Use product rule: \( \log_2[(x+3)(x-1)] = 3 \)

["# Solving ( \log_2[(x+3)(x-1)] = 3 ) Using the Product Rule", "Understanding logarithmic equations is essential for mastering algebra and logarithmic functions. One powerful technique is the product rule of logarithms, which allows us to simplify expressions involving the logarithm of a product. This article explores how to solve logarithmic equations like ( \log_2[(x+3)(x-1)] = 3 ) using the product rule, offering clear, step-by-step solving strategies and real-world applications.", "## What Is the Product Rule in Logarithms?", "The product rule states:", "[\n\log_b [M \cdot N] = \log_b M + \log_b N\n]", "This means the logarithm of a product is equal to the sum of the logarithms of the factors—as long as both ( M ) and ( N ) are positive, because logarithms are only defined for positive numbers. This rule is crucial when dealing with logarithmic expressions inside products, a common scenario in equation solving.", "## Step-by-Step Solution Using the Product Rule", "We are given:", "[\n\log_2[(x+3)(x-1)] = 3\n]", "### Step 1: Apply the Product Rule", "Use the product rule to rewrite the left-hand side:", "[\n\log_2(x+3) + \log_2(x-1) = 3\n]", "### Step 2: Rewrite in Exponential Form", "To eliminate the logarithm, recall that ( \log_b A = C ) is equivalent to ( A = b^C ). Applying this:", "[\n(x+3)(x-1) = 2^3 = 8\n]", "### Step 3: Simplify the Equation", "Expand the left-hand side:", "[\n(x+3)(x-1) = x^2 - x + 3x - 3 = x^2 + 2x - 3\n]", "Set equal to 8:", "[\nx^2 + 2x - 3 = 8\n]", "Subtract 8 from both sides:", "[\nx^2 + 2x - 11 = 0\n]", "### Step 4: Solve the Quadratic Equation", "Use the quadratic formula:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad \ ext{where } a=1, b=2, c=-11\n]", "[\nx = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-11)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 44}}{2} = \frac{-2 \pm \sqrt{48}}{2}\n]", "Simplify ( \sqrt{48} = 4\sqrt{3} ):", "[\nx = \frac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3}\n]", "### Step 5: Check Validity of Solutions", "Because we used logarithms, all solutions must ensure the arguments ( x+3 > 0 ) and ( x-1 > 0 ).", "- ( x = -1 + 2\sqrt{3} \approx -1 + 3.46 = 2.46 ):\n ( x+3 \approx 5.46 > 0 ), ( x-1 \approx 1.46 > 0 ) — valid.", "- ( x = -1 - 2\sqrt{3} \approx -1 - 3.46 = -4.46 ):\n ( x+3 \approx -1.46 < 0 ), ( x-1 \approx -5.46 < 0 ) — invalid.", "Thus, the only valid solution is:", "[\nx = -1 + 2\sqrt{3}\n]", "## Real-World Applications of the Product Rule in Logarithms", "The product rule isn’t just theoretical—it’s widely used in fields requiring logarithmic calculations:", "- Acoustics: Measuring sound intensity in decibels, where sound levels combine multiplicatively.\n- Finance: Calculating compounded growth when returns compound multiplicatively over time.\n- Chemistry: Analyzing pH values from hydrogen ion concentrations, involving logarithmic products.", "Mastering this rule empowers students and professionals to simplify and solve complex real-world problems involving exponential and logarithmic relationships.", "## Conclusion", "Using the product rule ( \log_b [M \cdot N] = \log_b M + \log_b N ) simplifies logarithmic equations involving products. Solving ( \log_2[(x+3)(x-1)] = 3 ) step-by-step demonstrates how to convert logarithmic equations into algebraic ones using exponents, verify solutions, and ensure domain validity. With practice, this method becomes a powerful tool for tackling logarithmic challenges across science, engineering, and finance.", "Keywords: log law, product rule logarithm, solve ( \log[(x+3)(x-1)] = 3 ), logarithmic equation, exponential form, algebra quick tips."]