Business Calculus Quiz: Solving Exponential And Log Equations

15 questions · exam conditions

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Solving Exponential And Log EquationsQuestion 1 of 15

A company's revenue growth follows the model $R(t) = 50,000 \cdot 1.08^t$ where $t$ is years since 2020. If the company wants to determine when their revenue will first exceed $100,000, which equation should they solve, and what is the approximate solution?

Solve

50,000 \cdot 1.08^t = 100,000

; solution is approximately

t = 8.75

years

Solve

50,000 \cdot 1.08^t > 100,000

; solution is approximately

t = 8.75

years

Solve

50,000 \cdot 1.08^t = 100,000

; solution is approximately

t = 9.01

years

Solve

50,000 \cdot 1.08^t = 100,000

; solution is approximately

t = 12.5

years

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Business Calculus Quiz

Business Calculus Quiz: Solving Exponential And Log Equations

Practice Solving Exponential And Log Equations 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 Solving Exponential And Log Equations, 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

Solve $50,000 \cdot 1.08^t = 100,000$ ; solution is approximately $t = 8.75$ years
Solve $50,000 \cdot 1.08^t > 100,000$ ; solution is approximately $t = 8.75$ years
Solve $50,000 \cdot 1.08^t = 100,000$ ; solution is approximately $t = 9.01$ years (correct answer)
Solve $50,000 \cdot 1.08^t = 100,000$ ; solution is approximately $t = 12.5$ years

Explanation: To find when revenue first exceeds $100,000, we solve for equality: $

50,000 \cdot 1.08^t = 100,000

. Dividing both sides by 50,000 gives

1.08^t = 2

. Taking the natural logarithm:

t \ln(1.08) = \ln(2)

, so

t = \frac{\ln(2)}{\ln(1.08)} \approx \frac{0.693}{0.077} \approx 9.01

$ years. Choice A uses the wrong value. Choice B incorrectly suggests solving an inequality rather than finding the boundary point. Choice D uses an incorrect calculation method.

Question 2

An initial investment in Fund A grows according to the model $A(t) = 10000 e^{0.05t}$ , while an initial investment in Fund B grows according to $B(t) = 15000 e^{0.03t}$ , where $t$ is the time in years. At approximately what time $t$ will the value of Fund A equal the value of Fund B?

$t = \frac{\ln(1.5)}{0.08}$
$t = \frac{\ln(1.5)}{0.02}$ (correct answer)
$t = \frac{1.5}{0.02}$
$t = \frac{\ln(0.67)}{0.02}$

Explanation: To find when the values are equal, set

A(t) = B(t)

10000 e^{0.05t} = 15000 e^{0.03t}

. Divide both sides by

10000

and by

e^{0.03t}

to isolate the exponential terms:

\frac{e^{0.05t}}{e^{0.03t}} = \frac{15000}{10000}

. Using the exponent rule

e^a/e^b = e^{a-b}

, we get

e^{0.05t - 0.03t} = 1.5

, which simplifies to

e^{0.02t} = 1.5

. To solve for

t

, take the natural logarithm of both sides:

\ln(e^{0.02t}) = \ln(1.5)

, which gives

0.02t = \ln(1.5)

. Finally, divide by

0.02

t = \frac{\ln(1.5)}{0.02}

Question 3

The number of units produced, $x$ (in thousands), is related to the advertising spending, $a$ (in thousands of dollars), by the equation $\log_2(x+3) + \log_2(x-1) = 5$ . If each unit sells for $20, what is the total revenue from the units produced?

$300,000
$140,000
$100,000 (correct answer)
$5,000

Explanation: First, solve the logarithmic equation for

x

. Using the product rule for logarithms,

\log_2((x+3)(x-1)) = 5

. Convert to exponential form:

(x+3)(x-1) = 2^5 = 32

. Expand the left side:

x^2 + 2x - 3 = 32

. Rearrange into a standard quadratic form:

x^2 + 2x - 35 = 0

. Factor the quadratic:

(x+7)(x-5) = 0

. The solutions are

x=-7

and

x=5

. Since

x

represents the number of units, it must be positive. Furthermore, the arguments of the logarithms,

(x+3)

and

(x-1)

, must be positive, which requires

x > 1

. Thus,

x=5

is the only valid solution. Since

x

is in thousands of units, this corresponds to 5,000 units. The total revenue is

5,000 \text{ units} \times \

20/\text{unit} = $100,000$.

