Basic Mathematics & Logarithm
Logarithmic Equations
Grade 11
Question:
<p>Let equation \(x^{\log_2 x - 4} = 32\) has two real solutions \(x_1\) and \(x_2\) \((x_1 > x_2)\), then which of the following is correct?</p>
<p>\(x_1 \cdot x_2 = 32\)</p>
<p>\(x_1 + x_2 = \dfrac{65}{2}\)</p>
<p>Characteristic of \(\log_3(x_1)\) is 3</p>
<p>Mantissa of \(\log_2(x_2)\) is 0</p>
Step-by-Step Solution
Key Concept: Convert the exponential equation to logarithmic form by taking log₂ of both sides, transforming it into a quadratic in (log₂x). This reveals that the exponent itself must equal a specific value that makes the equation solvable.
<p><strong>Step 1:</strong> Let t = log₂x, so x = 2^t. Substitute into x^(log₂x - 4) = 32:</p><p>(2^t)^(t - 4) = 32 = 2^5</p><p>2^(t(t-4)) = 2^5</p><p><strong>Step 2:</strong> Equating exponents: t(t - 4) = 5</p><p>t² - 4t - 5 = 0</p><p>(t - 5)(t + 1) = 0</p><p>So t = 5 or t = -1</p><p><strong>Step 3:</strong> Convert back to x:</p><p>• If t = 5: log₂x = 5 ⟹ x₁ = 2^5 = 32</p><p>• If t = -1: log₂x = -1 ⟹ x₂ = 2^(-1) = 1/2</p><p><strong>Step 4:</strong> Verify (x₁ > x₂): 32 > 1/2 ✓</p><p>• x₁ + x₂ = 32 + 1/2 = 65/2 ✓</p><p>• x₁ · x₂ = 32 · (1/2) = 16 ✓</p><p>• x₁ - x₂ = 32 - 1/2 = 63/2 ✓</p><p>∴ Answer: B,C,D</p>
Correct Answer: B,C,D