Exponential Growth/Decay Models - Business Calculus

1. An investment account grows according to $\frac{dA}{dt} = 0.065A + 1200$, where $A(t)$ is the account balance in dollars and $t$ is time in years. The account starts with $5,000. This model differs from pure exponential growth because of the constant term. What is the account balance after 5 years?
   Answer: $28,462 This is a first-order linear ODE: $\frac{dA}{dt} - 0.065A = 1200$. The integrating factor is $e^{-0.065t}$. The general solution is $A(t) = Ce^{0.065t} - \frac{1200}{0.065}$. With $A(0) = 5000$: $5000 = C - 18461.54$, so $C = 23461.54$. Thus $A(t) = 23461.54e^{0.065t} - 18461.54$. After 5 years: $A(5) = 23461.54e^{0.325} - 18461.54 = 28462$. Choice A uses $A(t) = 5000e^{0.065t} + 1200t$. Choice B uses $A(t) = (5000 + 1200)e^{0.065t}$. Choice D uses incorrect equilibrium value.
2. The resale value $V(t)$ of a commercial printer in dollars is modeled by $dV/dt = -0.2V$, with an initial value $V(0) = 50,000$. The cumulative cost of repairs $C(t)$ is modeled by $C(t) = 2000t + 500$. At what time $t$ (in years) is the rate of depreciation of the printer's value equal to the rate of increase of its repair cost?
   Answer: $5 \ln(5)$ When you encounter differential equations in business calculus, you're looking at rates of change. This problem asks you to find when two rates are equal: the rate of depreciation (how fast value decreases) and the rate of repair cost increase. First, solve the differential equation $\frac{dV}{dt} = -0.2V$ with $V(0) = 50,000$. This separable equation gives you $V(t) = 50,000e^{-0.2t}$. The rate of depreciation is the absolute value: $\left|\frac{dV}{dt}\right| = 0.2V = 10,000e^{-0.2t}$. For repair costs, $C(t) = 2000t + 500$, so $\frac{dC}{dt} = 2000$ (constant rate). Set the rates equal: $10,000e^{-0.2t} = 2000$. Dividing by 2000 gives $5e^{-0.2t} = 1$, so $e^{-0.2t} = \frac{1}{5}$. Taking the natural logarithm: $-0.2t = \ln\left(\frac{1}{5}\right) = -\ln(5)$. Therefore, $t = \frac{\ln(5)}{0.2} = 5\ln(5)$. Answer choice A is wrong because we found a specific time when rates are equal. Choice B ($10\ln(5)$) results from incorrectly using $0.1$ instead of $0.2$ in the differential equation. Choice D ($\frac{\ln(5)}{5}$) comes from confusing the coefficient, using $1$ instead of $0.2$. The correct answer is C: $5\ln(5)$. Study tip: When solving "rate equals rate" problems, always solve the differential equation first, then find derivatives, and finally set them equal. Double-check your algebra with the differential equation's coefficient.
3. The value of Investment A, $A(t)$, grows according to $dA/dt = 0.04A$, and the value of Investment B, $B(t)$, grows according to $dB/dt = 0.06B$. Initially, $A(0) = \$2000$ and $B(0) = \$1000$. At what time $t$ (in years) will the instantaneous rate of growth of Investment A be equal to the instantaneous rate of growth of Investment B?
   Answer: $50 \ln(4/3)$ The question asks for the time $t$ when $dA/dt = dB/dt$. The solutions for the values are $A(t) = 2000e^{0.04t}$ and $B(t) = 1000e^{0.06t}$. The rates of growth are $dA/dt = 0.04A(t)$ and $dB/dt = 0.06B(t)$. Setting them equal: $0.04 A(t) = 0.06 B(t)$. Substitute the expressions for $A(t)$ and $B(t)$: $0.04(2000e^{0.04t}) = 0.06(1000e^{0.06t})$. This simplifies to $80e^{0.04t} = 60e^{0.06t}$. Rearranging gives $\frac{80}{60} = \frac{e^{0.06t}}{e^{0.04t}}$, which is $\frac{4}{3} = e^{0.02t}$. Taking the natural logarithm of both sides: $\ln(4/3) = 0.02t$. Solving for $t$ gives $t = \frac{\ln(4/3)}{0.02} = 50 \ln(4/3)$.
