ACT Math Quiz: Logarithmic Functions
20 questions · exam conditions
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Logarithmic FunctionsQuestion 1 of 20

What is the value of xx that satisfies log⁡3(x+2)+log⁡3(x−4)=3\log_3(x + 2) + \log_3(x - 4) = 3?

55
77
99
1111
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ACT Math Quiz

ACT Math Quiz: Logarithmic Functions

Practice Logarithmic Functions in ACT Math 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 Logarithmic Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for ACT Math.

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

What is the value of xx that satisfies log⁡3(x+2)+log⁡3(x−4)=3\log_3(x + 2) + \log_3(x - 4) = 3?

  1. 55
  2. 77 (correct answer)
  3. 99
  4. 1111
Explanation: This is a logarithms question testing the product rule and extraneous solution detection. Choice B (7) is correct — apply the log product rule: log₃(x + 2) + log₃(x − 4) = log₃((x + 2)(x − 4)) = 3. Convert to exponential form: (x + 2)(x − 4) = 3³ = 27. Expand: x² − 2x − 8 = 27 → x² − 2x − 35 = 0 → (x − 7)(x + 5) = 0 → x = 7 or x = −5. Check: x = −5 makes log₃(−5 + 2) = log₃(−3), which is undefined (can't take log of a negative). So x = 7 is the only valid solution. Choice A (5) comes from a factoring error: perhaps solving x² − 2x − 35 = 0 as (x − 5)(x + 7) = 0. Choice C (9) comes from treating each log separately: log₃(x + 2) = 3 → x + 2 = 27 → x = 25... or log₃(x − 4) = 3 → x − 4 = 27 → x = 31. Choice D (11) comes from adding: (x + 2) + (x − 4) = 27 → 2x − 2 = 27 → x = 14.5, rounding or computing differently. Pro tip: After applying the log product rule, you'll get a quadratic. It will typically have two roots — always check BOTH in the original equation. A root that produces a negative or zero argument for any logarithm is extraneous and must be discarded.

Question 2

A worksheet asks you to simplify log⁡(2)+log⁡(50)\log(2)+\log(50) (base 10). Which single logarithm is equivalent?

  1. log⁡(52)\log(52)
  2. log⁡ ⁣(250)\log\!\left(\dfrac{2}{50}\right)
  3. log⁡(100)\log(100) (correct answer)
  4. log⁡(2⋅50)⋅log⁡(10)\log(2\cdot 50)\cdot\log(10)
Explanation: This problem uses the logarithm product property. The product property states that log_a(x) + log_a(y) = log_a(xy). Applying this property to log(2) + log(50), we get log(2) + log(50) = log(2 × 50) = log(100). Choice A incorrectly adds the arguments instead of multiplying them.

Question 3

What is log⁡10(1000)\log_{10}(1000)?

  1. 3 (correct answer)
  2. 2
  3. 10
  4. 1
Explanation: To evaluate this logarithm, we need to find what power 10 must be raised to get 1000. The logarithm property states that log_a(b) = c means a^c = b. We can rewrite 1000 as 10^3, so log₁₀(1000) = log₁₀(10310^3). Using the power rule for logarithms, log_a(xnx^n) = n·log_a(x), we get 3·log₁₀(10) = 3·1 = 3.

Question 4

Evaluate log⁡3(81)\log_3(81).

  1. 4 (correct answer)
  2. 3
  3. 2
  4. 5
Explanation: To evaluate this logarithm, we need to find what power 3 must be raised to get 81. The logarithm property states that log⁡a(b)=c\log_a(b) = c means ac=ba^c = b. We can rewrite 81 as 343^4 (since 3⋅3⋅3⋅3=813 \cdot 3 \cdot 3 \cdot 3 = 81). Therefore, log⁡3(81)=log⁡3(34)\log_3(81) = \log_3(3^4). Using the power rule, this equals 4⋅log⁡3(3)=4⋅1=44 \cdot \log_3(3) = 4 \cdot 1 = 4.

Question 5

A calculator app uses base-10 logs. If log⁡(x)=−2\log(x)= -2, what is the value of xx?

  1. −2-2
  2. 10−210^{-2} (correct answer)
  3. 22
  4. −102-10^{2}
Explanation: Given log(x) = -2 (base 10 implied), we need to find x. Using the definition log₁₀(x) = -2 means 10^(-2) = x. Therefore, x = 10^(-2) = 1/10² = 1/100 = 0.01. Choice A gives just -2, which is the logarithm value, not x itself.

Question 6

What is the value of xx if log⁡(x)=2\log(x) = 2?

  1. 100 (correct answer)
  2. 10
  3. 20
  4. 200
Explanation: To solve this equation, we need to convert from logarithmic to exponential form. The equation log(x) = 2 means "10 raised to what power equals x?" Since log without a specified base typically means log₁₀, we have 10² = x. Therefore, x = 100. Choice B would give 10¹ = 10, which doesn't satisfy the original equation.

