The equation \( \log_2(x - 1) + \log_2(x + 1) = 3 \) has a solution. What is it?

["# Solving ( \log_2(x - 1) + \log_2(x + 1) = 3 ): What Is the Solution?", "Logarithmic equations can seem challenging at first, but with careful manipulation, they become manageable. One classic equation is:", "[\n\log_2(x - 1) + \log_2(x + 1) = 3\n]", "In this article, we’ll explore how to solve this equation step-by-step and determine its unique solution.", "## Step 1: Use Logarithmic Properties to Combine Terms", "We start by applying the logarithmic product rule:\n[\n\log_b A + \log_b B = \log_b (A \cdot B)\n]\nSo, combine the logs on the left side:", "[\n\log_2\left((x - 1)(x + 1)\right) = 3\n]", "Simplify the expression inside the logarithm:", "[\n\log_2(x^2 - 1) = 3\n]", "## Step 2: Convert the Logarithmic Equation to Exponential Form", "Recall that ( \log_b A = C ) is equivalent to ( A = b^C ). Applying this:", "[\nx^2 - 1 = 2^3\n]\n[\nx^2 - 1 = 8\n]", "Solve for ( x^2 ):", "[\nx^2 = 9\n]", "So,", "[\nx = \pm 3\n]", "## Step 3: Check for Validity in the Original Equation", "Logarithmic functions are only defined for positive arguments, so we must ensure:", "- ( x - 1 > 0 \Rightarrow x > 1 )\n- ( x + 1 > 0 \Rightarrow x > -1 )", "The stricter condition ( x > 1 ) means only positive values up to validation matter.", "Check ( x = 3 ) and ( x = -3 ):", "- ( x = 3 ): valid, since ( 3 - 1 = 2 > 0 ) and ( 3 + 1 = 4 > 0 )\n- ( x = -3 ): invalid, because ( -3 - 1 = -4 < 0 ), making ( \log_2(-4) ) undefined", "## Step 4: Finalize the Solution", "The only valid solution is:", "[\nx = 3\n]", "Verification:\nSubstitute back:\n[\n\log_2(3 - 1) + \log_2(3 + 1) = \log_2(2) + \log_2(4) = 1 + 2 = 3 \quad \ ext{✓}\n]", "## Conclusion", "The logarithmic equation ( \log_2(x - 1) + \log_2(x + 1) = 3 ) has a unique solution at ( x = 3 ), satisfying all domain and algebraic constraints.", "---", "What is the solution?\n[\n\boxed{3}\n]"]