\( \log_2(x(x - 2)) = 3 \) → \( x(x - 2) = 2^3 = 8 \).

["# Solving ( \log_2(x(x - 2)) = 3 ): A Step-by-Step Explanation", "Understanding logarithmic equations is essential for mastering algebra and advancing in mathematical problem-solving. One common challenge is solving equations of the form ( \log_b(f(x)) = c ), particularly when the logarithm visits bases like 2, 10, or ( e ). In this article, we’ll explore how to solve ( \log_2(x(x - 2)) = 3 ) step-by-step—demystifying the transformation from logarithmic to exponential form—and arrive at the solution ( x(x - 2) = 8 ), followed by valid solutions and practical applications.", "---", "## Understanding Logarithmic Equations", "The equation ( \log_b(a) = c ) translates directly to the exponential form ( a = b^c ). This conversion relies on the fundamental definition of logarithms:\nIf ( \log_b(x) = c ), then ( x = b^c ).", "Applying this principle is key when solving logarithmic equations.", "---", "## Step 1: Convert the Logarithmic Equation", "We start with:\n[\n\log_2(x(x - 2)) = 3\n]", "Using the logarithmic-equality rule, rewrite it in exponential form:\n[\nx(x - 2) = 2^3\n]", "Since ( 2^3 = 8 ), this simplifies directly to:\n[\nx(x - 2) = 8\n]", "---", "## Step 2: Expand and Rearrange into a Quadratic Equation", "Distribute ( x ) across ( (x - 2) ):\n[\nx^2 - 2x = 8\n]", "Move all terms to one side to form a standard quadratic equation:\n[\nx^2 - 2x - 8 = 0\n]", "---", "## Step 3: Solve the Quadratic Equation", "We solve ( x^2 - 2x - 8 = 0 ) using factoring:\nLook for two numbers multiplying to ( -8 ) and adding to ( -2 ). Those numbers are ( -4 ) and ( +2 ).\nThus,\n[\n(x - 4)(x + 2) = 0\n]", "Setting each factor to zero gives:\n[\nx = 4 \quad \ ext{or} \quad x = -2\n]", "---", "## Step 4: Verify Solutions in the Original Equation", "Before finalizing, always check solutions in the original logarithmic equation to avoid extraneous results. Recall that the logarithm is only defined when its argument is positive:\n[\nx(x - 2) > 0\n]", "Check ( x = 4 ):\n( 4(4 - 2) = 4 \ imes 2 = 8 > 0 ) → valid", "Check ( x = -2 ):\n( -2(-2 - 2) = -2(-4) = 8 > 0 ) → valid", "Both values satisfy the domain condition.", "---", "## Step 5: Final Answer and Solutions", "The solutions to ( \log_2(x(x - 2)) = 3 ) are:\n[\n\boxed{x = 4} \quad \ ext{and} \quad \boxed{x = -2}\n]", "Both are mathematically valid, but context matters in real-world applications—such as when modeling physical phenomena where negative values may be nonsensical.", "---", "## Why This Transformation Matters", "Converting ( \log_b(f(x)) = c ) to ( f(x) = b^c ) is a cornerstone of logarithmic problem-solving. It unlocks access to standard algebraic techniques, enabling efficient solution strategies across education, engineering, and data science.", "---", "## Practical Applications", "- Exponential Growth Models: Understanding intensity or decay rates often involves such transformations.\n- Information Theory: Logarithms underpin entropy calculations and data compression, where base-2 logs frequently appear.\n- Computer Science: Analyzing algorithm complexity and binary tree depths uses base-2 logarithms.", "---", "## Summary", "The equation ( \log_2(x(x - 2)) = 3 ) simplifies neatly to ( x(x - 2) = 8 ), a quadratic equation solved via factoring. Verification ensures both ( x = 4 ) and ( x = -2 ) are valid. Mastering this transformation enhances your ability to tackle logarithmic challenges across disciplines.", "---", "Keywords: ( \log_2(x(x - 2)) = 3 ), solving logarithmic equations, converting log to exponential form, quadratic equation, domain verification, algebra tutorial, logarithms applied.", "---", "Efficiently solving equations like ( \log_2(x(x - 2)) = 3 ) not only strengthens algebraic intuition but also prepares learners for advanced mathematical and computational challenges. Use this step-by-step guide to build confidence and precision in logarithmic reasoning."]