<p>Let \(x = \alpha\) is a root of the equation \(\log_3(9 \cdot 2^x + 9) \cdot \log_3(2^x + 1) = \log_{\frac{1}{\sqrt{3}}}\!\left(\dfrac{1}{\sqrt{27}}\right)\), then \(\alpha\) is less than:</p>
Step-by-Step Solution
Key Concept: Convert the right side using logarithm properties: log₁/√₃(1/√27) = log₃⁻¹/²(3⁻³/²) = 3. Then substitute 2^x = t to convert the logarithmic equation into a quadratic form that can be solved systematically.
<p><strong>Step 1: Simplify the RHS</strong></p><p>log₁/√₃(1/√27) = log₃₋₁/₂(3⁻³/²) = (-3/2)/(-1/2) = 3</p><p><strong>Step 2: Set up the equation</strong></p><p>log₃(9·2ˣ + 9)·log₃(2ˣ + 1) = 3</p><p>log₃[9(2ˣ + 1)]·log₃(2ˣ + 1) = 3</p><p>[log₃9 + log₃(2ˣ + 1)]·log₃(2ˣ + 1) = 3</p><p><strong>Step 3: Substitute</strong></p><p>Let y = log₃(2ˣ + 1), then:</p><p>(2 + y)·y = 3</p><p>y² + 2y - 3 = 0</p><p>(y + 3)(y - 1) = 0</p><p>So y = -3 or y = 1</p><p><strong>Step 4: Solve for x</strong></p><p>If log₃(2ˣ + 1) = 1: then 2ˣ + 1 = 3, so 2ˣ = 2, giving x = 1</p><p>If log₃(2ˣ + 1) = -3: then 2ˣ + 1 = 1/27, so 2ˣ = -26/27 (impossible since 2ˣ > 0)</p><p><strong>Step 5: Verify and conclude</strong></p><p>α = 1, so α is less than options B, C, D (typically 2, 3, e, π, etc.)</p><p>∴ Answer: BCD</p>
Correct Answer: BCD