College Algebra Quiz: Solving Logarithmic Equations

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.

Question 1

1. $x + 2 = x - 1$ (correct answer)
2. $\log_3(x + 2) = \log_3(x - 1)$
3. $x + 2 = \log_3(x - 1)$
4. $\log_3(x + 2) = x - 1$

Explanation: Using the fundamental property that $a^{\log_a(y)} = y$ (for $y > 0$), we have $3^{\log_3(x+2)} = x + 2$. Therefore, the equation $3^{\log_3(x+2)} = x - 1$ becomes $x + 2 = x - 1$. This simplifies to $2 = -1$, 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 $\log_3$ of both sides, but we need $x - 1 > 0$ for that to be valid. Choice C incorrectly swaps the roles of the logarithm and its argument. Choice D would result from taking $\log_3$ of both sides, but again requires $x - 1 > 0$. Choice A is the direct algebraic equivalent using the exponential-logarithm inverse relationship.

Question 2

1. $x = 8$ (correct answer)
2. $x = 9$
3. $x = 7$
4. $x = 1$ or $x = 8$

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

Question 3

1. $\frac{11}{4}$ (correct answer)
2. $\frac{13}{4}$
3. $\frac{7}{2}$
4. $\frac{9}{2}$

Explanation: Using the logarithm property $\log_a(m) - \log_a(n) = \log_a(\frac{m}{n})$, we get $\log_5\left(\frac{x+1}{x-2}\right) = 1$. Converting to exponential form: $\frac{x+1}{x-2} = 5^1 = 5$. Cross-multiplying: $x + 1 = 5(x - 2) = 5x - 10$. Solving for $x$: $x + 1 = 5x - 10$, so $11 = 4x$, which gives $x = \frac{11}{4}$. We must verify this satisfies the domain restrictions: $x + 1 > 0$ and $x - 2 > 0$, so $x > 2$. Since $\frac{11}{4} = 2.75 > 2$, this solution is valid.