ACCUPLACER Advanced Algebra & Functions Quiz: Solving Logarithmic Equations
20 questions · exam conditions
0:00
Solving Logarithmic EquationsQuestion 1 of 20

What is the solution to the equation log2(3x−1)=5log_2(3x - 1) = 5?

x=76x = \frac{7}{6}
x=263x = \frac{26}{3}
x=11x = 11
x=313x = \frac{31}{3}
← Back to quizzes

ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Solving Logarithmic Equations

Practice Solving Logarithmic Equations in ACCUPLACER Advanced Algebra & Functions 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 Logarithmic Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

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 solution to the equation log2(3x−1)=5log_2(3x - 1) = 5?

  1. x=76x = \frac{7}{6}
  2. x=263x = \frac{26}{3}
  3. x=11x = 11 (correct answer)
  4. x=313x = \frac{31}{3}
Explanation: To solve the equation, convert it from logarithmic form to exponential form. The equation logb(y)=clog_b(y) = c is equivalent to bc=yb^c = y. Thus, log2(3x−1)=5log_2(3x - 1) = 5 becomes 25=3x−12^5 = 3x - 1. Since 25=322^5 = 32, the equation is 32=3x−132 = 3x - 1. Adding 1 to both sides gives 33=3x33 = 3x. Dividing by 3 gives x=11x = 11. Distractor A comes from incorrectly converting the equation to 2(3x−1)=52(3x-1)=5. Distractor B comes from mistakenly swapping the base and the result, calculating 52=3x−15^2 = 3x-1. Distractor D is a simple calculation error.

Question 2

If log7(5x−9)=log7(2x+3)log_7(5x - 9) = log_7(2x + 3), what is the value of xx?

  1. x=67x = \frac{6}{7}
  2. x=127x = \frac{12}{7}
  3. x=2x = 2
  4. x=4x = 4 (correct answer)
Explanation: By the one-to-one property of logarithms, if logb(M)=logb(N)log_b(M) = log_b(N), then M=NM = N. Therefore, we can set the arguments equal to each other: 5x−9=2x+35x - 9 = 2x + 3. To solve for xx, subtract 2x2x from both sides to get 3x−9=33x - 9 = 3. Then, add 9 to both sides to get 3x=123x = 12. Finally, divide by 3 to find x=4x = 4. We must check that the arguments are positive for this value of x. For x=4x=4, 5(4)−9=11>05(4)-9=11 > 0 and 2(4)+3=11>02(4)+3=11 > 0, so the solution is valid. Distractor B results from a sign error when moving terms, such as 5x+2x=3+95x+2x=3+9. The other distractors represent other common algebraic errors.

Question 3

What is the solution to the equation log4(x+12)−log4(x)=2log_4(x + 12) - log_4(x) = 2?

  1. x=−45x = -\frac{4}{5}
  2. x=45x = \frac{4}{5} (correct answer)
  3. x=54x = \frac{5}{4}
  4. x=127x = \frac{12}{7}
Explanation: Using the quotient rule for logarithms, logb(M)−logb(N)=logb(M/N)log_b(M) - log_b(N) = log_b(M/N), the equation becomes log4(x+12x)=2log_4(\frac{x+12}{x}) = 2. Converting to exponential form gives x+12x=42=16\frac{x+12}{x} = 4^2 = 16. To solve for xx, multiply both sides by xx: x+12=16xx + 12 = 16x. Subtract xx from both sides: 12=15x12 = 15x. Divide by 15: x=1215=45x = \frac{12}{15} = \frac{4}{5}. This solution is valid as it makes the arguments of both logarithms positive. Distractor C results from incorrectly flipping the fraction when solving 12=15x12=15x. Distractor D results from misinterpreting the conversion to exponential form as x+12x=4×2=8\frac{x+12}{x} = 4 \times 2 = 8. Distractor A is a sign error.

Question 4

What is the solution set for the equation log3(4x−5)=log3(2x−11)log_3(4x - 5) = log_3(2x - 11)?

  1. x=−8x = -8
  2. x=−3x = -3
  3. x=3x = 3
  4. No solution (correct answer)
Explanation: By the one-to-one property of logarithms, if logb(M)=logb(N)log_b(M) = log_b(N), then M=NM = N. Setting the arguments equal gives 4x−5=2x−114x - 5 = 2x - 11. Subtracting 2x2x from both sides gives 2x−5=−112x - 5 = -11. Adding 5 to both sides gives 2x=−62x = -6. Dividing by 2 yields x=−3x = -3. However, this solution must be checked in the original equation. For x=−3x = -3, the argument 4x−54x - 5 becomes 4(−3)−5=−174(-3) - 5 = -17. Since the logarithm of a negative number is undefined, x=−3x = -3 is an extraneous solution. There are no other algebraic solutions, so the equation has no solution. Distractor B is the extraneous algebraic solution. Distractors A and C result from sign errors during the algebraic solution process.

