College Algebra Quiz: Solving Exponential Equations

15 questions · exam conditions

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Solving Exponential EquationsQuestion 1 of 15

A savings account has $P(t)=10000e^{0.025t}$ (t in years); when does it reach $12000$ ?

t\approx 7.29

years

t\approx 18.2

years

t\approx 3.65

years

t\approx 9.99

years

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College Algebra Quiz

College Algebra Quiz: Solving Exponential Equations

Practice Solving Exponential Equations in College Algebra 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 Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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 savings account has $P(t)=10000e^{0.025t}$ (t in years); when does it reach $12000$ ?

$t\approx 7.29$ years (correct answer)
$t\approx 18.2$ years
$t\approx 3.65$ years
$t\approx 9.99$ years

Explanation: This question tests solving exponential equations related to real-world contexts in college algebra. Exponential equations model scenarios where quantities grow or decay at a constant rate over time, using the formula P = P_0 * e^(rt) where P is the final amount, P_0 the initial amount, r the rate, and t the time. In the given scenario, a savings account starts with $10000 and grows at a rate of 0.025 per year, students are expected to solve for t when it reaches $12000 using the provided values. The correct answer correctly applies the exponential formula to determine the time t ≈ 7.29 years, demonstrating understanding of growth processes. A common distractor arises from misinterpreting the growth rate, such as using 0.05 instead of 0.025, which leads to incorrect results. To help students, emphasize the importance of accurately interpreting the exponential formula and understanding the context of the problem. Practice solving similar equations with varying rates and initial values to build confidence.

Question 2

A radioactive substance decays according to $A(t) = A_0 e^{-0.1t}$ , where $A(t)$ is the amount remaining after $t$ years. If 75% of the substance has decayed, how many years have elapsed?

Approximately 13.86 years, using natural logarithms to solve (correct answer)
Approximately 16.09 years, using natural logarithms to solve
Approximately 10.39 years, using natural logarithms to solve
Approximately 18.42 years, using natural logarithms to solve

Explanation: If 75% has decayed, then 25% remains. So

A(t) = 0.25A_0

. The equation becomes

0.25A_0 = A_0 e^{-0.1t}

. Dividing by

A_0

0.25 = e^{-0.1t}

. Taking the natural logarithm of both sides:

\ln(0.25) = -0.1t

. So

t = \frac{\ln(0.25)}{-0.1} = \frac{-\ln(4)}{-0.1} = \frac{\ln(4)}{0.1} = 10\ln(4) \approx 10(1.386) = 13.86

years. Choice B comes from using

\ln(0.75)

instead of

\ln(0.25)

. Choice C results from the error

t = \frac{\ln(0.25)}{0.1}

(missing the negative). Choice D uses an incorrect logarithm value.

Question 3

A bacteria culture grows according to the equation $N(t) = 500 \cdot 3^{t/4}$ , where $N(t)$ is the number of bacteria after $t$ hours. After how many hours will the population reach 13,500 bacteria?

12 hours after the initial measurement (correct answer)
16 hours after the initial measurement
10 hours after the initial measurement
14 hours after the initial measurement

Explanation: We need to solve

500 \cdot 3^{t/4} = 13500

. Dividing both sides by 500:

3^{t/4} = 27

. Since

27 = 3^3

, we have

3^{t/4} = 3^3

. Therefore

\frac{t}{4} = 3

, so

t = 12

. Choice B comes from incorrectly setting

3^{t/4} = 3^4

when

27 = 3^4

is wrong. Choice C results from the error

\frac{t}{4} = 2.5

when thinking

27 = 3^{2.5}

. Choice D comes from solving

3^{t/4} = 3^{3.5}

incorrectly.

Question 4

A bacteria culture follows $P(t)=250e^{0.16t}$ (t in hours); when does it reach $500$ ?