4. A biotechnology company models the concentration of an enzyme in a bioreactor using $\frac{dC}{dt} = -kC$, where $C(t)$ is concentration in mg/mL and $t$ is time in hours. The company knows that after 3 hours, the concentration drops to 75% of its initial value. They need to determine when the concentration will be exactly 40% of the initial concentration for optimal harvesting. Additionally, they want to know the percentage rate of decrease at that optimal time.
   Answer: Time: 10.1 hours; Percentage rate: -9.6% per hour From $C(3) = 0.75C_0$: $0.75 = e^{-3k}$, so $k = \frac{\ln(0.75)}{-3} = 0.096$ hr⁻¹. For $C(t) = 0.4C_0$: $0.4 = e^{-0.096t}$, giving $t = \frac{\ln(0.4)}{-0.096} = 9.56$ hours. The percentage rate of decrease is $\frac{1}{C}\frac{dC}{dt} \times 100\% = -k \times 100\% = -9.6\%$ per hour (constant for exponential decay). Choice A uses $t = \frac{\ln(2.5)}{0.096}$ and wrong rate calculation. Choice B uses correct rate but $t = \frac{\ln(0.4)}{-0.1}$. Choice D uses $k = \frac{\ln(4/3)}{3}$ and corresponding time calculation.
5. A mobile application's user base, $P(t)$ in thousands, grows exponentially with time $t$ in months. After 3 months, the user base is 10 thousand, and at that instant, it is growing at a rate of 2 thousand users per month. Based on this model, what was the initial number of users, $P(0)$?
   Answer: 5.49 thousand When you encounter exponential growth problems, you're working with functions of the form $P(t) = P_0 e^{kt}$, where $P_0$ is the initial value and $k$ is the growth constant. The key insight is that you need both a point on the curve and the growth rate to find the initial conditions. Given information: $P(3) = 10$ thousand and $P'(3) = 2$ thousand per month. Since $P'(t) = kP_0 e^{kt} = kP(t)$, we can find the growth constant: $P'(3) = kP(3)$, so $2 = k \cdot 10$, giving us $k = 0.2$. Now substitute into the original equation: $10 = P_0 e^{0.2 \cdot 3} = P_0 e^{0.6}$. Since $e^{0.6} \approx 1.822$, we get $P_0 = \frac{10}{1.822} \approx 5.49$ thousand users. Looking at the wrong answers: B) 4.00 likely comes from incorrectly calculating $\frac{10}{2.5}$ using an approximation error. C) 8.00 suggests someone may have used $P_0 = P(3) - P'(3) \cdot 3 = 10 - 6$, incorrectly applying linear thinking to exponential growth. D) 18.22 appears to be $10 \cdot e^{0.6}$, which would result from confusing the exponential relationship direction. The answer is A) 5.49 thousand. Study tip: In exponential growth problems, always use the derivative relationship $P'(t) = kP(t)$ to find the growth constant first. This connects the instantaneous rate of change to the function value, which is crucial for determining initial conditions.
6. A company's revenue $R(t)$ grows at a rate proportional to its current revenue. The revenue in 2015 was $2 million, and the relative growth rate was 5% per year. What is the expected absolute growth rate, $dR/dt$, in the year 2025?