Question 4

The population $P(t)$ of a certain bacteria culture (in thousands) at time $t$ (in hours) follows a model where $t$ must satisfy the equation $e^{2t} - 6e^t + 5 = 0$ . The company harvests the bacteria at the latest valid time predicted by the model. What is the population, given by $P(t) = 100e^t$ , at that time?

100,000
500,000 (correct answer)
600,000
$\ln(5)$

Explanation: The equation

e^{2t} - 6e^t + 5 = 0

is a quadratic in form. Let

y = e^t

. The equation becomes

y^2 - 6y + 5 = 0

. Factoring the quadratic gives

(y-1)(y-5) = 0

, so the solutions for

y

are

y=1

and

y=5

. Substitute back

e^t

for

y

e^t = 1

e^t = 5

. Solving for

t

gives

t = \ln(1) = 0

t = \ln(5)

. The latest valid time is

t = \ln(5)

. The question asks for the population

P(t) = 100e^t

at this time. At

t = \ln(5)

, we have

e^t = 5

. Therefore,

P(\ln(5)) = 100(5) = 500

. Since

P(t)

is in thousands, the population is 500,000.

Question 5

The relationship between advertising spending $A$ and sales $S$ is modeled in two different ways by two marketing agencies. Agency 1 uses $S_1(A) = 100 + 5 \log_2(A)$ . Agency 2 uses $S_2(A) = 150 + k \log_4(A)$ . If both models predict the same sales when spending $A=16$ , what is the value of the constant $k$ ?

$-15$ (correct answer)
$-5$
$15$
$-7.5$

Explanation: First, calculate the sales predicted by Agency 1 when

A=16

S_1(16) = 100 + 5 \log_2(16)

. Since

2^4 = 16

\log_2(16) = 4

. So,

S_1(16) = 100 + 5(4) = 100 + 20 = 120

. Next, set Agency 2's prediction equal to this value:

S_2(16) = 120

. The equation is

120 = 150 + k \log_4(16)

. Since

4^2=16

\log_4(16) = 2

. The equation becomes

120 = 150 + k(2)

. Now, solve for

k

120 - 150 = 2k

, which gives

-30 = 2k

, so

k = -15

Question 6

A company's daily profit $P(x)$ from selling $x$ units is given by $P(x) = 1000(1 - e^{-0.02x})$ . The company's operations are considered sustainable only if the profit exceeds a fixed daily cost of $500. For which range of units $x$ are the operations sustainable?

$x < 50 \ln(2)$
$x > \frac{\ln(0.5)}{0.02}$
$x > 50 \ln(2)$ (correct answer)
$x > 50 \ln(1.5)$

Explanation: To find the range of

x

for sustainable operations, we must solve the inequality

P(x) > 500

1000(1 - e^{-0.02x}) > 500

. Divide by 1000:

1 - e^{-0.02x} > 0.5

. Subtract 1 from both sides:

-e^{-0.02x} > -0.5

. Multiply by -1 and reverse the inequality sign:

e^{-0.02x} < 0.5

. Take the natural logarithm of both sides:

\ln(e^{-0.02x}) < \ln(0.5)

, which simplifies to

-0.02x < \ln(0.5)

. Using the log property

\ln(0.5) = \ln(1/2) = -\ln(2)

, the inequality is

-0.02x < -\ln(2)

. To solve for

x

, divide by -0.02 and reverse the inequality sign again:

x > \frac{-\ln(2)}{-0.02} = \frac{\ln(2)}{0.02}

. Since

1/0.02 = 50

, the solution is

x > 50 \ln(2)

Question 7

The decay of a radioactive substance used in medical imaging is modeled by $A(t) = A_0 e^{-kt}$ . A hospital's policy requires disposing of the substance when it has decayed to 15% of its original amount. If the substance's half-life is 8 days, what is the disposal time, rounded to the nearest day?

