College Algebra Quiz: Solving Logarithmic Equations
3 questions · exam conditions
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Solving Logarithmic EquationsQuestion 1 of 3

Which equation is equivalent to 3log⁡3(x+2)=x−13^{\log_3(x+2)} = x - 1?

x+2=x−1x + 2 = x - 1
log⁡3(x+2)=log⁡3(x−1)\log_3(x + 2) = \log_3(x - 1)
x+2=log⁡3(x−1)x + 2 = \log_3(x - 1)
log⁡3(x+2)=x−1\log_3(x + 2) = x - 1
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College Algebra Quiz

College Algebra Quiz: Solving Logarithmic Equations

Practice Solving Logarithmic 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 Logarithmic 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

Which equation is equivalent to 3log⁡3(x+2)=x−13^{\log_3(x+2)} = x - 1?

  1. x+2=x−1x + 2 = x - 1 (correct answer)
  2. log⁡3(x+2)=log⁡3(x−1)\log_3(x + 2) = \log_3(x - 1)
  3. x+2=log⁡3(x−1)x + 2 = \log_3(x - 1)
  4. log⁡3(x+2)=x−1\log_3(x + 2) = x - 1
Explanation: Using the fundamental property that alog⁡a(y)=ya^{\log_a(y)} = y (for y>0y > 0), we have 3log⁡3(x+2)=x+23^{\log_3(x+2)} = x + 2. Therefore, the equation 3log⁡3(x+2)=x−13^{\log_3(x+2)} = x - 1 becomes x+2=x−1x + 2 = x - 1. This simplifies to 2=−12 = -1, which is impossible, indicating that the original equation has no solution. However, the question asks for the equivalent form, not the solution. Choice B would be correct if we took log⁡3\log_3 of both sides, but we need x−1>0x - 1 > 0 for that to be valid. Choice C incorrectly swaps the roles of the logarithm and its argument. Choice D would result from taking log⁡3\log_3 of both sides, but again requires x−1>0x - 1 > 0. Choice A is the direct algebraic equivalent using the exponential-logarithm inverse relationship.

Question 2

Solve the equation log⁡2(x)+log⁡2(x−7)=3\log_2(x) + \log_2(x - 7) = 3. What is the positive solution?

  1. x=8x = 8 (correct answer)
  2. x=9x = 9
  3. x=7x = 7
  4. x=1x = 1 or x=8x = 8
Explanation: Using the logarithm property log⁡a(m)+log⁡a(n)=log⁡a(mn)\log_a(m) + \log_a(n) = \log_a(mn), we get log⁡2[x(x−7)]=3\log_2[x(x-7)] = 3. Converting to exponential form: x(x−7)=23=8x(x-7) = 2^3 = 8. Expanding: x2−7x=8x^2 - 7x = 8, so x2−7x−8=0x^2 - 7x - 8 = 0. Factoring: we need two numbers that multiply to -8 and add to -7. These are -8 and 1, so (x−8)(x+1)=0(x - 8)(x + 1) = 0. This gives x=8x = 8 or x=−1x = -1. We must check the domain: both x>0x > 0 and x−7>0x - 7 > 0 (so x>7x > 7) are required for the logarithms to be defined. Only x=8x = 8 satisfies x>7x > 7. Verification: log⁡2(8)+log⁡2(1)=3+0=3\log_2(8) + \log_2(1) = 3 + 0 = 3 ✓. Choice B is wrong because log⁡2(9)+log⁡2(2)=log⁡2(18)≠3\log_2(9) + \log_2(2) = \log_2(18) \neq 3. Choice C is wrong because x=7x = 7 makes log⁡2(x−7)=log⁡2(0)\log_2(x-7) = \log_2(0) undefined. Choice D includes x=1x = 1, but this makes x−7=−6<0x - 7 = -6 < 0, so log⁡2(x−7)\log_2(x-7) is undefined.

Question 3

If log⁡5(x+1)−log⁡5(x−2)=1\log_5(x + 1) - \log_5(x - 2) = 1, then xx equals:

  1. 114\frac{11}{4} (correct answer)
  2. 134\frac{13}{4}
  3. 72\frac{7}{2}
  4. 92\frac{9}{2}
Explanation: Using the logarithm property log⁡a(m)−log⁡a(n)=log⁡a(mn)\log_a(m) - \log_a(n) = \log_a(\frac{m}{n}), we get log⁡5(x+1x−2)=1\log_5\left(\frac{x+1}{x-2}\right) = 1. Converting to exponential form: x+1x−2=51=5\frac{x+1}{x-2} = 5^1 = 5. Cross-multiplying: x+1=5(x−2)=5x−10x + 1 = 5(x - 2) = 5x - 10. Solving for xx: x+1=5x−10x + 1 = 5x - 10, so 11=4x11 = 4x, which gives x=114x = \frac{11}{4}. We must verify this satisfies the domain restrictions: x+1>0x + 1 > 0 and x−2>0x - 2 > 0, so x>2x > 2. Since 114=2.75>2\frac{11}{4} = 2.75 > 2, this solution is valid.