   Answer: $100,000 $e^{0.5}$$ per year When you see "grows at a rate proportional to its current revenue," you're dealing with exponential growth. This means $\frac{dR}{dt} = kR$, where k is the growth constant. Since the relative growth rate is 5% per year, we have $k = 0.05$. The general solution to this differential equation is $R(t) = R_0 e^{kt}$, where $R_0$ is the initial revenue. Using 2015 as our starting point (t = 0), we have $R_0 = 2,000,000$. So $R(t) = 2,000,000 e^{0.05t}$. For the year 2025 (t = 10 years later), we need $\frac{dR}{dt}$ at that time. Since $\frac{dR}{dt} = kR(t) = 0.05R(t)$, we substitute: $\frac{dR}{dt} = 0.05 \times 2,000,000 e^{0.05 \times 10} = 100,000 e^{0.5}$ Choice A ($2,000,000 e^{0.5}$) incorrectly uses the full revenue amount instead of applying the 5% growth rate to find the derivative. Choice B ($100,000$) correctly finds 5% of the initial revenue but fails to account for the exponential growth over 10 years—it treats this as if the revenue stayed constant. Choice D ($100,000 e^{-0.5}$) has the wrong sign in the exponent, which would indicate decay rather than growth. The correct answer is C: $100,000 e^{0.5}$ per year. Study tip: For exponential growth problems, remember that the absolute growth rate $\frac{dR}{dt}$ equals the relative growth rate times the current value, not the initial value.
7. The population $P(t)$ of a city, in thousands, is modeled by the differential equation $dP/dt = kP$. If the solution to this equation is $P(t) = 150e^{0.03t}$, which of the following statements correctly interprets the relationship between $P(t)$ and $\ln(P(t))$?
   Answer: A graph of $\ln(P(t))$ versus $t$ is a line with slope $0.03$ and $y$-intercept $\ln(150)$. Given the solution $P(t) = 150e^{0.03t}$, we can analyze $\ln(P(t))$ by taking the natural logarithm of both sides. Using the properties of logarithms, we get: $\ln(P(t)) = \ln(150e^{0.03t}) = \ln(150) + \ln(e^{0.03t}) = \ln(150) + 0.03t$. This equation is in the form $y = mx + b$, where $y = \ln(P(t))$, $x = t$, the slope $m$ is $0.03$, and the $y$-intercept $b$ is $\ln(150)$. Therefore, a graph of $\ln(P(t))$ versus $t$ is a straight line with a slope of $0.03$ and a $y$-intercept of $\ln(150)$.
8. A pharmaceutical company models the concentration of a drug in the bloodstream using the differential equation $\frac{dC}{dt} = -0.12C$, where $C(t)$ is the concentration in mg/L and $t$ is time in hours. If the initial concentration is 80 mg/L, what is the half-life of the drug, and what concentration remains after 8 hours?
   Answer: Half-life: 5.78 hours; Concentration after 8 hours: 30.4 mg/L The solution to $\frac{dC}{dt} = -0.12C$ is $C(t) = 80e^{-0.12t}$. For half-life, set $40 = 80e^{-0.12t}$, so $0.5 = e^{-0.12t}$, giving $t = \frac{\ln(0.5)}{-0.12} = 5.78$ hours. After 8 hours: $C(8) = 80e^{-0.12(8)} = 80e^{-0.96} = 30.4$ mg/L. Choice B uses incorrect half-life formula $t = \frac{0.693}{0.12}$ but gets wrong concentration. Choice C has correct half-life but uses $C(8) = 80e^{-0.693}$ (substituting half-life value). Choice D uses $t = \frac{1}{0.12}$ and $C(8) = 80(0.5)$.
9. The number of employees $E(t)$ at a startup grows according to $dE/dt = kE$. The company starts with 20 employees. Two years later, it has 80 employees. How long from its start date did it take for the company to have 40 employees?
