<p>Find the number of integral values of \(x\) satisfying the equation: \[|\log_2^2 x - 5\log_2 x + 6| = |\log_2^2 x - 7\log_2 x + 12| - |2\log_2 x - 6|.\]</p>
Step-by-Step Solution
Key Concept: Substitute t = log₂x to convert into an algebraic equation, then analyze when the absolute value equation |A| = |B| - |C| holds by considering the signs of expressions A, B-C and factoring the quadratics.
<p><strong>Step 1:</strong> Let t = log₂x. The equation becomes:<br>|t² - 5t + 6| = |t² - 7t + 12| - |2t - 6|</p><p><strong>Step 2:</strong> Factor the quadratics:<br>t² - 5t + 6 = (t - 2)(t - 3)<br>t² - 7t + 12 = (t - 3)(t - 4)<br>2t - 6 = 2(t - 3)</p><p><strong>Step 3:</strong> Rearrange: |t² - 5t + 6| + |2t - 6| = |t² - 7t + 12|</p><p><strong>Step 4:</strong> For this to hold, we need (t² - 5t + 6) and (2t - 6) to have the same sign (or one is zero), and their sum equals ±(t² - 7t + 12).</p><p><strong>Step 5:</strong> Case Analysis:<br>• When t ≤ 2: Both (t-2)(t-3) ≥ 0 and (t-3) ≤ 0, so |t² - 5t + 6| + |2t - 6| = (t² - 5t + 6) + (6 - 2t) = t² - 7t + 12 = |t² - 7t + 12| ✓<br>• When 2 < t < 3: (t² - 5t + 6) < 0, (2t - 6) < 0, equation fails<br>• When t ≥ 3: (t-2)(t-3) ≥ 0 and (t-3) ≥ 0, equation becomes (t² - 5t + 6) + (2t - 6) = (t² - 7t + 12), which gives -7t + 2t = -12 + 6, so -5t = -6, thus t = 6/5 (rejected as t ≥ 3)<br>• When t = 3: All terms = 0, equation holds ✓</p><p><strong>Step 6:</strong> Valid range: t ≤ 2 or t = 3<br>Since t = log₂x: x ≤ 2² = 4 or x = 2³ = 8<br>With x > 0: x ∈ (0, 4] ∪ {8}</p><p><strong>Step 7:</strong> Integral values: x ∈ {1, 2, 3, 4, 8}</p><p>∴ Answer: <strong>5</strong></p>
Correct Answer: 5