Question 7

A student solves the equation ln⁡(x)=0\ln(x)=0. What is the value of xx?

  1. 00
  2. 11 (correct answer)
  3. ee
  4. −1-1
Explanation: This problem uses the fundamental property that ln(1) = 0 and the inverse relationship between natural logarithm and exponential functions. The equation ln(x) = 0 means that e^0 = x. Since e^0 = 1 for any base, we have x = 1. Choice A incorrectly gives the exponent value, while choice C gives the base of natural logarithm.

Question 8

If log⁡2(x)+log⁡2(4)=5\log_2(x) + \log_2(4) = 5, what is the value of xx?

  1. 4
  2. 8 (correct answer)
  3. 16
  4. 32
Explanation: This is a logarithms question testing the product rule. Choice B (8) is correct — apply the log product rule: log₂(x) + log₂(4) = log₂(4x) = 5. Convert to exponential form: 4x = 2⁵ = 32. Solve: x = 8. Since log₂(4) = 2, you can also solve as: log₂(x) = 5 − 2 = 3 → x = 2³ = 8. Choice A (4) results from computing 2³ incorrectly — arriving at the right exponent of 3 but evaluating 2³ as 4 (possibly confusing 2² = 4 with 2³ = 8). Choice C (16) results from an off-by-one exponent error after correctly applying the product rule: correctly getting log₂(x) = 3, but then computing x = 2⁴ = 16 instead of 2³ = 8. Choice D (32) ignores the log₂(4) term entirely, solving log₂(x) = 5 → x = 2⁵ = 32. Pro tip: The log product rule states log_b(M) + log_b(N) = log_b(MN). Use it to combine the two log terms before converting to exponential form — this is almost always faster than working with them separately.

Question 9

What is ln⁡(e2)\ln(e^2)?

  1. 22 (correct answer)
  2. 11
  3. ee
  4. e2e^2
Explanation: To evaluate this natural logarithm, we use the power rule for logarithms. The property states that log⁡a(xn)=n⋅log⁡a(x)\log_a(x^n) = n \cdot \log_a(x). Since ln means log_e, we have ln⁡(e2)\ln(e^2). Using the power rule: ln⁡(e2)=2⋅ln⁡(e)=2⋅1=2\ln(e^2) = 2 \cdot \ln(e) = 2 \cdot 1 = 2. This demonstrates that the natural logarithm and exponential functions with base e are inverse operations.

Question 10

A student is rewriting a logarithmic equation in exponential form. Which equation is equivalent to log⁡2(32)=5\log_{2}(32)=5?

  1. 232=52^{32}=5
  2. 52=325^{2}=32
  3. 25=322^{5}=32 (correct answer)
  4. 325=232^{5}=2
Explanation: To convert log₂(32) = 5 to exponential form, we use the definition that log_a(b) = c means a^c = b. Here, a = 2, b = 32, and c = 5. Therefore, the exponential form is 2⁵ = 32. We can verify: 2⁵ = 2 × 2 × 2 × 2 × 2 = 32. Choice A incorrectly swaps the base and result.

Question 11

Which equation is equivalent to log⁡10(100)=2\log_{10}(100) = 2?

  1. 2100=102^{100} = 10
  2. 210=1002^{10} = 100
  3. 103=10010^3 = 100
  4. 102=10010^2 = 100 (correct answer)
Explanation: To convert from logarithmic to exponential form, we use the fundamental property that log_a(b) = c is equivalent to a^c = b. Given log₁₀(100) = 2, the base is 10, the exponent is 2, and the result is 100. Therefore, the equivalent exponential equation is 10^2 = 100. Choice B incorrectly switches the base and exponent positions.

Question 12

Which expression is equivalent to log⁡(xy)\log(xy) using logarithm properties?

  1. log⁡(x)+log⁡(y)\log(x) + \log(y) (correct answer)
  2. log⁡(x)⋅log⁡(y)\log(x) \cdot \log(y)
  3. log⁡(x)log⁡(y)\frac{\log(x)}{\log(y)}
  4. log⁡(x)−log⁡(y)\log(x) - \log(y)
Explanation: This problem requires the product rule for logarithms. The logarithm property for products states that log⁡a(xy)=log⁡a(x)+log⁡a(y)\log_a(xy) = \log_a(x) + \log_a(y). This means when we have the logarithm of a product, we can separate it into the sum of individual logarithms. Therefore, log⁡(xy)=log⁡(x)+log⁡(y)\log(xy) = \log(x) + \log(y). Choice B incorrectly suggests multiplying logarithms instead of adding them.

Question 13

log⁡7(49)\log_7(49) equals what?