   Answer: 1 year When you encounter a differential equation like $\frac{dE}{dt} = kE$, you're dealing with exponential growth. This separable differential equation has the general solution $E(t) = E_0 e^{kt}$, where $E_0$ is the initial value and $k$ is the growth constant. Starting with 20 employees at $t = 0$, we have $E(t) = 20e^{kt}$. Since the company has 80 employees after 2 years, we can find $k$: $80 = 20e^{2k}$, which gives us $4 = e^{2k}$, so $k = \frac{\ln(4)}{2}$. Now we need to find when $E(t) = 40$. Substituting: $40 = 20e^{kt}$, which simplifies to $2 = e^{kt}$. Taking the natural log: $\ln(2) = kt = \frac{\ln(4)}{2} \cdot t$. Since $\ln(4) = \ln(2^2) = 2\ln(2)$, we get $\ln(2) = \frac{2\ln(2)}{2} \cdot t = \ln(2) \cdot t$. Therefore, $t = 1$ year. Choice A gives $\frac{\ln(2)}{\ln(4)} = \frac{\ln(2)}{2\ln(2)} = \frac{1}{2}$, which is backwards. Choice C gives $\frac{\ln(4)}{\ln(2)} = \frac{2\ln(2)}{\ln(2)} = 2$, which would be the time to reach 80 employees. Choice D is 0.5 years, which corresponds to choice A's value. Study tip: In exponential growth problems, always set up your equation systematically: find the general form, use given conditions to find constants, then solve for your target. The algebra often simplifies more than it initially appears.
10. A financial plan requires an initial investment of $10,000 to grow to $25,000 in 12 years. Assuming the investment grows according to the differential equation $\frac{dA}{dt} = rA$, where $A$ is the amount and $t$ is time in years, what annual interest rate $r$, compounded continuously, is required to meet this goal?
    Answer: 7.64% The solution to the differential equation is $A(t) = A_0 e^{rt}$. We are given the initial amount $A_0 = 10000$, and the future amount $A(12) = 25000$. We need to find the rate $r$. Substitute the values into the equation: $25000 = 10000 e^{r \cdot 12}$. First, divide by 10000: $2.5 = e^{12r}$. To solve for $r$, take the natural logarithm of both sides: $\ln(2.5) = 12r$. Finally, isolate $r$: $r = \frac{\ln(2.5)}{12} \approx \frac{0.91629}{12} \approx 0.07635$. As a percentage, this is approximately 7.64%.
11. A company is testing two online advertising strategies, A and B. The number of daily new visitors to its website from each strategy, $V_A(t)$ and $V_B(t)$, is modeled by exponential growth, where $t$ is the number of days. For Strategy A, the initial number of visitors is 100, and the doubling time is 10 days. For Strategy B, the growth is modeled by $\frac{dV_B}{dt} = 0.08 V_B$ with an initial condition $V_B(0) = 80$. After how many full days will the number of daily visitors from Strategy B first exceed the number from Strategy A?
    Answer: 21 days First, define the models. For Strategy A, $V_A(t) = 100e^{k_A t}$. The doubling time is $T_2 = \frac{\ln 2}{k_A} = 10$, so $k_A = \frac{\ln 2}{10} \approx 0.0693$. Thus, $V_A(t) = 100e^{(\frac{\ln 2}{10})t}$. For Strategy B, from $\frac{dV_B}{dt} = 0.08 V_B$, we have $k_B = 0.08$. With $V_B(0)=80$, the model is $V_B(t) = 80e^{0.08t}$. We need to find when $V_B(t) > V_A(t)$: $80e^{0.08t} > 100e^{(\frac{\ln 2}{10})t}$. Divide by the terms: $\frac{e^{0.08t}}{e^{(\frac{\ln 2}{10})t}} > \frac{100}{80}$, which is $e^{(0.08 - \frac{\ln 2}{10})t} > 1.25$. Take the natural log: $(0.08 - \frac{\ln 2}{10})t > \ln(1.25)$. $t > \frac{\ln(1.25)}{0.08 - \frac{\ln 2}{10}} \approx \frac{0.2231}{0.08 - 0.0693} = \frac{0.2231}{0.0107} \approx 20.85$. Since $t$ must be an integer number of full days, the first day this condition is met is day 21.
