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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.

Question 1 / 15

0 of 15 answered

A savings account has P(t)=2500e0.03tP(t)=2500e^{0.03t}P(t)=2500e0.03t (t in years); when does it reach 500050005000?

Select an answer to continue

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)=2500e0.03tP(t)=2500e^{0.03t}P(t)=2500e0.03t (t in years); when does it reach 500050005000?

  1. t≈11.6t\approx 11.6t≈11.6 years
  2. t≈23.1t\approx 23.1t≈23.1 years (correct answer)
  3. t≈17.3t\approx 17.3t≈17.3 years
  4. t≈5.78t\approx 5.78t≈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 2500andgrowsatarateof0.03peryear,studentsareexpectedtosolvefortwhenitreaches2500 and grows at a rate of 0.03 per year, students are expected to solve for t when it reaches 2500andgrowsatarateof0.03peryear,studentsareexpectedtosolvefortwhenitreaches5000 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 2

A bacteria culture follows P(t)=500e0.30tP(t)=500e^{0.30t}P(t)=500e0.30t (t in hours); when does it reach 400040004000?

  1. t≈13.9t\approx 13.9t≈13.9 hours
  2. t≈6.93t\approx 6.93t≈6.93 hours (correct answer)
  3. t≈2.31t\approx 2.31t≈2.31 hours
  4. t≈5.55t\approx 5.55t≈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 3

A bacteria culture follows P(t)=900e0.10tP(t)=900e^{0.10t}P(t)=900e0.10t (t in hours); when does it reach 270027002700?

  1. t≈10.99t\approx 10.99t≈10.99 hours (correct answer)
  2. t≈6.93t\approx 6.93t≈6.93 hours
  3. t≈21.97t\approx 21.97t≈21.97 hours
  4. t≈3.47t\approx 3.47t≈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 4

A savings account has P(t)=10000e0.025tP(t)=10000e^{0.025t}P(t)=10000e0.025t (t in years); when does it reach 120001200012000?

  1. t≈7.29t\approx 7.29t≈7.29 years (correct answer)
  2. t≈18.2t\approx 18.2t≈18.2 years
  3. t≈3.65t\approx 3.65t≈3.65 years
  4. t≈9.99t\approx 9.99t≈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 10000andgrowsatarateof0.025peryear,studentsareexpectedtosolvefortwhenitreaches10000 and grows at a rate of 0.025 per year, students are expected to solve for t when it reaches 10000andgrowsatarateof0.025peryear,studentsareexpectedtosolvefortwhenitreaches12000 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 5

A bacteria culture follows P(t)=200e0.18tP(t)=200e^{0.18t}P(t)=200e0.18t (t in hours); when does it reach 100010001000?

  1. t≈3.58t\approx 3.58t≈3.58 hours
  2. t≈8.94t\approx 8.94t≈8.94 hours (correct answer)
  3. t≈12.4t\approx 12.4t≈12.4 hours
  4. t≈27.8t\approx 27.8t≈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 6

A radioactive substance decays according to A(t)=A0e−0.1tA(t) = A_0 e^{-0.1t}A(t)=A0​e−0.1t, where A(t)A(t)A(t) is the amount remaining after ttt years. If 75% of the substance has decayed, how many years have elapsed?

  1. Approximately 13.86 years, using natural logarithms to solve (correct answer)
  2. Approximately 16.09 years, using natural logarithms to solve
  3. Approximately 10.39 years, using natural logarithms to solve
  4. Approximately 18.42 years, using natural logarithms to solve

Explanation: If 75% has decayed, then 25% remains. So A(t)=0.25A0A(t) = 0.25A_0A(t)=0.25A0​. The equation becomes 0.25A0=A0e−0.1t0.25A_0 = A_0 e^{-0.1t}0.25A0​=A0​e−0.1t. Dividing by A0A_0A0​: 0.25=e−0.1t0.25 = e^{-0.1t}0.25=e−0.1t. Taking the natural logarithm of both sides: ln⁡(0.25)=−0.1t\ln(0.25) = -0.1tln(0.25)=−0.1t. So t=ln⁡(0.25)−0.1=−ln⁡(4)−0.1=ln⁡(4)0.1=10ln⁡(4)≈10(1.386)=13.86t = \frac{\ln(0.25)}{-0.1} = \frac{-\ln(4)}{-0.1} = \frac{\ln(4)}{0.1} = 10\ln(4) \approx 10(1.386) = 13.86t=−0.1ln(0.25)​=−0.1−ln(4)​=0.1ln(4)​=10ln(4)≈10(1.386)=13.86 years. Choice B comes from using ln⁡(0.75)\ln(0.75)ln(0.75) instead of ln⁡(0.25)\ln(0.25)ln(0.25). Choice C results from the error t=ln⁡(0.25)0.1t = \frac{\ln(0.25)}{0.1}t=0.1ln(0.25)​ (missing the negative). Choice D uses an incorrect logarithm value.

Question 7

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

  1. 12 hours after the initial measurement (correct answer)
  2. 16 hours after the initial measurement
  3. 10 hours after the initial measurement
  4. 14 hours after the initial measurement

Explanation: We need to solve 500⋅3t/4=13500500 \cdot 3^{t/4} = 13500500⋅3t/4=13500. Dividing both sides by 500: 3t/4=273^{t/4} = 273t/4=27. Since 27=3327 = 3^327=33, we have 3t/4=333^{t/4} = 3^33t/4=33. Therefore t4=3\frac{t}{4} = 34t​=3, so t=12t = 12t=12. Choice B comes from incorrectly setting 3t/4=343^{t/4} = 3^43t/4=34 when 27=3427 = 3^427=34 is wrong. Choice C results from the error t4=2.5\frac{t}{4} = 2.54t​=2.5 when thinking 27=32.527 = 3^{2.5}27=32.5. Choice D comes from solving 3t/4=33.53^{t/4} = 3^{3.5}3t/4=33.5 incorrectly.

