<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: Evaluate the RHS constant first using logarithm properties, then use substitution t = 2^x to convert the transcendental equation into an algebraic form. Recognize that log₃(9·2^x + 9) = log₃(9(2^x + 1)) = 2 + log₃(2^x + 1).
<p><strong>Step 1: Simplify the RHS</strong></p><p>log₁/√₃(1/∛27) = log₃₋₁/₂(27⁻¹/³) = log₃₋₁/₂(3⁻¹)</p><p>Using change of base: log₃₋₁/₂(3⁻¹) = (log 3⁻¹)/(log 3⁻¹/²) = (-log 3)/(-½ log 3) = 2</p><p><strong>Step 2: Simplify the LHS</strong></p><p>log₃(9·2ˣ + 9) = log₃[9(2ˣ + 1)] = log₃(9) + log₃(2ˣ + 1) = 2 + log₃(2ˣ + 1)</p><p><strong>Step 3: Set up equation with substitution</strong></p><p>[2 + log₃(2ˣ + 1)]·log₃(2ˣ + 1) = 2</p><p>Let u = log₃(2ˣ + 1):</p><p>(2 + u)·u = 2</p><p>u² + 2u - 2 = 0</p><p><strong>Step 4: Solve for u</strong></p><p>u = (-2 ± √(4 + 8))/2 = (-2 ± 2√3)/2 = -1 ± √3</p><p>Since 2ˣ + 1 > 1, we need log₃(2ˣ + 1) > 0, so u = -1 + √3 ≈ 0.732</p><p><strong>Step 5: Solve for x</strong></p><p>log₃(2ˣ + 1) = √3 - 1</p><p>2ˣ + 1 = 3^(√3 - 1) = 3^√3/3</p><p>2ˣ = 3^(√3 - 1) - 1 ≈ 3^0.732 - 1 ≈ 2.28 - 1 = 1.28</p><p>x = log₂(3^(√3 - 1) - 1) ≈ 0.35</p><p>Since √3 ≈ 1.732, we have α < 1, α < 2, and α < 3</p><p>∴ Answer: <strong>BCD</strong></p>
Correct Answer: BCD