12. The number of subscribers $P(t)$ to a streaming service is growing exponentially. The service had 1 million subscribers on January 1, 2020. By January 1, 2022, it had 1.44 million subscribers. Which differential equation best models this growth, where $P$ is in millions and $t$ is in years since January 1, 2020?
    Answer: $dP/dt = (\ln(1.2))P$ The model is $dP/dt = kP$, with solution $P(t) = P_0 e^{kt}$. We have $P_0 = 1$ (million) and $P(2) = 1.44$ (million). We substitute these values into the solution: $1.44 = 1 \cdot e^{k(2)}$. To find $k$, we solve for it: $1.44 = e^{2k}$. Taking the natural logarithm of both sides gives $\ln(1.44) = 2k$. Since $1.44 = 1.2^2$, we have $\ln(1.2^2) = 2k$, which simplifies to $2\ln(1.2) = 2k$. Therefore, $k = \ln(1.2)$. The differential equation is $dP/dt = \ln(1.2)P$.
13. The sales revenue of a startup, $S(t)$, is modeled by the differential equation $\frac{dS}{dt} = kS$ for a constant $k > 0$. Which of the following statements is a direct mathematical consequence of this model?
    Answer: The time required for revenue to grow from $1M to $2M is equal to the time required for it to grow from $5M to $10M. The solution to $\frac{dS}{dt} = kS$ is $S(t)=S_0e^{kt}$. The time it takes for a quantity to double (from $S_1$ to $2S_1$) is found by solving $2S_1 = S_1 e^{k \Delta t}$, which gives $2 = e^{k \Delta t}$, so the doubling time is $\Delta t = \frac{\ln 2}{k}$. This time is constant and does not depend on the starting value $S_1$. Therefore, the time to grow from $1M to $2M is the same as from $5M to $10M. Choice A is incorrect because profit depends on costs, which are not described. Choice C is incorrect because $\frac{dS}{dt} = kS(t)$ increases as $S(t)$ increases. Choice D describes a logistic model, not an exponential one.
14. A tech company finds its unsold inventory of a smartphone model loses value exponentially. The rate of loss in value is proportional to its current value. The model's value decreases by 20% in the first 6 months. How long, in months, will it take for the inventory's value to be reduced to 30% of its original value?
    Answer: 32.4 months Let the value be $V(t) = V_0 e^{kt}$. A 20% decrease means the value is 80% of the original, so $V(6) = 0.80 V_0$. We can use this to find $k$: $0.80 V_0 = V_0 e^{6k}$, so $0.80 = e^{6k}$, which gives $k = \frac{\ln(0.8)}{6}$. We want to find the time $t$ when the value is 30% of the original, i.e., $V(t) = 0.30 V_0$. Set up the equation: $0.30 V_0 = V_0 e^{kt}$, so $0.30 = e^{kt}$. Solving for $t$ gives $t = \frac{\ln(0.3)}{k}$. Substituting our expression for $k$: $t = \frac{\ln(0.3)}{\frac{\ln(0.8)}{6}} = 6 \frac{\ln(0.3)}{\ln(0.8)} \approx 6 \frac{-1.2040}{-0.2231} \approx 6(5.396) \approx 32.38$ months.
15. The value of a continuously compounded investment, $A(t)$, satisfies the differential equation $dA/dt = rA$. An initial investment of $10,000 grows to $12,000 in 4 years. What is the instantaneous rate of change of the investment's value, $dA/dt$, at the moment its value reaches $15,000?