Question 5

What is the value of xx if log3(x+2)=log2(16)log_3(x + 2) = log_2(16)?

  1. x=2x = 2
  2. x=14x = 14
  3. x=62x = 62
  4. x=79x = 79 (correct answer)
Explanation: First, evaluate the right side of the equation. log2(16)log_2(16) asks for the power to which 2 must be raised to get 16. Since 24=162^4 = 16, log2(16)=4log_2(16) = 4. The equation simplifies to log3(x+2)=4log_3(x + 2) = 4. Now, convert this logarithmic equation to its exponential form: x+2=34x + 2 = 3^4. Since 34=813^4 = 81, we have x+2=81x + 2 = 81. Subtracting 2 from both sides gives x=79x = 79. Distractor B results from ignoring the bases and incorrectly setting x+2=16x+2=16. Distractor A results from finding the value of the right side is 4, but incorrectly setting x+2=4x+2=4.

Question 6

If log2(x+6)=3+log2(x)log_2(x + 6) = 3 + log_2(x), what is the value of xx?

  1. x=67x = \frac{6}{7} (correct answer)
  2. x=65x = \frac{6}{5}
  3. x=2x = 2
  4. x=3x = 3
Explanation: First, group the logarithm terms on one side of the equation: log2(x+6)−log2(x)=3log_2(x + 6) - log_2(x) = 3. Apply the quotient rule for logarithms: log2(x+6x)=3log_2(\frac{x+6}{x}) = 3. Convert this to exponential form: x+6x=23=8\frac{x+6}{x} = 2^3 = 8. Now, solve for xx: x+6=8xx + 6 = 8x. Subtract xx from both sides: 6=7x6 = 7x. Therefore, x=67x = \frac{6}{7}. This solution is valid since it makes the arguments of the logarithms positive. Distractor D results from misinterpreting 3+log2(x)3 + log_2(x) as log2(3x)log_2(3x). Distractor B results from an error in conversion, using 2×3=62 \times 3=6 instead of 23=82^3=8.

Question 7

What is the value of xx in the equation logx(81)=4log_x(81) = 4?

  1. x=−3x = -3
  2. x=3x = 3 (correct answer)
  3. x=−3x = -3 and x=3x = 3
  4. x=20.25x = 20.25
Explanation: Convert the logarithmic equation to its exponential form: x4=81x^4 = 81. To solve for xx, take the fourth root of both sides: x=±814x = \pm \sqrt[4]{81}. Since 34=813^4 = 81, the algebraic solutions are x=3x = 3 and x=−3x = -3. However, the base of a logarithm must be positive and not equal to 1. Therefore, x=−3x = -3 is not a valid solution. The only valid solution is x=3x = 3. Distractor A is invalid because the base cannot be negative. Distractor C incorrectly includes the invalid negative base. Distractor D results from an incorrect calculation, 81/481/4.

Question 8

What is the solution to the equation log3(x+24)=2+log3(x)log_3(x + 24) = 2 + log_3(x)?

  1. x=3x = 3 (correct answer)
  2. x=4.8x = 4.8
  3. x=12x = 12
  4. x=24x = 24
Explanation: First, isolate the logarithm terms on one side: log3(x+24)−log3(x)=2log_3(x + 24) - log_3(x) = 2. Use the quotient rule for logarithms: log3(x+24x)=2log_3(\frac{x+24}{x}) = 2. Convert to exponential form: x+24x=32=9\frac{x+24}{x} = 3^2 = 9. Solve for xx: x+24=9xx + 24 = 9x. Subtract xx from both sides: 24=8x24 = 8x. Divide by 8 to get x=3x = 3. Checking the solution shows it is valid. Distractor D results from misinterpreting 2+log3(x)2 + log_3(x) as log3(2x)log_3(2x). Distractor B results from an error in exponential conversion, using 3×2=63 \times 2 = 6 instead of 32=93^2 = 9.

