Business Calculus Quiz: Exponential Growth Decay Models

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Exponential Growth Decay ModelsQuestion 1 of 20

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?

Half-life: 5.78 hours; Concentration after 8 hours: 30.4 mg/L

Half-life: 6.93 hours; Concentration after 8 hours: 34.2 mg/L

Half-life: 5.78 hours; Concentration after 8 hours: 34.2 mg/L

Half-life: 8.33 hours; Concentration after 8 hours: 40.0 mg/L

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Business Calculus Quiz

Business Calculus Quiz: Exponential Growth Decay Models

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

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

Half-life: 5.78 hours; Concentration after 8 hours: 30.4 mg/L (correct answer)
Half-life: 6.93 hours; Concentration after 8 hours: 34.2 mg/L
Half-life: 5.78 hours; Concentration after 8 hours: 34.2 mg/L
Half-life: 8.33 hours; Concentration after 8 hours: 40.0 mg/L

Explanation: The solution to

\frac{dC}{dt} = -0.12C

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)

Question 2

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?

$25,846
$32,154
$28,462 (correct answer)
$21,538

Explanation: 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.

Question 3

The number of active users for a new mobile app, $N(t)$ , is modeled by the differential equation $dN/dt = kN$ , where $t$ is in months. If the number of users triples in the first 6 months, which expression represents the time it takes for the number of users to increase by 50% from any given point in time?

$6 \frac{\ln(1.5)}{\ln(3)}$ (correct answer)
$6 \frac{\ln(3)}{\ln(1.5)}$
$3$
$\frac{\ln(1.5)}{6}$

Explanation: The solution to the differential equation is

N(t) = N_0 e^{kt}

. We are given that

N(6) = 3N_0

. Substituting this gives

3N_0 = N_0 e^{6k}

, so

3 = e^{6k}

, which means

k = \frac{\ln(3)}{6}

. We want to find the time

\Delta t

it takes for the user base to increase by 50%, which means

N(t_0 + \Delta t) = 1.5 N(t_0)

. Using the growth model,

N_0 e^{k(t_0+\Delta t)} = 1.5 N_0 e^{kt_0}

. Dividing both sides by

N_0 e^{kt_0}

gives

e^{k\Delta t} = 1.5

. Solving for

\Delta t

, we get

\Delta t = \frac{\ln(1.5)}{k}

. Substituting the value of

k

we found:

\Delta t = \frac{\ln(1.5)}{\ln(3)/6} = 6 \frac{\ln(1.5)}{\ln(3)}

Question 4

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?

$50 \ln(2)$
$50 \ln(4/3)$ (correct answer)
$25 \ln(2)$
$25 \ln(3/4)$

Explanation: 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)

Question 5

The value $V(t)$ of a piece of equipment is depreciating exponentially. At $t=2$ years, its value is $40,000 and it is depreciating at a rate of $2,000 per year. When will the equipment's value fall to $10,000?

$2 + 20 \ln(4)$ years (correct answer)
$20 \ln(4)$ years
17 years
$20 \ln(4) - 2$ years

Explanation: The model is

dV/dt = kV

. We are given

V(2) = 40,000

and

dV/dt|_{t=2} = -2,000

. Substitute these into the differential equation to find

k

-2000 = k(40000)

, which gives

k = -0.05

. The value can be modeled starting from

t=2

V(t) = V(2)e^{k(t-2)} = 40000e^{-0.05(t-2)}

. We want to find

t

when

V(t)=10,000

. So,

10000 = 40000e^{-0.05(t-2)}

. This simplifies to

0.25 = e^{-0.05(t-2)}

. Taking the natural log,

\ln(0.25) = -0.05(t-2)

, which is

-\ln(4) = -0.05(t-2)

. Thus,

\ln(4) = 0.05(t-2)

. Solving for

t

t-2 = \frac{\ln(4)}{0.05} = 20\ln(4)

, so

t = 2 + 20\ln(4)

Question 6

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?

$15000 \ln(1.2)$ per year
$2500 \ln(1.2)$ per year
$500$ per year
$3750 \ln(1.2)$ per year (correct answer)

Explanation: 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

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.

Question 7

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?

The rates are never equal
$10 \ln(5)$
$5 \ln(5)$ (correct answer)
$\frac{\ln(5)}{5}$

Explanation: 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.

Question 8

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?

$2,000,000 e^{0.5}$ per year
$100,000$ per year
$100,000 e^{0.5}$ per year (correct answer)
$100,000 e^{-0.5}$ per year

Explanation: 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.

Question 9

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 $\ln(P(t))$ ?

A graph of $\ln(P(t))$ versus $t$ is a line with slope $0.03$ and $y$ -intercept $150$ .
A graph of $\ln(P(t))$ versus $t$ is an exponential curve with growth factor $e^{0.03}$ .
A graph of $\ln(P(t))$ versus $t$ is a line with slope $0.03$ and $y$ -intercept $\ln(150)$ . (correct answer)
A graph of $\ln(P(t))$ versus $t$ is a line with slope $\ln(150)$ and $y$ -intercept $0.03$ .

Explanation: 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

0.03

, and the

y

-intercept

b

\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)

Question 10

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?

$\frac{\ln(2)}{\ln(4)}$ years
1 year (correct answer)
$\frac{\ln(4)}{\ln(2)}$ years
0.5 years

Explanation: 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.

Question 11

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?

Time: 3.2 weeks; Rate of change: -9.6 defective items per day per week
Time: 22.4 weeks; Rate of change: -1.12 defective items per day per week
Time: 3.2 weeks; Rate of change: -1.6 defective items per day per week
Time: 22.4 weeks; Rate of change: -1.6 defective items per day per week (correct answer)

Explanation: 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.

