Basic Mathematics & Logarithm
AM-GM Inequality and Logarithm
Grade 11
Question:
<p>Let <em>a</em>, <em>b</em> and <em>c</em> be non-negative real numbers satisfying \(a + b + c = 9\). If the maximum value of the expression \(a^2 b^3 c^4\) can be expressed as \(2^x 3^y\), where <em>x</em> and <em>y</em> are natural numbers, then the value of \(\log_{10}(x^y)\) is:</p>
<p>2</p>
<p>3</p>
<p>4</p>
<p>6</p>
Step-by-Step Solution
Key Concept: Use AM-GM inequality with weighted terms matching the exponents in the expression. The maximum of a^2 b^3 c^4 occurs when the terms in AM-GM are proportional to their exponents (2:3:4), giving a:b:c = 2:3:4.
<p><strong>Step 1:</strong> Apply weighted AM-GM inequality. For maximum of a²b³c⁴ subject to a+b+c=9, use the condition that at maximum, the weighted contributions are equal:</p><p>a/2 = b/3 = c/4 = k (say)</p><p><strong>Step 2:</strong> From the constraint a+b+c=9: 2k+3k+4k=9 → 9k=9 → k=1</p><p>Therefore: a=2, b=3, c=4</p><p><strong>Step 3:</strong> Calculate maximum value: a²b³c⁴ = 2² × 3³ × 4⁴ = 4 × 27 × 256</p><p><strong>Step 4:</strong> Express 4⁴ as powers of 2 and 3: 4⁴ = (2²)⁴ = 2⁸</p><p>So: a²b³c⁴ = 2² × 3³ × 2⁸ = 2¹⁰ × 3³</p><p><strong>Step 5:</strong> Therefore x=10 and y=3, so x^y = 10³ = 1000</p><p><strong>Step 6:</strong> Calculate log₁₀(x^y) = log₁₀(1000) = log₁₀(10³) = 3</p><p>∴ Answer: D (3)</p>
Correct Answer: D