$t\approx 2.17$ hours
$t\approx 4.33$ hours (correct answer)
$t\approx 8.66$ hours
$t\approx 13.9$ hours

Explanation: This question tests solving exponential equations related to real-world contexts in college algebra. Exponential equations model scenarios where quantities grow or decay at a constant rate over time, using the formula P = P_0 * e^(rt) where P is the final amount, P_0 the initial amount, r the rate, and t the time. In the given scenario, a bacteria culture starts with 250 and grows at a rate of 0.16 per hour, students are expected to solve for t when it reaches 500 using the provided values. The correct answer correctly applies the exponential formula to determine the time t ≈ 4.33 hours, demonstrating understanding of growth processes. A common distractor arises from misinterpreting the growth rate, such as using 0.08 instead of 0.16, which leads to incorrect results. To help students, emphasize the importance of accurately interpreting the exponential formula and understanding the context of the problem. Practice solving similar equations with varying rates and initial values to build confidence.

Question 5

A savings account has $P(t)=2500e^{0.03t}$ (t in years); when does it reach $5000$ ?

$t\approx 11.6$ years
$t\approx 23.1$ years (correct answer)
$t\approx 17.3$ years
$t\approx 5.78$ years

Explanation: This question tests solving exponential equations related to real-world contexts in college algebra. Exponential equations model scenarios where quantities grow or decay at a constant rate over time, using the formula P = P_0 * e^(rt) where P is the final amount, P_0 the initial amount, r the rate, and t the time. In the given scenario, a savings account starts with $2500 and grows at a rate of 0.03 per year, students are expected to solve for t when it reaches $5000 using the provided values. The correct answer correctly applies the exponential formula to determine the time t ≈ 23.1 years, demonstrating understanding of growth processes. A common distractor arises from misinterpreting the growth rate, such as using 0.06 instead of 0.03, which leads to incorrect results. To help students, emphasize the importance of accurately interpreting the exponential formula and understanding the context of the problem. Practice solving similar equations with varying rates and initial values to build confidence.

Question 6

A bacteria culture follows $P(t)=500e^{0.30t}$ (t in hours); when does it reach $4000$ ?

$t\approx 13.9$ hours
$t\approx 6.93$ hours (correct answer)
$t\approx 2.31$ hours
$t\approx 5.55$ hours

Explanation: This question tests solving exponential equations related to real-world contexts in college algebra. Exponential equations model scenarios where quantities grow or decay at a constant rate over time, using the formula P = P_0 * e^(rt) where P is the final amount, P_0 the initial amount, r the rate, and t the time. In the given scenario, a bacteria culture starts with 500 and grows at a rate of 0.30 per hour, students are expected to solve for t when it reaches 4000 using the provided values. The correct answer correctly applies the exponential formula to determine the time t ≈ 6.93 hours, demonstrating understanding of growth processes. A common distractor arises from misinterpreting the growth rate, such as using 0.15 instead of 0.30, which leads to incorrect results. To help students, emphasize the importance of accurately interpreting the exponential formula and understanding the context of the problem. Practice solving similar equations with varying rates and initial values to build confidence.

Question 7

A bacteria culture follows $P(t)=900e^{0.10t}$ (t in hours); when does it reach $2700$ ?

$t\approx 10.99$ hours (correct answer)
$t\approx 6.93$ hours
$t\approx 21.97$ hours
$t\approx 3.47$ hours

Explanation: This question tests solving exponential equations related to real-world contexts in college algebra. Exponential equations model scenarios where quantities grow or decay at a constant rate over time, using the formula P = P_0 * e^(rt) where P is the final amount, P_0 the initial amount, r the rate, and t the time. In the given scenario, a bacteria culture starts with 900 and grows at a rate of 0.10 per hour, students are expected to solve for t when it reaches 2700 using the provided values. The correct answer correctly applies the exponential formula to determine the time t ≈ 10.99 hours, demonstrating understanding of growth processes. A common distractor arises from misinterpreting the growth rate, such as using 0.05 instead of 0.10, which leads to incorrect results. To help students, emphasize the importance of accurately interpreting the exponential formula and understanding the context of the problem. Practice solving similar equations with varying rates and initial values to build confidence.

Question 8

A bacteria culture follows $P(t)=200e^{0.18t}$ (t in hours); when does it reach $1000$ ?