14 days
22 days (correct answer)
30 days
43 days

Explanation: First, use the half-life information to find the decay constant

k

. Half-life means that at

t=8

A(8) = 0.5 A_0

. So,

0.5 A_0 = A_0 e^{-k(8)}

. Dividing by

A_0

gives

0.5 = e^{-8k}

. Taking the natural logarithm,

\ln(0.5) = -8k

. So,

k = \frac{\ln(0.5)}{-8} = \frac{-\ln(2)}{-8} = \frac{\ln(2)}{8}

. Next, find the time

t

when the substance has decayed to 15% of its original amount, i.e., when

A(t) = 0.15 A_0

. The equation is

0.15 A_0 = A_0 e^{-kt}

. Dividing by

A_0

gives

0.15 = e^{-kt}

. Taking the natural logarithm,

\ln(0.15) = -kt

. Solving for

t

gives

t = \frac{\ln(0.15)}{-k}

. Substitute the expression for

k

t = \frac{\ln(0.15)}{-(\ln(2)/8)} = \frac{-8 \ln(0.15)}{\ln(2)}

. Using a calculator,

t \approx \frac{-8(-1.897)}{0.693} \approx 21.9

days. Rounded to the nearest day, the disposal time is 22 days.

Question 8

A product must be heated to a temperature of 180°C. It starts at 20°C in an oven set to 200°C. The temperature $T$ after $t$ minutes follows Newton's Law of Heating: $T(t) = T_a + (T_0 - T_a)e^{-kt}$ , where $T_a$ is the ambient temperature and $T_0$ is the initial temperature. If the product reaches 100°C after 10 minutes, approximately how long will it take in total to reach 180°C?

20.0 minutes
28.5 minutes
37.4 minutes (correct answer)
45.1 minutes

Explanation: First, set up the model with the given values:

T_a = 200

and

T_0 = 20

. So,

T(t) = 200 + (20 - 200)e^{-kt} = 200 - 180e^{-kt}

. Use the information

T(10) = 100

to find

k

100 = 200 - 180e^{-10k} \implies -100 = -180e^{-10k} \implies e^{-10k} = 100/180 = 5/9

. So,

-10k = \ln(5/9)

, which gives

k = -\frac{\ln(5/9)}{10} = \frac{\ln(9/5)}{10} = \frac{\ln(1.8)}{10}

. Now, find the time

t

when the temperature reaches 180°C:

180 = 200 - 180e^{-kt} \implies -20 = -180e^{-kt} \implies e^{-kt} = 20/180 = 1/9

. Take the natural logarithm:

-kt = \ln(1/9) = -\ln(9)

. So,

t = \frac{\ln(9)}{k}

. Substitute the value of

k

t = \frac{\ln(9)}{\frac{\ln(1.8)}{10}} = \frac{10 \ln(9)}{\ln(1.8)} \approx \frac{10(2.1972)}{0.5878} \approx 37.38

minutes.

Question 9

The number of active users for two competing social media platforms, A and B, are modeled by $U_A(t) = 10^6 \cdot \log_3(t)$ and $U_B(t) = 10^6 \cdot \log_9(4t)$ , respectively, where $t$ is the number of months since launch ( $t \ge 1$ ). At what value of $t$ will the number of active users be equal for both platforms?

