Solution: Combine logs: $ \log_2\left(\frac{x + 4}{x - 1}\right) = 3 $. Rewrite in exponential form: $ \frac{x + 4}{x - 1} = 2^3 = 8 $. Solve: $ x + 4 = 8x - 8 \Rightarrow -7x = -12 \Rightarrow x = \frac{12}{7} $. Verify $ x > 1 $ to ensure validity.

["Solving Logarithmic Equations: A Step-by-Step Guide to $ \log_2\left(\frac{x + 4}{x - 1}\right) = 3 $", "Logarithmic equations appear frequently in algebra and advanced math, yet solving them can be intuitive once you follow a clear process. This article explores how to solve the equation:", "$$\n\log_2\left(\frac{x + 4}{x - 1}\right) = 3\n$$", "We’ll walk through converting the logarithmic form to exponential form, solving the resulting linear equation, and verifying the solution for mathematical consistency. This method not only finds the answer but also reinforces key logarithmic principles.", "---", "### Step 1: Eliminate the Logarithm Using Exponential Form", "The logarithmic equation $ \log_b(A) = C $ is equivalent to the exponential form $ A = b^C $. Applying this to our equation:", "$$\n\log_2\left(\frac{x + 4}{x - 1}\right) = 3 \quad \Rightarrow \quad \frac{x + 4}{x - 1} = 2^3\n$$", "Since $ 2^3 = 8 $, we rewrite:", "$$\n\frac{x + 4}{x - 1} = 8\n$$", "---", "### Step 2: Solve the Rational Equation", "Now, solve $ \frac{x + 4}{x - 1} = 8 $. Multiply both sides by $ x - 1 $ to eliminate the denominator (noting $ x <br/>\ne 1 $, because the original logarithm requires a valid, positive argument):", "$$\nx + 4 = 8(x - 1)\n$$", "Expand the right-hand side:", "$$\nx + 4 = 8x - 8\n$$", "Bring all terms to one side:", "$$\nx - 8x = -8 - 4\n$$\n$$\n-7x = -12\n$$", "Solve for $ x $:", "$$\nx = \frac{12}{7}\n$$", "---", "### Step 3: Verify the Solution Satisfies All Constraints", "Before accepting the solution, check that it fits the domain of the logarithm. The argument $ \frac{x + 4}{x - 1} $ must be positive, because the logarithm is defined only for positive real numbers.", "Compute:", "- $ x = \frac{12}{7} \approx 1.714 $\n- $ x + 4 = \frac{12}{7} + 4 = \frac{40}{7} > 0 $\n- $ x - 1 = \frac{12}{7} - 1 = \frac{5}{7} > 0 $", "Since both numerator and denominator are positive, the fraction is positive:", "$$\n\frac{x + 4}{x - 1} = \frac{40/7}{5/7} = 8 > 0\n$$", "Thus, the logarithm is defined, and $ x = \frac{12}{7} $ is valid.", "Also, note $ x > 1 $ ensures the denominator $ x - 1 <br/>\ne 0 $ and avoids division by zero.", "---", "### Final Answer", "$$\n\boxed{x = \frac{12}{7}}\n$$", "---", "Why This Method Works\nConverting logarithmic equations to exponential form transforms the problem into an algebraic one, which is often easier to solve. By carefully checking domain restrictions, we ensure the solution is not only mathematically correct but also valid within the context of real numbers and logarithmic function properties.", "This approach applies broadly to logarithmic equations involving rational expressions—making it a foundational skill for calculus, engineering, and computer science applications."]