  1. 1
  2. 3
  3. 2 (correct answer)
  4. 4
Explanation: To evaluate this logarithm, we use the definition: log⁡7(49)\log_7(49) asks "To what power must we raise 7 to get 49?" Since 727^2 = 7×77 \times 7 = 49, we have log⁡7(49)\log_7(49) = 2. Choice B would be incorrect because 737^3 = 343, not 49.

Question 14

Solve for xx if log⁡3(x)=4\log_3(x) = 4.

  1. 81 (correct answer)
  2. 12
  3. 9
  4. 3
Explanation: To solve this equation, we need to convert from logarithmic to exponential form. The logarithm property states that loga(b)=clog_a(b) = c means ac=ba^c = b. Given log3(x)=4log_3(x) = 4, we convert to exponential form: 34=x3^4 = x. Calculating 34=3⋅3⋅3⋅3=813^4 = 3 \cdot 3 \cdot 3 \cdot 3 = 81, so x=81x = 81.

Question 15

Which expression is equivalent to log⁡(xy)\log(\frac{x}{y})?

  1. log⁡(y)−log⁡(x)\log(y) - \log(x)
  2. log⁡(x)+log⁡(y)\log(x) + \log(y)
  3. log⁡(x)⋅log⁡(y)\log(x) \cdot \log(y)
  4. log⁡(x)−log⁡(y)\log(x) - \log(y) (correct answer)
Explanation: This problem uses the quotient rule for logarithms. The quotient rule states that log(x/y) = log(x) - log(y). This property allows us to separate the logarithm of a fraction into the difference of two logarithms. Choice B incorrectly suggests adding the logarithms, which would apply to a product, not a quotient.

Question 16

A student evaluates log⁡9(3)\log_9(3). What is the value of log⁡9(3)\log_9(3)?

  1. 33
  2. 22
  3. 12\dfrac{1}{2} (correct answer)
  4. 13\dfrac{1}{3}
Explanation: This problem uses the relationship between logarithms with different bases that are powers of each other. Since 9 = 3^2, we need to find what power of 9 equals 3. This means 9^x = 3, or (3^2)^x = 3^1, which gives us 3^(2x) = 3^1. Therefore, 2x = 1, so x = 1/2. Choice B incorrectly gives the reciprocal relationship.

Question 17

For positive numbers xx and yy, which expression is equivalent to log⁡(xy)\log(xy) (base 10)?

  1. log⁡(x)+log⁡(y)\log(x)+\log(y) (correct answer)
  2. log⁡(x)−log⁡(y)\log(x)-\log(y)
  3. log⁡(x)⋅log⁡(y)\log(x)\cdot\log(y)
  4. log⁡ ⁣(xy)\log\!\left(\dfrac{x}{y}\right)
Explanation: We need the product rule for logarithms to simplify log(xy). The product rule states that log_a(xy) = log_a(x) + log_a(y) for positive x and y. Therefore, log(xy) = log(x) + log(y). Choice C incorrectly multiplies the logarithms, while choice B would apply to log(x/y).

Question 18

In a chemistry calculation using natural logarithms, evaluate ln⁡(e5)\ln(e^5).

  1. e5e^5
  2. 5e5e
  3. 55 (correct answer)
  4. ln⁡(5e)\ln(5e)
Explanation: To evaluate ln(e⁵), we use the property that logarithms and exponentials are inverse functions. Specifically, log_a(axa^x) = x for any valid base a. Since ln is the natural logarithm with base e, we have ln(e⁵) = 5. This follows directly from the definition: if ln(e⁵) = x, then e^x = e⁵, so x = 5.

Question 19

A student solves a simple logarithmic equation. If log⁡5(x)=3\log_{5}(x)=3, what is xx?

  1. 1515
  2. 125125 (correct answer)
  3. 35\dfrac{3}{5}
  4. 51/35^{1/3}
Explanation: To solve log₅(x) = 3, we use the definition that log_a(b) = c means a^c = b. Here, log₅(x) = 3 means 5³ = x. Therefore, x = 5³ = 5 × 5 × 5 = 125. Choice C incorrectly gives 3/5, while choice D would be the solution to log₅(x) = 1/3.

Question 20

A researcher is simplifying a natural-log expression with positive values. Which expression is equivalent to ln⁡ ⁣(xy)\ln\!\left(\dfrac{x}{y}\right)?

  1. ln⁡(x)+ln⁡(y)\ln(x)+\ln(y)
  2. ln⁡(x)−ln⁡(y)\ln(x)-\ln(y) (correct answer)
  3. ln⁡(x)ln⁡(y)\dfrac{\ln(x)}{\ln(y)}
  4. ln⁡(xy)\ln(xy)
Explanation: To express ln(x/y) in terms of ln(x) and ln(y), we use the quotient rule for logarithms. The quotient rule states that log_a(x/y) = log_a(x) - log_a(y). Since this applies to natural logarithms as well, ln(x/y) = ln(x) - ln(y). Choice A would be correct for ln(xy), not ln(x/y).