Solve for \( x \): \( \log_2(x + 3) + \log_2(x - 1) = 3 \)

["Solve for ( x ): ( \log_2(x + 3) + \log_2(x - 1) = 3 )", "Understanding logarithmic equations is essential for mastering algebra and applied math. One common problem students encounter is solving equations that combine logarithmic terms. In this article, we’ll step through how to solve the equation:", "[\n\log_2(x + 3) + \log_2(x - 1) = 3\n]", "and explain why certain values are valid or invalid. We’ll also cover key concepts like domain restrictions, combining logs, and solving linear logarithmic equations — all valuable skills for students, mathematicians, and science enthusiasts.", "---", "### Step 1: Apply Logarithmic Properties", "To simplify the left-hand side, use the product rule of logarithms:", "[\n\log_b A + \log_b B = \log_b (A \cdot B)\n]", "Applying this property here:", "[\n\log_2(x + 3) + \log_2(x - 1) = \log_2\left( (x + 3)(x - 1) \right)\n]", "So the equation becomes:", "[\n\log_2\left( (x + 3)(x - 1) \right) = 3\n]", "---", "### Step 2: Remove the Logarithm by Exponentiating Both Sides", "To eliminate the logarithm, rewrite the equation in exponential form. Since the base is 2 and the log equals 3:", "[\n(x + 3)(x - 1) = 2^3 = 8\n]", "---", "### Step 3: Expand and Solve the Quadratic Equation", "Multiply out the left-hand expression:", "[\n(x + 3)(x - 1) = x^2 - x + 3x - 3 = x^2 + 2x - 3\n]", "Set this equal to 8:", "[\nx^2 + 2x - 3 = 8\n]", "Bring all terms to one side:", "[\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}\n]", "For ( a = 1 ), ( b = 2 ), ( c = -11 ):", "[\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:", "[\n\sqrt{48} = \sqrt{16 \ imes 3} = 4\sqrt{3}\n]", "So,", "[\nx = \frac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3}\n]", "This gives two potential solutions:", "[\nx = -1 + 2\sqrt{3} \quad \ ext{or} \quad x = -1 - 2\sqrt{3}\n]", "---", "### Step 5: Apply Domain Restrictions (Critical Step!)", "Logarithmic functions ( \log_2(x + 3) ) and ( \log_2(x - 1) ) are only defined when their arguments are positive:", "- ( x + 3 > 0 \Rightarrow x > -3 )\n- ( x - 1 > 0 \Rightarrow x > 1 )", "Thus, the domain is ( x > 1 )", "Now test both solutions:", "- ( x = -1 + 2\sqrt{3} \approx -1 + 2(1.732) = -1 + 3.464 = 2.464 > 1 ) → valid\n- ( x = -1 - 2\sqrt{3} \approx -1 - 3.464 = -4.464 ) → not greater than 1 → invalid", "---", "### Final Answer", "The only valid solution is:", "[\n\boxed{x = -1 + 2\sqrt{3}}\n]", "---", "### Why This Solution Matters", "This problem illustrates essential principles:", "- Using logarithmic identities (product rule)\n- Accurately combining logs and simplifying expressions\n- Solving resulting algebraic equations (quadratics)\n- Most importantly: Always checking domain restrictions to exclude extraneous solutions introduced by algebra", "Mastery of these steps ensures correctness and deepens understanding—key for excelling in math competitions, engineering, and scientific research.", "---", "### Want to Master More Logarithmic Problems?", "Check out advanced strategies for solving logarithmic equations, including strategies for more complex bases and systems involving logarithmic and exponential functions.", "---", "Keywords: solve logarithmic equation, solve for ( x ) in log-based equation, logarithmic equation with product rule, domain restrictions, quadratic logarithmic solutions, math helpful videos, algebra practice problems, logarithms for high school, mathematical problem-solving, collapsable summary.", "---", "Meta Description:\nLearn how to solve ( \log_2(x + 3) + \log_2(x - 1) = 3 ) step-by-step. Includes domain checks, quadratic solving, and key logarithmic rules. Perfect for math students and tutors."]