$t=2$
$t=4$ (correct answer)
$t=8$
$t=16$

Explanation: Set the number of users equal:

10^6 \cdot \log_3(t) = 10^6 \cdot \log_9(4t)

. The

10^6

factor cancels, leaving

\log_3(t) = \log_9(4t)

. To solve this, convert both logarithms to the same base using the change of base formula,

\log_b(x) = \frac{\log_a(x)}{\log_a(b)}

. Let's use base 3. The equation becomes

\log_3(t) = \frac{\log_3(4t)}{\log_3(9)}

. Since

\log_3(9) = 2

, this simplifies to

\log_3(t) = \frac{\log_3(4t)}{2}

. Multiply by 2:

2\log_3(t) = \log_3(4t)

. Use the power rule for logarithms:

\log_3(t^2) = \log_3(4t)

. Now that the bases and coefficients are the same, we can equate the arguments:

t^2 = 4t

. This gives

t^2 - 4t = 0

, or

t(t-4) = 0

. The solutions are

t=0

and

t=4

. Since the model is valid for

t \ge 1

, the only valid solution is

t=4

Question 10

A company measures its market influence on a logarithmic scale called the Market Influence Index (MII), given by $M = \log(S/S_0)$ , where $S$ is the company's annual sales and $S_0$ is a baseline sales figure. If Company X has an MII of 4.5 and Company Y has an MII of 6.0, how many times larger are Company Y's sales than Company X's sales?

1.5 times larger
4.48 times larger
31.6 times larger (correct answer)
1.33 times larger

Explanation: For Company X,

4.5 = \log(S_X/S_0)

, which means

S_X = S_0 \cdot 10^{4.5}

. For Company Y,

6.0 = \log(S_Y/S_0)

, which means

S_Y = S_0 \cdot 10^{6.0}

. To find how many times larger Company Y's sales are, we compute the ratio

S_Y/S_X

\frac{S_Y}{S_X} = \frac{S_0 \cdot 10^{6.0}}{S_0 \cdot 10^{4.5}} = 10^{6.0 - 4.5} = 10^{1.5}

. Calculating the value,

10^{1.5} = 10^1 \cdot 10^{0.5} = 10\sqrt{10} \approx 10 \cdot 3.162 \approx 31.6

Question 11

A pharmaceutical company models drug concentration in the bloodstream as $C(t) = 15e^{-0.23t}$ mg/L, where $t$ is hours after injection. When will the concentration drop to exactly 25% of its initial value?

After approximately $t = 6.02$ hours, using the equation $15e^{-0.23t} = 3.75$ (correct answer)
After approximately $t = 4.33$ hours, using the equation $15e^{-0.23t} = 11.25$
After approximately $t = 6.02$ hours, using the equation $e^{-0.23t} = 0.25$
After approximately $t = 8.67$ hours, using the equation $15e^{-0.23t} = 0.25$

Explanation: Initial concentration is

C(0) = 15e^0 = 15

mg/L. Twenty-five percent of this is

0.25 \times 15 = 3.75

mg/L. We solve

15e^{-0.23t} = 3.75

. Dividing by 15:

e^{-0.23t} = 0.25

. Taking natural logarithm:

-0.23t = \ln(0.25) = \ln(1/4) = -\ln(4)

. Thus

t = \frac{\ln(4)}{0.23} \approx \frac{1.386}{0.23} \approx 6.02

hours. Choice B incorrectly calculates 75% instead of 25%. Choice C sets up the wrong equation by omitting the coefficient 15. Choice D uses 0.25 as the absolute concentration rather than 25% of the initial value.

Question 12

If $x$ and $y$ are positive numbers such that $\ln(x) - \ln(y) = 2$ and $\log_5(x) + \log_5(y) = 4$ , what is the value of $x$ ?

$25e$ (correct answer)
$\frac{625}{e}$
$25\sqrt{e}$
$625e$

Explanation: We have a system of two equations with two variables. From the first equation, using the quotient rule for logarithms:

\ln(x/y) = 2

. Converting to exponential form gives

x/y = e^2

, so

x = ye^2

. From the second equation, using the product rule for logarithms:

\log_5(xy) = 4

. Converting to exponential form gives

xy = 5^4 = 625

. Now substitute the expression for

x

from the first equation into the second:

(ye^2)y = 625

. This simplifies to

y^2e^2 = 625

. Taking the square root of both sides gives

ye = \sqrt{625} = 25

(since

y>0

). From this, we find

y = 25/e

. Finally, substitute this value of

y

back into the expression for

x

x = (25/e)e^2 = 25e

Question 13

The solution to the equation $2^{x+2} = 3^{x-1}$ can be written in the form $x = \frac{\ln(A)}{\ln(B)}$ . What are the values of $A$ and $B$ ?

