Basic Mathematics & Logarithm
Logarithmic Equations
Grade 11
Question:
<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, which means x²-2x must equal the base 2x²+2x+3.
<p><strong>Step 1:</strong> Use the property that log_a(b) = 0 ⟺ b = 1.</p><p>Therefore: log_(2x²+2x+3)(x²-2x) = 1</p><p><strong>Step 2:</strong> This means: x²-2x = 2x²+2x+3</p><p>Rearranging: -x²-4x-3 = 0, or x²+4x+3 = 0</p><p>(x+1)(x+3) = 0 ⟹ x = -1 or x = -3</p><p><strong>Step 3:</strong> Check domain restrictions:</p><p>For x = -1:</p><p>• x²-2x = 1-(-2) = 3 > 0 ✓</p><p>• 2x²+2x+3 = 2-2+3 = 3 > 0 and ≠ 1 ✓</p><p>• x²+2x+5 = 1-2+5 = 4 > 0 and ≠ 1 ✓</p><p>For x = -3:</p><p>• x²-2x = 9-(-6) = 15 > 0 ✓</p><p>• 2x²+2x+3 = 18-6+3 = 15 > 0 and ≠ 1 ✓</p><p>• x²+2x+5 = 9-6+5 = 8 > 0 and ≠ 1 ✓</p><p><strong>Step 4:</strong> Both values satisfy all domain constraints.</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2