$t\approx 3.58$ hours
$t\approx 8.94$ hours (correct answer)
$t\approx 12.4$ hours
$t\approx 27.8$ hours

Explanation: This question tests solving exponential equations related to real-world contexts in college algebra. Exponential equations model scenarios where quantities grow or decay at a constant rate over time, using the formula P = P_0 * e^(rt) where P is the final amount, P_0 the initial amount, r the rate, and t the time. In the given scenario, a bacteria culture starts with 200 and grows at a rate of 0.18 per hour, students are expected to solve for t when it reaches 1000 using the provided values. The correct answer correctly applies the exponential formula to determine the time t ≈ 8.94 hours, demonstrating understanding of growth processes. A common distractor arises from misinterpreting the growth rate, such as using 0.09 instead of 0.18, which leads to incorrect results. To help students, emphasize the importance of accurately interpreting the exponential formula and understanding the context of the problem. Practice solving similar equations with varying rates and initial values to build confidence.

Question 9

A bacteria culture follows $P(t)=1000e^{0.22t}$ (t in hours); when does it reach $3000$ ?

$t\approx 9.99$ hours
$t\approx 4.99$ hours (correct answer)
$t\approx 13.6$ hours
$t\approx 3.15$ hours

Explanation: This question tests solving exponential equations related to real-world contexts in college algebra. Exponential equations model scenarios where quantities grow or decay at a constant rate over time, using the formula P = P_0 * e^(rt) where P is the final amount, P_0 the initial amount, r the rate, and t the time. In the given scenario, a bacteria culture starts with 1000 and grows at a rate of 0.22 per hour, students are expected to solve for t when it reaches 3000 using the provided values. The correct answer correctly applies the exponential formula to determine the time t ≈ 4.99 hours, demonstrating understanding of growth processes. A common distractor arises from misinterpreting the growth rate, such as using 0.11 instead of 0.22, which leads to incorrect results. To help students, emphasize the importance of accurately interpreting the exponential formula and understanding the context of the problem. Practice solving similar equations with varying rates and initial values to build confidence.

Question 10

A savings account has $P(t)=1500e^{0.08t}$ (t in years); when does it reach $3000$ ?

$t\approx 4.33$ years
$t\approx 8.66$ years (correct answer)
$t\approx 11.6$ years
$t\approx 17.3$ years

Explanation: This question tests solving exponential equations related to real-world contexts in college algebra. Exponential equations model scenarios where quantities grow or decay at a constant rate over time, using the formula P = P_0 * e^(rt) where P is the final amount, P_0 the initial amount, r the rate, and t the time. In the given scenario, a savings account starts with $1500 and grows at a rate of 0.08 per year, students are expected to solve for t when it reaches $3000 using the provided values. The correct answer correctly applies the exponential formula to determine the time t ≈ 8.66 years, demonstrating understanding of growth processes. A common distractor arises from misinterpreting the growth rate, such as using 0.04 instead of 0.08, which leads to incorrect results. To help students, emphasize the importance of accurately interpreting the exponential formula and understanding the context of the problem. Practice solving similar equations with varying rates and initial values to build confidence.

Question 11

If $4^{x-3} = 2^{x+1}$ , what is the value of $x$ ?

$x = 4$
$x = 7$ (correct answer)
$x = 5$
$x = 6$

Explanation: Since

4 = 2^2

, we can rewrite the equation as

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

. Using the power rule, this becomes

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

, or

2^{2x-6} = 2^{x+1}

. Since the bases are equal, the exponents must be equal:

2x-6 = x+1

. Solving for

x

2x - x = 1 + 6

, so

x = 7

. Choice A results from solving

x-3 = x+1

incorrectly. Choice C comes from the error

2x-6 = x-1

. Choice D results from

2x = x+6

without properly distributing the exponent.

Question 12

A radioactive sample has $A(t)=60e^{-0.20t}$ (t in days); when does it reach $30$ grams?

