<p>Number of real values of \(x\), such that \(\log_{(x^2+2x+5)}\left(\log_{(2x^2+2x+3)}(x^2 - 2x)\right) = 0\) is:</p>
Step-by-Step Solution
Key Concept: A logarithm equals 0 if and only if its argument equals 1, so we need log₍₂ₓ²₊₂ₓ₊₃₎(x² - 2x) = 1. This means x² - 2x must equal the base 2x² + 2x + 3.
<p><strong>Step 1:</strong> For log₍ₐ₎(b) = 0, we need b = 1.</p><p>So log₍₂ₓ²₊₂ₓ₊₃₎(x² - 2x) = 1</p><p><strong>Step 2:</strong> This means x² - 2x = 2x² + 2x + 3</p><p>Simplifying: -x² - 4x - 3 = 0, or x² + 4x + 3 = 0</p><p>(x + 1)(x + 3) = 0, so x = -1 or x = -3</p><p><strong>Step 3:</strong> Check validity conditions:</p><p>For x = -1: x² - 2x = 1 + 2 = 3 > 0 ✓</p><p>For x = -3: x² - 2x = 9 + 6 = 15 > 0 ✓</p><p><strong>Step 4:</strong> Check base 2x² + 2x + 3:</p><p>Δ = 4 - 24 = -20 < 0, so 2x² + 2x + 3 > 0 always ✓</p><p>Also 2x² + 2x + 3 ≥ 2 > 1 always ✓</p><p><strong>Step 5:</strong> Check base x² + 2x + 5:</p><p>Δ = 4 - 20 = -16 < 0, so x² + 2x + 5 > 0 always ✓</p><p>Minimum value = 4, so always > 1 ✓</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2