$A=12$ , $B=1.5$ (correct answer)
$A=6$ , $B=1$
$A=5$ , $B=1$
$A=0.75$ , $B=6$

Explanation: To solve

2^{x+2} = 3^{x-1}

, take the natural logarithm of both sides:

\ln(2^{x+2}) = \ln(3^{x-1})

. Using the power rule for logarithms, this becomes

(x+2)\ln(2) = (x-1)\ln(3)

. Distribute

\ln(2)

and

\ln(3)

x\ln(2) + 2\ln(2) = x\ln(3) - \ln(3)

. Rearrange the terms to group

x

on one side:

2\ln(2) + \ln(3) = x\ln(3) - x\ln(2)

. Factor out

x

2\ln(2) + \ln(3) = x(\ln(3) - \ln(2))

. Use log rules to simplify the expressions in parentheses:

\ln(2^2) + \ln(3) = x(\ln(3/2))

, which is

\ln(4) + \ln(3) = x\ln(1.5)

. Further simplification gives

\ln(4 \cdot 3) = x\ln(1.5)

, so

\ln(12) = x\ln(1.5)

. Solving for

x

yields

x = \frac{\ln(12)}{\ln(1.5)}

. Comparing this to the form

x = \frac{\ln(A)}{\ln(B)}

, we have

A=12

and

B=1.5

Question 14

If $\log_2(x) + \log_4(x) = 6$ , what is the value of $x$ ?

$x = 16$ , after converting to a common base and solving the resulting equation (correct answer)
$x = 32$ , after converting to a common base and solving the resulting equation
$x = 64$ , after converting to a common base and solving the resulting equation
$x = 128$ , after converting to a common base and solving the resulting equation

Explanation: Convert

\log_4(x)

to base 2 using the change of base formula:

\log_4(x) = \frac{\log_2(x)}{\log_2(4)} = \frac{\log_2(x)}{2}

. The equation becomes:

\log_2(x) + \frac{\log_2(x)}{2} = 6

. Factoring:

\log_2(x)\left(1 + \frac{1}{2}\right) = 6

, so

\log_2(x) \cdot \frac{3}{2} = 6

. Therefore:

\log_2(x) = 6 \cdot \frac{2}{3} = 4

. Converting to exponential form:

x = 2^4 = 16

. Choice B would result from solving

\log_2(x) = 5

. Choice C would result from solving

\log_2(x) = 6

. Choice D would result from solving

\log_2(x) = 7

Question 15

A bacteria population grows according to $P(t) = 1200 \cdot 3^{t/4}$ where $t$ is in hours. After how many hours will the population reach exactly 32,400?

$t = 12$ hours, since $3^{12/4} = 3^3 = 27$ and $1200 \times 27 = 32,400$ (correct answer)
$t = 16$ hours, since $3^{16/4} = 3^4 = 81$ and $1200 \times 81 = 97,200$
$t = 10.75$ hours, using logarithms to solve $3^{t/4} = 27$
$t = 8$ hours, since $3^{8/4} = 3^2 = 9$ and $1200 \times 9 = 10,800$

Explanation: We need to solve

1200 \cdot 3^{t/4} = 32,400

. Dividing both sides by 1200:

3^{t/4} = \frac{32,400}{1200} = 27

. Since

27 = 3^3

, we have

3^{t/4} = 3^3

, which means

\frac{t}{4} = 3

, so

t = 12

hours. Choice A correctly identifies this solution. Choice B calculates

3^4

but this gives the wrong population value. Choice C uses an incorrect logarithmic approach. Choice D uses

3^2

which gives a population of only 10,800, not 32,400.