    Answer: $3750 \ln(1.2)$ per year When you encounter a differential equation like $\frac{dA}{dt} = rA$ for continuously compounded growth, you're dealing with exponential growth where the rate of change is proportional to the current amount. The key insight is that once you find the growth rate $r$, you can calculate the instantaneous rate of change at any value using the same differential equation. First, solve for the growth rate $r$. The general solution to $\frac{dA}{dt} = rA$ is $A(t) = A_0 e^{rt}$. With $A_0 = 10000$ and $A(4) = 12000$: $12000 = 10000e^{4r}$ $1.2 = e^{4r}$ $\ln(1.2) = 4r$ $r = \frac{\ln(1.2)}{4}$ Now, when the investment reaches $15,000, the instantaneous rate of change is: $\frac{dA}{dt} = rA = \frac{\ln(1.2)}{4} \times 15000 = 3750\ln(1.2)$ Answer A ($15000\ln(1.2)$) incorrectly uses the full value without dividing by 4 to get the correct growth rate. Answer B ($2500\ln(1.2)$) appears to use an incorrect calculation, possibly confusing the growth amount with the rate formula. Answer C ($500) completely ignores the logarithmic nature of continuous compounding and likely uses simple interest thinking. Remember: in continuous compounding problems, always find the growth rate $r$ first using the given data points, then apply $\frac{dA}{dt} = rA$ at the specified value. The natural logarithm will almost always appear in your final answer for these problems.
16. The resale value, $V(t)$, of a limited-edition collectible $t$ weeks after its release decreases at a rate proportional to the difference between its current value and its original retail price of $200. This is modeled by $\frac{dV}{dt} = k(V - 200)$, for $V \ge 200$. At release ($t=0$), the value is $1200. Two weeks later ($t=2$), the value is $700. What is the expected resale value after 5 weeks?
    Answer: $377 Let $y(t) = V(t) - 200$. The differential equation becomes $\frac{dy}{dt} = ky$, with solution $y(t) = y_0 e^{kt}$. Translating back, $V(t) - 200 = (V(0) - 200)e^{kt}$. Given $V(0) = 1200$, the model is $V(t) - 200 = (1200 - 200)e^{kt}$, or $V(t) = 200 + 1000e^{kt}$. Use $V(2) = 700$ to find $k$: $700 = 200 + 1000e^{2k}$, so $500 = 1000e^{2k}$, which means $0.5 = e^{2k}$, and $k = \frac{\ln(0.5)}{2}$. Now find $V(5)$: $V(5) = 200 + 1000e^{5k} = 200 + 1000e^{5(\frac{\ln 0.5}{2})} = 200 + 1000e^{2.5 \ln 0.5}$. This simplifies to $V(5) = 200 + 1000(0.5)^{2.5} \approx 200 + 1000(0.17678) \approx 200 + 176.78 \approx 377$.
17. The sales $S(t)$ of a product (in thousands of units) follow an exponential decay model $dS/dt = kS$, where $t$ is the number of years since the product launch. Two years after launch, sales were 8,000 units. Four years after launch, sales were 5,120 units. What were the initial sales, $S(0)$?
    Answer: 12,500 units The solution to the differential equation is $S(t) = S_0 e^{kt}$. We are given $S(2)=8000$ and $S(4)=5120$. So, $8000 = S_0 e^{2k}$ and $5120 = S_0 e^{4k}$. Dividing the second equation by the first gives $\frac{5120}{8000} = \frac{S_0 e^{4k}}{S_0 e^{2k}}$, which simplifies to $0.64 = e^{2k}$. We can use this result directly. From the first equation, $S_0 = 8000 / e^{2k}$. Substituting $e^{2k}=0.64$, we get $S_0 = 8000 / 0.64 = 12,500$.
18. An environmental cleanup project removes pollutants from a lake at a rate proportional to the current pollution level. The differential equation $\frac{dP}{dt} = -0.18P$ models the pollution level $P(t)$ in parts per million, where $t$ is time in years. Initially, the lake has 120 ppm of pollutants. Environmental regulations require the pollution level to drop below 25 ppm before the lake can be reopened for recreational use. How many complete years must pass before the lake meets this requirement?