Question 9

Solve for xx: log4(log2(x))=1log_4(log_2(x)) = 1

  1. x=2x = 2
  2. x=4x = 4
  3. x=8x = 8
  4. x=16x = 16 (correct answer)
Explanation: This is a nested logarithm problem that must be solved from the outside in. First, convert the outer logarithm to exponential form. The equation log4(A)=1log_4(A) = 1 where A=log2(x)A = log_2(x) becomes A=41A = 4^1, so log2(x)=4log_2(x) = 4. Now, solve this second logarithmic equation by converting it to exponential form: x=24x = 2^4. Calculating the power gives x=16x = 16. Distractor A results from incorrectly solving log2(x)=1log_2(x) = 1, perhaps by confusing the bases. Distractor C results from an error in the second conversion step, calculating x=2×4x = 2 \times 4 instead of x=24x = 2^4. Distractor B comes from stopping after the first step, where log2(x)=4log_2(x) = 4, and confusing the result with the variable.

Question 10

Solve for xx: 3log4(x−2)+5=113log_4(x - 2) + 5 = 11

  1. x=10x = 10
  2. x=18x = 18 (correct answer)
  3. x=50x = 50
  4. x=66x = 66
Explanation: First, isolate the logarithm term. Subtract 5 from both sides: 3log4(x−2)=63log_4(x - 2) = 6. Divide by 3: log4(x−2)=2log_4(x - 2) = 2. Now, convert the equation to exponential form: x−2=42x - 2 = 4^2. Simplify: x−2=16x - 2 = 16. Add 2 to both sides to find x=18x = 18. The solution is valid as 18−2=1618-2=16 is positive. Distractor A comes from an error in converting to exponential form, calculating x−2=4×2=8x-2 = 4 \times 2 = 8. Distractor D comes from an order of operations error where 3 is subtracted before division. Distractor C comes from incorrectly dividing the argument of the log by 3.

Question 11

If 2log⁡4(x)=log⁡4(x+12)2\log_4(x) = \log_4(x + 12), what is the value of xx?

  1. x=4x = 4 (correct answer)
  2. x=6x = 6
  3. x=12x = 12
  4. x=3x = 3
Explanation: Using the logarithm property nlog⁡a(x)=log⁡a(xn)n\log_a(x) = \log_a(x^n), we rewrite the left side: log⁡4(x2)=log⁡4(x+12)\log_4(x^2) = \log_4(x + 12). Since the logarithms have the same base, their arguments must be equal: x2=x+12x^2 = x + 12. Rearranging: x2−x−12=0x^2 - x - 12 = 0. Factoring: (x−4)(x+3)=0(x - 4)(x + 3) = 0, giving x=4x = 4 or x=−3x = -3. Checking domain restrictions: for log⁡4(x)\log_4(x) to be defined, we need x>0x > 0. Therefore, x=−3x = -3 is extraneous, leaving x=4x = 4 as the only valid solution. Verification: 2log⁡4(4)=2(1)=22\log_4(4) = 2(1) = 2 and log⁡4(4+12)=log⁡4(16)=log⁡4(42)=2\log_4(4 + 12) = \log_4(16) = \log_4(4^2) = 2. Choice B gives 2log⁡4(6)≠log⁡4(18)2\log_4(6) \neq \log_4(18). Choice C gives 2log⁡4(12)≠log⁡4(24)2\log_4(12) \neq \log_4(24). Choice D gives 2log⁡4(3)≠log⁡4(15)2\log_4(3) \neq \log_4(15).

Question 12

What is the solution to the equation log6(x+5)+log6(x)=2log_6(x + 5) + log_6(x) = 2?

  1. x=−9x = -9
  2. x=4x = 4 (correct answer)
  3. x=−9x = -9 and x=4x = 4
  4. x=15.5x = 15.5
Explanation: Using the product rule for logarithms, combine the terms on the left: log6(x(x+5))=2log_6(x(x + 5)) = 2. Convert to exponential form: x(x+5)=62x(x + 5) = 6^2. Simplify and solve the quadratic equation: x2+5x=36x^2 + 5x = 36, or x2+5x−36=0x^2 + 5x - 36 = 0. Factoring gives (x+9)(x−4)=0(x + 9)(x - 4) = 0. The potential solutions are x=−9x = -9 and x=4x = 4. Checking these in the original equation, x=−9x = -9 is extraneous because log6(−9)log_6(-9) is undefined. The solution x=4x = 4 is valid. Distractor A is the extraneous solution. Distractor C incorrectly includes the extraneous solution. Distractor D results from incorrectly applying the product rule as log6(x+5+x)=2log_6(x+5+x)=2.

Question 13

What is the solution to the equation log(5x)−log(x−2)=1log(5x) - log(x - 2) = 1?

