Solving Logarithmic Equations - College Algebra

1. If $\log_5(x + 1) - \log_5(x - 2) = 1$, then $x$ equals:
   Answer: $\frac{11}{4}$ 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.
2. Solve the equation $\log_2(x) + \log_2(x - 7) = 3$. What is the positive solution?
   Answer: $x = 8$ 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.
3. Which equation is equivalent to $3^{\log_3(x+2)} = x - 1$?
   Answer: $x + 2 = x - 1$ 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.

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