Question 12

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, $S(0)$ ?

12,500 units (correct answer)
10,880 units
10,000 units
6,400 units

Explanation: 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

Question 13

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?

$dP/dt = 0.22P$
$dP/dt = (\ln(1.44))P$
$dP/dt = 0.2P$
$dP/dt = (\ln(1.2))P$ (correct answer)

Explanation: 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

Question 14

A manufacturing company's profit $P(t)$ (in millions of dollars) has been growing exponentially for 5 years, modeled by $dP/dt = kP$ . For the next 5 years, due to increased competition, the profit growth is projected to follow a new exponential model, but with a growth constant that is half of the original. If the initial profit was $P(0) = \$ 2 million and the profit after 5 years was $P(5) = \$ 4 million, what is the projected profit at $t=10$ years?

$6$ million
$4\sqrt{2}$ million (correct answer)
$8$ million
$4\ln(2)$ million

Explanation: When you encounter exponential growth problems with changing rates, break them into separate time periods and solve sequentially. For the first 5 years, the profit follows

\frac{dP}{dt} = kP

, which has the general solution

P(t) = P_0 e^{kt}

. With

P(0) = 2

and

P(5) = 4

, you get

4 = 2e^{5k}

, so

e^{5k} = 2

and

k = \frac{\ln(2)}{5}

. For years 5-10, the growth constant becomes

\frac{k}{2} = \frac{\ln(2)}{10}

. Starting from

P(5) = 4

as your new initial condition, the profit for

t \geq 5

follows

P(t) = 4e^{\frac{\ln(2)}{10}(t-5)}

. At

t = 10

P(10) = 4e^{\frac{\ln(2)}{10} \cdot 5} = 4e^{\frac{\ln(2)}{2}}

. Since

e^{\frac{\ln(2)}{2}} = (e^{\ln(2)})^{1/2} = 2^{1/2} = \sqrt{2}

, you get

P(10) = 4\sqrt{2}

. Choice A ($6 million) incorrectly assumes linear growth in the second period. Choice C ($8 million) mistakenly applies the original growth rate

k

for the full 10 years. Choice D ($4\ln(2) million) confuses the growth constant with the actual profit value. The correct answer is B. Study tip: In multi-period exponential problems, always use the ending value of one period as the starting value for the next, and carefully track how the growth parameters change between periods.

Question 15

A company purchases a machine whose value, $V(t)$ in dollars, depreciates over time $t$ (in years) according to the differential equation $\frac{dV}{dt} = kV$ . After 2 years, the machine is worth $72,000. After 4 years, it is worth $32,000. What was the initial purchase price of the machine?

$108,000
$112,000
$144,000
$162,000 (correct answer)

Explanation: The solution to the differential equation is

V(t) = V_0 e^{kt}

, where

V_0

is the initial price. We are given two data points:

V(2) = 72000

and

V(4) = 32000

. This gives a system of two equations:

$72000 = V_0 e^{2k}$
$32000 = V_0 e^{4k}$ Dividing equation (2) by equation (1) gives: $\frac{32000}{72000} = \frac{V_0 e^{4k}}{V_0 e^{2k}}$ , which simplifies to $\frac{4}{9} = e^{2k}$ . Now, substitute this result back into equation (1): $72000 = V_0 (e^{2k}) = V_0 (\frac{4}{9})$ . Solving for $V_0$ : $V_0 = 72000 \cdot \frac{9}{4} = 18000 \cdot 9 = 162,000$ . The initial purchase price was $162,000.

Question 16

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 ( $t=12$ )?

0.208 million/month
0.304 million/month (correct answer)
0.405 million/month
4.500 million/month

Explanation: 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.

Question 17

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?

9.6 months
21.0 months
32.4 months (correct answer)
35.0 months

Explanation: 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.

Question 18

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?

3.31%
7.64% (correct answer)
7.90%
12.50%

Explanation: 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%.

Question 19

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?

The company's profit is also increasing exponentially at the same rate.
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. (correct answer)
The absolute increase in revenue per year, $\frac{dS}{dt}$ , is constant over time.
The model predicts that sales will eventually level off as the company's market becomes saturated.

Explanation: The solution to

\frac{dS}{dt} = kS

S(t)=S_0e^{kt}

. The time it takes for a quantity to double (from

S_1

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.

Question 20

A product's sales, $S(t)$ in units per month, initially grow exponentially with a relative growth rate of 15% per month ( $k=0.15$ ). After 12 months, a competitor launches a similar product, and the relative growth rate of sales drops to 5% per month ( $k=0.05$ ). If initial sales were 1,000 units/month ( $S(0)=1000$ ), what are the sales 18 months after launch?

2,460 units
6,400 units
8,166 units (correct answer)
14,841 units

Explanation: This problem has two phases. Phase 1 (

0 \le t \le 12

S(t) = 1000e^{0.15t}

. First, find the sales at

t=12

S(12) = 1000e^{0.15 \cdot 12} = 1000e^{1.8}

. Phase 2 (

t > 12

): The growth rate changes to

k=0.05

. The new model starts with the initial value

S(12)

. For the time period from

t=12

t=18

, the duration is

18-12=6

months. The sales at

t=18

are calculated as:

S(18) = S(12) \cdot e^{0.05 \cdot 6}

. Substituting the value for

S(12)

S(18) = (1000e^{1.8}) \cdot e^{0.3} = 1000e^{1.8+0.3} = 1000e^{2.1}

. Calculating the value:

S(18) \approx 1000(8.166) \approx 8166

units.