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: for a+b+c=9, maximize a^2·b^3·c^4 by setting the weighted coefficients equal, then express the maximum in terms of powers of 2 and 3.
<p><strong>Step 1: Apply weighted AM-GM</strong></p><p>For maximizing a²b³c⁴ subject to a+b+c=9, use AM-GM with weights 2, 3, 4:</p><p>$$\frac{2a + 3b + 4c}{2+3+4} \geq \sqrt[9]{a^2 b^3 c^4}$$</p><p><strong>Step 2: Equality condition in AM-GM</strong></p><p>Equality holds when terms are equal: $\frac{a}{2} = \frac{b}{3} = \frac{c}{4} = k$</p><p>So $a = 2k, b = 3k, c = 4k$</p><p>From constraint: $2k + 3k + 4k = 9 \Rightarrow 9k = 9 \Rightarrow k = 1$</p><p>Therefore: $a = 2, b = 3, c = 4$</p><p><strong>Step 3: Calculate maximum value</strong></p><p>$$a^2b^3c^4 = 2^2 \cdot 3^3 \cdot 4^4 = 4 \cdot 27 \cdot 256$$</p><p>$$= 2^2 \cdot 3^3 \cdot (2^2)^4 = 2^2 \cdot 3^3 \cdot 2^8 = 2^{10} \cdot 3^3$$</p><p><strong>Step 4: Identify x and y</strong></p><p>From $2^{10} \cdot 3^3$: $x = 10, y = 3$</p><p><strong>Step 5: Calculate final answer</strong></p><p>$$\log_{10}(x^y) = \log_{10}(10^3) = 3\log_{10}(10) = 3$$</p><p>∴ Answer: <strong>D (3)</strong></p>
Correct Answer: D