{t/5}. This simplifies to 0.1=(0.4)t/50.1 = (0.4)^{t/5}. Taking the natural logarithm of both sides: ln⁡(0.1)=(t/5)ln⁡(0.4)\ln(0.1) = (t/5) \ln(0.4). Solving for tt: t=5ln⁡(0.1)ln⁡(0.4)≈5−2.3026−0.9163≈12.56t = 5 \frac{\ln(0.1)}{\ln(0.4)} \approx 5 \frac{-2.3026}{-0.9163} \approx 12.56 years. The question asks for the additional years after year 5, so we calculate 12.56−5=7.5612.56 - 5 = 7.56 years. The closest answer is 7.6 years.

Question 12

A company's annual profit P(t)P(t) (in millions of dollars) from a specific product line is modeled by the equation P(t)=4.8e−0.12t+1.5P(t) = 4.8 e^{-0.12t} + 1.5, where tt is the number of years after the product's launch. Which statement accurately describes the business implications of this model?

  1. The product's initial annual profit is $4.8 million, and it will eventually become unprofitable.
  2. The product's annual profit is expected to decline over time but will stabilize at a long-term level of $1.5 million. (correct answer)
  3. The annual profit is declining by a constant 12% each year.
  4. The product will become unprofitable because its profits approach a long-term value of zero.
Explanation: The model consists of a decaying exponential term, 4.8e−0.12t4.8 e^{-0.12t}, and a constant term, 1.51.5. The initial profit at launch (t=0t=0) is P(0)=4.8e0+1.5=4.8+1.5=6.3P(0) = 4.8e^0 + 1.5 = 4.8 + 1.5 = 6.3 million. As time tt increases (t→∞t \to \infty), the exponential term e−0.12te^{-0.12t} approaches 0. Therefore, the profit function P(t)P(t) approaches 4.8(0)+1.5=1.54.8(0) + 1.5 = 1.5. This means the profit declines from its initial value but levels off, or stabilizes, at a long-term value of $1.5 million per year. It does not become unprofitable.

Question 13

A pharmaceutical company is tracking the concentration of two drugs in a patient's bloodstream. Drug A has an initial concentration of 80 mg/L and decays at 12% per hour. Drug B has an initial concentration of 40 mg/L and decays at 8% per hour. The drugs interact when their combined concentration falls below 60 mg/L.

Based on the information in the passage, after how many hours will the drug interaction begin?

  1. 4.8 hours
  2. 5.4 hours
  3. 6.1 hours (correct answer)
  4. 6.7 hours
Explanation: Drug A concentration: A(t) = 80e^(-0.12t). Drug B concentration: B(t) = 40e^(-0.08t). Combined concentration: C(t) = 80e^(-0.12t) + 40e^(-0.08t). We need C(t) = 60, so 80e^(-0.12t) + 40e^(-0.08t) = 60. Dividing by 20: 4e^(-0.12t) + 2e^(-0.08t) = 3. This requires numerical solution. Testing t = 6: C(6) = 80e^(-0.72) + 40e^(-0.48) = 80(0.4868) + 40(0.6188) = 38.94 + 24.75 = 63.69 mg/L. Testing t = 6.5: C(6.5) = 80e^(-0.78) + 40e^(-0.52) = 80(0.4584) + 40(0.5945) = 36.67 + 23.78 = 60.45 mg/L. Testing t = 6.1: C(6.1) = 80e^(-0.732) + 40e^(-0.488) = 80(0.4814) + 40(0.6141) = 38.51 + 24.56 = 63.07 mg/L. Using more precise numerical methods, the combined concentration reaches 60 mg/L at approximately t = 6.1 hours.

Question 14

An investment account earns interest compounded continuously. An initial deposit triples in value in 15 years. What initial investment PP is required to have a balance of $162,000 in 30 years?

  1. $18,000 (correct answer)
  2. $27,000
  3. $40,500
  4. $54,000
Explanation: The model for continuous compounding is A(t)=PektA(t) = P e^{kt}. If the investment triples in 15 years, then A(15)=3PA(15) = 3P. So, 3P=Pek⋅153P = P e^{k \cdot 15}, which simplifies to 3=e15k3 = e^{15k}. We want to find the value at t=30t=30. A(30)=Pek⋅30=P(e15k)2A(30) = P e^{k \cdot 30} = P (e^{15k})^2. Since we know e15k=3e^{15k} = 3, we can substitute this into the equation: A(30)=P(3)2=9PA(30) = P(3)^2 = 9P. We are given that the balance at 30 years is $162,000. So, 9P=162,0009P = 162,000. Solving for PP gives P=162,000/9=18,000P = 162,000 / 9 = 18,000.