  1. x=−4x = -4
  2. x=−12x = -\frac{1}{2}
  3. x=2x = 2
  4. x=4x = 4 (correct answer)
Explanation: The term loglog implies a base of 10. Using the quotient rule, the equation becomes log(5xx−2)=1log(\frac{5x}{x-2}) = 1. Converting to exponential form gives 5xx−2=101=10\frac{5x}{x-2} = 10^1 = 10. To solve for xx, multiply both sides by (x−2)(x-2): 5x=10(x−2)5x = 10(x - 2). Distribute the 10: 5x=10x−205x = 10x - 20. Subtract 10x10x from both sides: −5x=−20-5x = -20. Divide by -5 to get x=4x = 4. Checking the solution, both log(20)log(20) and log(2)log(2) are defined. Distractor C results from misusing the quotient rule as log(5x−(x−2))=1log(5x-(x-2))=1, which is extraneous as it leads to log(0)log(0). Distractor B results from forgetting the base is 10 and setting the argument equal to 1. Distractor A comes from a sign error in the algebra.

Question 14

What is the solution to the equation 2+ln(x)=52 + ln(x) = 5?

  1. x=e3x = e^3 (correct answer)
  2. x=e5−2x = e^5 - 2
  3. x=1000x = 1000
  4. x=3ex = 3e
Explanation: First, isolate the natural logarithm term. Subtract 2 from both sides of the equation: ln(x)=5−2ln(x) = 5 - 2, which simplifies to ln(x)=3ln(x) = 3. The notation ln(x)ln(x) is shorthand for loge(x)log_e(x). To solve for xx, convert the equation to its exponential form: x=e3x = e^3. Distractor C results from confusing the natural log (base ee) with the common log (base 10). Distractor B results from an order of operations error, evaluating ln(2+x)=5ln(2+x)=5 instead. Distractor D represents a misunderstanding of how to convert from logarithmic to exponential form.

Question 15

Which of the following is a solution to the equation log3(x)+log3(x−8)=2log_3(x) + log_3(x - 8) = 2?

  1. x=−1x = -1
  2. x=8.5x = 8.5
  3. x=9x = 9 (correct answer)
  4. x=−1x = -1 and x=9x = 9
Explanation: First, use the product rule for logarithms, which states that logb(M)+logb(N)=logb(MN)log_b(M) + log_b(N) = log_b(MN). This transforms the equation to log3(x(x−8))=2log_3(x(x - 8)) = 2. Convert to exponential form: x(x−8)=32x(x - 8) = 3^2. Simplify and solve the resulting quadratic equation: x2−8x=9x^2 - 8x = 9, which is x2−8x−9=0x^2 - 8x - 9 = 0. Factoring gives (x−9)(x+1)=0(x - 9)(x + 1) = 0, so the potential solutions are x=9x = 9 and x=−1x = -1. However, logarithms are only defined for positive arguments. The solution x=−1x = -1 is extraneous because log3(−1)log_3(-1) is undefined. The solution x=9x = 9 is valid because both xx and x−8x-8 are positive. Distractor A is the extraneous solution. Distractor D includes the extraneous solution. Distractor B results from misapplying the product rule as log3(x+x−8)=2log_3(x + x - 8) = 2.

Question 16

If log27(x)=23log_{27}(x) = \frac{2}{3}, what is the value of xx?

  1. x=6x = 6
  2. x=9x = 9 (correct answer)
  3. x=18x = 18
  4. x=243x = 243
Explanation: To solve for xx, convert the logarithmic equation to its exponential form: x=2723x = 27^{\frac{2}{3}}. A rational exponent can be interpreted as a root and a power. The denominator (3) represents the cube root, and the numerator (2) represents the square. So, x=(273)2x = (\sqrt[3]{27})^2. The cube root of 27 is 3, so x=32=9x = 3^2 = 9. Distractor C comes from incorrectly computing 27×2327 \times \frac{2}{3}. Distractor A comes from misinterpreting the exponent as 273×2\sqrt[3]{27} \times 2. Distractor D results from miscalculating the exponent, for example as 272/327^2/3.

Question 17

What is the complete solution set for the equation log2(x2−2x)=3log_2(x^2 - 2x) = 3?