Question 8

A bacteria culture follows P(t)=1000e0.22tP(t)=1000e^{0.22t}P(t)=1000e0.22t (t in hours); when does it reach 300030003000?

  1. t≈9.99t\approx 9.99t≈9.99 hours
  2. t≈4.99t\approx 4.99t≈4.99 hours (correct answer)
  3. t≈13.6t\approx 13.6t≈13.6 hours
  4. t≈3.15t\approx 3.15t≈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 9

A savings account has P(t)=1500e0.08tP(t)=1500e^{0.08t}P(t)=1500e0.08t (t in years); when does it reach 300030003000?

  1. t≈4.33t\approx 4.33t≈4.33 years
  2. t≈8.66t\approx 8.66t≈8.66 years (correct answer)
  3. t≈11.6t\approx 11.6t≈11.6 years
  4. t≈17.3t\approx 17.3t≈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 1500andgrowsatarateof0.08peryear,studentsareexpectedtosolvefortwhenitreaches1500 and grows at a rate of 0.08 per year, students are expected to solve for t when it reaches 1500andgrowsatarateof0.08peryear,studentsareexpectedtosolvefortwhenitreaches3000 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 10

If 4x−3=2x+14^{x-3} = 2^{x+1}4x−3=2x+1, what is the value of xxx?

  1. x=4x = 4x=4
  2. x=7x = 7x=7 (correct answer)
  3. x=5x = 5x=5
  4. x=6x = 6x=6

Explanation: Since 4=224 = 2^24=22, we can rewrite the equation as (22)x−3=2x+1(2^2)^{x-3} = 2^{x+1}(22)x−3=2x+1. Using the power rule, this becomes 22(x−3)=2x+12^{2(x-3)} = 2^{x+1}22(x−3)=2x+1, or 22x−6=2x+12^{2x-6} = 2^{x+1}22x−6=2x+1. Since the bases are equal, the exponents must be equal: 2x−6=x+12x-6 = x+12x−6=x+1. Solving for xxx: 2x−x=1+62x - x = 1 + 62x−x=1+6, so x=7x = 7x=7. Choice A results from solving x−3=x+1x-3 = x+1x−3=x+1 incorrectly. Choice C comes from the error 2x−6=x−12x-6 = x-12x−6=x−1. Choice D results from 2x=x+62x = x+62x=x+6 without properly distributing the exponent.

Question 11

A radioactive sample has A(t)=60e−0.20tA(t)=60e^{-0.20t}A(t)=60e−0.20t (t in days); when does it reach 303030 grams?

  1. t≈1.73t\approx 1.73t≈1.73 days
  2. t≈3.47t\approx 3.47t≈3.47 days (correct answer)
  3. t≈6.93t\approx 6.93t≈6.93 days
  4. t≈13.9t\approx 13.9t≈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 12

A bacteria culture follows P(t)=250e0.16tP(t)=250e^{0.16t}P(t)=250e0.16t (t in hours); when does it reach 500500500?

  1. t≈2.17t\approx 2.17t≈2.17 hours
  2. t≈4.33t\approx 4.33t≈4.33 hours (correct answer)
  3. t≈8.66t\approx 8.66t≈8.66 hours
  4. t≈13.9t\approx 13.9t≈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 13

A savings account has P(t)=3000e0.06tP(t)=3000e^{0.06t}P(t)=3000e0.06t (t in years); when does it reach 450045004500?

  1. t≈6.76t\approx 6.76t≈6.76 years (correct answer)
  2. t≈4.05t\approx 4.05t≈4.05 years
  3. t≈11.3t\approx 11.3t≈11.3 years
  4. t≈2.70t\approx 2.70t≈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 3000andgrowsatarateof0.06peryear,studentsareexpectedtosolvefortwhenitreaches3000 and grows at a rate of 0.06 per year, students are expected to solve for t when it reaches 3000andgrowsatarateof0.06peryear,studentsareexpectedtosolvefortwhenitreaches4500 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)=800e0.12tP(t)=800e^{0.12t}P(t)=800e0.12t (t in hours); when does it reach 160016001600?

  1. t≈2.89t\approx 2.89t≈2.89 hours
  2. t≈5.78t\approx 5.78t≈5.78 hours (correct answer)
  3. t≈11.6t\approx 11.6t≈11.6 hours
  4. t≈8.66t\approx 8.66t≈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)=800e0.05tP(t)=800e^{0.05t}P(t)=800e0.05t (t in years); when does it reach 100010001000?

  1. t≈4.46t\approx 4.46t≈4.46 years (correct answer)
  2. t≈2.23t\approx 2.23t≈2.23 years
  3. t≈3.57t\approx 3.57t≈3.57 years
  4. t≈1.12t\approx 1.12t≈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 800andgrowsatarateof0.05peryear,studentsareexpectedtosolvefortwhenitreaches800 and grows at a rate of 0.05 per year, students are expected to solve for t when it reaches 800andgrowsatarateof0.05peryear,studentsareexpectedtosolvefortwhenitreaches1000 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.