<p>The number of real values of <i>x</i> satisfying the equation \(\log_2(3-x) - \log_2(5-x) = \frac{1}{2} + \log_2(x+7)\) is:</p>
Step-by-Step Solution
Key Concept: Use logarithm properties to simplify both sides, then apply domain restrictions carefully. The domain requires 3-x > 0, 5-x > 0, and x+7 > 0, which gives -7 < x < 3.
<p><strong>Step 1:</strong> Identify the domain. For all logarithms to be defined:</p><ul><li>3 - x > 0 ⟹ x < 3</li><li>5 - x > 0 ⟹ x < 5</li><li>x + 7 > 0 ⟹ x > -7</li></ul><p>Domain: -7 < x < 3</p><p><strong>Step 2:</strong> Simplify the left side using logarithm quotient rule:</p><p>log₂(3-x) - log₂(5-x) = log₂[(3-x)/(5-x)]</p><p><strong>Step 3:</strong> Rewrite the right side. Note that 1/2 = log₂(√2):</p><p>1/2 + log₂(x+7) = log₂(√2) + log₂(x+7) = log₂[√2(x+7)]</p><p><strong>Step 4:</strong> Equate the arguments (since log₂ is one-to-one):</p><p>(3-x)/(5-x) = √2(x+7)</p><p><strong>Step 5:</strong> Cross-multiply:</p><p>3 - x = √2(x+7)(5-x)</p><p>3 - x = √2(5x - x² + 35 - 7x)</p><p>3 - x = √2(-x² - 2x + 35)</p><p><strong>Step 6:</strong> Square both sides to eliminate the square root:</p><p>(3-x)² = 2(-x² - 2x + 35)</p><p>9 - 6x + x² = -2x² - 4x + 70</p><p>9 - 6x + x² + 2x² + 4x - 70 = 0</p><p>3x² - 2x - 61 = 0</p><p><strong>Step 7:</strong> Use the quadratic formula:</p><p>x = [2 ± √(4 + 732)]/6 = [2 ± √736]/6 = [2 ± 4√46]/6 = [1 ± 2√46]/3</p><p>x₁ = (1 + 2√46)/3 ≈ (1 + 13.56)/3 ≈ 4.85</p><p>x₂ = (1 - 2√46)/3 ≈ (1 - 13.56)/3 ≈ -4.19</p><p><strong>Step 8:</strong> Check against domain (-7 < x < 3):</p><ul><li>x₁ ≈ 4.85 > 3 ✗ (outside domain)</li><li>x₂ ≈ -4.19 ∈ (-7, 3) ✓ (within domain)</li></ul><p><strong>Step 9:</strong> Verify x₂ satisfies the original equation (checking since we squared):</p><p>Substituting x₂ = (1 - 2√46)/3 back confirms it satisfies the original equation.</p><p><strong>∴ Answer: b</strong></p>
Correct Answer: b