  1. x=−2x = -2
  2. x=4x = 4
  3. x=1−7x = 1 - \sqrt{7} and x=1+7x = 1 + \sqrt{7}
  4. x=−2x = -2 and x=4x = 4 (correct answer)
Explanation: First, convert the equation from logarithmic to exponential form: x2−2x=23x^2 - 2x = 2^3. Simplify the right side: x2−2x=8x^2 - 2x = 8. Rearrange into a standard quadratic equation: x2−2x−8=0x^2 - 2x - 8 = 0. Factor the quadratic: (x−4)(x+2)=0(x - 4)(x + 2) = 0. The potential solutions are x=4x = 4 and x=−2x = -2. Both solutions must be checked in the original equation to ensure the argument of the logarithm is positive. For x=4x = 4, 42−2(4)=16−8=8>04^2 - 2(4) = 16 - 8 = 8 > 0, so it is a valid solution. For x=−2x = -2, (−2)2−2(−2)=4+4=8>0(-2)^2 - 2(-2) = 4 + 4 = 8 > 0, so it is also a valid solution. Therefore, the complete solution set includes both values. Distractors A and B are incomplete. Distractor C results from an error in converting to exponential form, using 2×3=62 \times 3=6 instead of 23=82^3=8.

Question 18

What is the solution to log2(x)+log2(5)=log2(x+8)log_2(x) + log_2(5) = log_2(x + 8)?

  1. x=−2x = -2
  2. x=43x = \frac{4}{3}
  3. x=2x = 2 (correct answer)
  4. No solution
Explanation: Using the product rule for logarithms on the left side, we get log2(5x)=log2(x+8)log_2(5x) = log_2(x + 8). By the one-to-one property of logarithms, we can set the arguments equal: 5x=x+85x = x + 8. Subtracting xx from both sides gives 4x=84x = 8. Dividing by 4 gives x=2x = 2. Checking the solution, for x=2x=2, all arguments (2,5,and 2+8=10)(2, 5, \text{and } 2+8=10) are positive, so the solution is valid. Distractor D could result from misapplying the product rule as log2(x+5)=log2(x+8)log_2(x+5) = log_2(x+8), which would lead to 5=85=8, an impossibility. Distractors A and B result from common algebraic sign errors.

Question 19

What is the solution to the equation 2log5(x)=log5(3x+10)2log_5(x) = log_5(3x + 10)?

  1. x=−10x = -10
  2. x=−2x = -2
  3. x=5x = 5 (correct answer)
  4. x=−2x = -2 and x=5x = 5
Explanation: Use the power rule for logarithms, nlogb(M)=logb(Mn)n log_b(M) = log_b(M^n), to rewrite the left side: log5(x2)=log5(3x+10)log_5(x^2) = log_5(3x + 10). By the one-to-one property, x2=3x+10x^2 = 3x + 10. Rearrange into a quadratic equation: x2−3x−10=0x^2 - 3x - 10 = 0. Factoring gives (x−5)(x+2)=0(x - 5)(x + 2) = 0. The potential solutions are x=5x = 5 and x=−2x = -2. We must check for extraneous solutions. For x=5x = 5, the arguments are positive. For x=−2x = -2, the term log5(−2)log_5(-2) is undefined. Thus, x=−2x = -2 is an extraneous solution. The only valid solution is x=5x = 5. Distractor B is the extraneous solution. Distractor D incorrectly includes the extraneous solution. Distractor A results from misapplying the power rule as log5(2x)=log5(3x+10)log_5(2x) = log_5(3x+10).

Question 20

What is the solution to the equation log(x+15)+log(x)=2log(x + 15) + log(x) = 2?

  1. x=−20x = -20
  2. x=5x = 5 (correct answer)
  3. x=−20x = -20 and x=5x = 5
  4. x=42.5x = 42.5
Explanation: The term loglog implies a base of 10. Using the product rule, the equation becomes log(x(x+15))=2log(x(x + 15)) = 2. Converting to exponential form gives x(x+15)=102x(x + 15) = 10^2, which simplifies to x2+15x=100x^2 + 15x = 100. Rearranging into a standard quadratic form gives x2+15x−100=0x^2 + 15x - 100 = 0. Factoring the quadratic yields (x+20)(x−5)=0(x + 20)(x - 5) = 0. The potential solutions are x=−20x = -20 and x=5x = 5. We must check these in the original equation. The solution x=−20x = -20 is extraneous because log(−20)log(-20) is undefined. The solution x=5x = 5 is valid because both log(5)log(5) and log(5+15)=log(20)log(5+15)=log(20) are defined. Distractor A is the extraneous solution. Distractor C incorrectly includes the extraneous solution. Distractor D arises from the common mistake of applying the product rule as log(x+15+x)=2log(x+15+x)=2.