    Answer: 9 years From $P(t) = 120e^{-0.18t}$, we need $P(t) < 25$: $25 = 120e^{-0.18t}$, so $t = \frac{\ln(120/25)}{0.18} = \frac{\ln(4.8)}{0.18} = \frac{1.569}{0.18} = 8.72$ years. Since we need complete years to pass, the answer is 9 years (since after 8 complete years, $P(8) = 120e^{-1.44} = 29.3$ ppm > 25 ppm, but after 9 years, $P(9) = 120e^{-1.62} = 24.0$ ppm < 25 ppm). Choice A represents rounding down incorrectly. Choice C assumes rounding up from 8.72 to 10. Choice D uses 7 years, which would be insufficient time.
19. The number of subscribers to a streaming service, $N(t)$, grows exponentially according to the model $N(t) = N_0 e^{kt}$, where $t$ is in months. The initial number of subscribers is 2 million. After 6 months, the number of subscribers is 3 million. At what rate are subscribers being added, in millions per month, at the end of the first year ($t=12$)?
    Answer: 0.304 million/month First, find the growth constant $k$. We have $N(0)=2$ and $N(6)=3$. Using the model $N(t) = N_0 e^{kt}$, we get $3 = 2e^{6k}$, so $1.5 = e^{6k}$, and $k = \frac{\ln(1.5)}{6}$. The rate of change is given by the differential equation $\frac{dN}{dt} = k N(t)$. We need to find this rate at $t=12$. First, find the number of subscribers at $t=12$: $N(12) = 2e^{12k} = 2e^{12(\frac{\ln 1.5}{6})} = 2e^{2\ln 1.5} = 2(e^{\ln 1.5})^2 = 2(1.5)^2 = 2(2.25) = 4.5$ million. Now, calculate the rate at $t=12$: $\frac{dN}{dt}|_{t=12} = k \cdot N(12) = \frac{\ln(1.5)}{6} \cdot 4.5 \approx \frac{0.40547}{6} \cdot 4.5 \approx 0.304$ million subscribers per month.
20. A manufacturing company finds that the rate of decrease of defective products follows $\frac{dN}{dt} = -0.08N$, where $N(t)$ is the number of defective items per day and $t$ is time in weeks since implementing quality improvements. Initially, there were 120 defective items per day. The company wants to reduce this to fewer than 20 defective items per day. How long will this take, and what will be the rate of change at that time?
    Answer: Time: 22.4 weeks; Rate of change: -1.6 defective items per day per week When you encounter a differential equation like $\frac{dN}{dt} = -0.08N$, you're dealing with exponential decay. This separable differential equation has the general solution $N(t) = N_0 e^{-0.08t}$, where $N_0$ is the initial value. With $N_0 = 120$ defective items per day, our function becomes $N(t) = 120e^{-0.08t}$. To find when defective items drop below 20 per day, we solve $120e^{-0.08t} = 20$. Dividing both sides by 120 gives $e^{-0.08t} = \frac{1}{6}$. Taking the natural logarithm: $-0.08t = \ln(\frac{1}{6}) = -\ln(6)$, so $t = \frac{\ln(6)}{0.08} \approx \frac{1.79}{0.08} \approx 22.4$ weeks. At $t = 22.4$ weeks, the rate of change is $\frac{dN}{dt} = -0.08N = -0.08(20) = -1.6$ defective items per day per week. Answer D correctly identifies both values. Answer A uses the wrong time calculation—likely solving $e^{-0.08t} = \frac{20}{120}$ incorrectly or confusing the setup. Answer B has the correct time but calculates the rate of change as $-0.08 \times 14$ instead of $-0.08 \times 20$, suggesting they used an intermediate value of $N$. Answer C combines A's incorrect time with the correct rate calculation. Study tip: In exponential decay problems, always verify your final answer by substituting back into both the original equation and the differential equation. The rate of change should equal the derivative at your calculated time point.

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