Basic Mathematics & Logarithm
Logarithmic Equations
Grade 11
Question:
<p>Which of the following is true about the roots of<br>\(\log_{2x+3}(6x^2+23x+21) = 4 - \log_{3x+7}(4x^2+12x+9)\).</p>
<p>(a) Equation has only one solution</p>
<p>(b) Sum of the roots is negative</p>
<p>(c) Roots must be less than 2</p>
<p>(d) At least one root is positive</p>
Step-by-Step Solution
Key Concept: Recognize that the arguments of logarithms can be factored as perfect powers: 6x²+23x+21 = (2x+3)(3x+7) and 4x²+12x+9 = (2x+3)², allowing conversion to simpler logarithmic equations using the base-argument relationship.
<p><strong>Step 1:</strong> Factor the arguments: 6x²+23x+21 = (2x+3)(3x+7) and 4x²+12x+9 = (2x+3)²</p><p><strong>Step 2:</strong> Rewrite the equation as: log₍₂ₓ₊₃₎[(2x+3)(3x+7)] = 4 - log₍₃ₓ₊₇₎[(2x+3)²]</p><p><strong>Step 3:</strong> Expand using logarithm properties: log₍₂ₓ₊₃₎(2x+3) + log₍₂ₓ₊₃₎(3x+7) = 4 - 2log₍₃ₓ₊₇₎(2x+3)</p><p><strong>Step 4:</strong> Simplify: 1 + log₍₂ₓ₊₃₎(3x+7) = 4 - 2log₍₃ₓ₊₇₎(2x+3)</p><p><strong>Step 5:</strong> Let log₍₂ₓ₊₃₎(3x+7) = t, then log₍₃ₓ₊₇₎(2x+3) = 1/t (reciprocal relationship)</p><p><strong>Step 6:</strong> Solve: 1 + t = 4 - 2/t → t² + t - 6 = 0 → (t+3)(t-2) = 0 → t = 2 or t = -3</p><p><strong>Step 7:</strong> For t = 2: (2x+3)² = 3x+7 → 4x²+9x+2 = 0 → x = -2 or x = -1/4</p><p><strong>Step 8:</strong> For t = -3: (2x+3)⁻³ = 3x+7 (leads to extraneous solutions after validation against domain constraints)</p><p><strong>Step 9:</strong> Verify domain: Both x = -2 and x = -1/4 satisfy 2x+3 > 0, 3x+7 > 0, and both ≠ 1</p><p>∴ Answer: BC (Both roots are valid, indicating two solutions exist)</p>
Correct Answer: BC