$t\approx 1.73$ days
$t\approx 3.47$ days (correct answer)
$t\approx 6.93$ days
$t\approx 13.9$ days

Explanation: This question tests solving exponential equations related to real-world contexts in college algebra. Exponential equations model scenarios where quantities grow or decay at a constant rate over time, using the formula P = P_0 * e^(rt) where P is the final amount, P_0 the initial amount, r the rate, and t the time. In the given scenario, a radioactive sample starts with 60 grams and decays at a rate of 0.20 per day, students are expected to solve for t when it reaches 30 grams using the provided values. The correct answer correctly applies the exponential formula to determine the time t ≈ 3.47 days, demonstrating understanding of decay processes. A common distractor arises from misinterpreting the decay rate, such as using 0.10 instead of 0.20, which leads to incorrect results. To help students, emphasize the importance of accurately interpreting the exponential formula and understanding the context of the problem. Practice solving similar equations with varying rates and initial values to build confidence.

Question 13

A savings account has $P(t)=3000e^{0.06t}$ (t in years); when does it reach $4500$ ?

$t\approx 6.76$ years (correct answer)
$t\approx 4.05$ years
$t\approx 11.3$ years
$t\approx 2.70$ years

Explanation: This question tests solving exponential equations related to real-world contexts in college algebra. Exponential equations model scenarios where quantities grow or decay at a constant rate over time, using the formula P = P_0 * e^(rt) where P is the final amount, P_0 the initial amount, r the rate, and t the time. In the given scenario, a savings account starts with $3000 and grows at a rate of 0.06 per year, students are expected to solve for t when it reaches $4500 using the provided values. The correct answer correctly applies the exponential formula to determine the time t ≈ 6.76 years, demonstrating understanding of growth processes. A common distractor arises from misinterpreting the growth rate, such as using 0.10 instead of 0.06, which leads to incorrect results. To help students, emphasize the importance of accurately interpreting the exponential formula and understanding the context of the problem. Practice solving similar equations with varying rates and initial values to build confidence.

Question 14

A bacteria culture follows $P(t)=800e^{0.12t}$ (t in hours); when does it reach $1600$ ?

$t\approx 2.89$ hours
$t\approx 5.78$ hours (correct answer)
$t\approx 11.6$ hours
$t\approx 8.66$ hours

Explanation: This question tests solving exponential equations related to real-world contexts in college algebra. Exponential equations model scenarios where quantities grow or decay at a constant rate over time, using the formula P = P_0 * e^(rt) where P is the final amount, P_0 the initial amount, r the rate, and t the time. In the given scenario, a bacteria culture starts with 800 and grows at a rate of 0.12 per hour, students are expected to solve for t when it reaches 1600 using the provided values. The correct answer correctly applies the exponential formula to determine the time t ≈ 5.78 hours, demonstrating understanding of growth processes. A common distractor arises from misinterpreting the growth rate, such as using 0.06 instead of 0.12, which leads to incorrect results. To help students, emphasize the importance of accurately interpreting the exponential formula and understanding the context of the problem. Practice solving similar equations with varying rates and initial values to build confidence.

Question 15

A savings account has $P(t)=800e^{0.05t}$ (t in years); when does it reach $1000$ ?

$t\approx 4.46$ years (correct answer)
$t\approx 2.23$ years
$t\approx 3.57$ years
$t\approx 1.12$ years

Explanation: This question tests solving exponential equations related to real-world contexts in college algebra. Exponential equations model scenarios where quantities grow or decay at a constant rate over time, using the formula P = P_0 * e^(rt) where P is the final amount, P_0 the initial amount, r the rate, and t the time. In the given scenario, a savings account starts with $800 and grows at a rate of 0.05 per year, students are expected to solve for t when it reaches $1000 using the provided values. The correct answer correctly applies the exponential formula to determine the time t ≈ 4.46 years, demonstrating understanding of growth processes. A common distractor arises from misinterpreting the growth rate, such as using 0.10 instead of 0.05, which leads to incorrect results. To help students, emphasize the importance of accurately interpreting the exponential formula and understanding the context of the problem. Practice solving similar equations with varying rates and initial values to build confidence.