<p>Find the number of real ordered pair(s) (<i>x</i>, <i>y</i>) for which: <br/>16<sup>x<sup>2</sup>+y</sup> + 16<sup>x+y<sup>2</sup></sup> = 1</p>
Step-by-Step Solution
Key Concept: Recognize that 16^x + 16^y = 1 can be rewritten by setting u = 16^x and v = 16^y, where both u and v are positive. The constraint u + v = 1 with u, v > 0 severely limits possible solutions, forcing us to find when equality holds under AM-GM or symmetry conditions.
<p><strong>Step 1:</strong> Let u = 16^x and v = 16^y. Then u > 0 and v > 0 for all real x, y.</p><p><strong>Step 2:</strong> The equation becomes: u + v = 1 where u, v ∈ (0, 1).</p><p><strong>Step 3:</strong> By AM-GM inequality: (u + v)/2 ≥ √(uv), so 1/2 ≥ √(uv), giving uv ≤ 1/4.</p><p><strong>Step 4:</strong> Equality in AM-GM holds only when u = v. If u = v and u + v = 1, then 2u = 1, so u = v = 1/2.</p><p><strong>Step 5:</strong> Therefore 16^x = 1/2 and 16^y = 1/2, giving x = log₁₆(1/2) = -1/4 and y = -1/4.</p><p><strong>Step 6:</strong> Verification: 16^(-1/4) + 16^(-1/4) = (2^4)^(-1/4) + (2^4)^(-1/4) = 2^(-1) + 2^(-1) = 1/2 + 1/2 = 1 ✓</p><p><strong>Step 7:</strong> Since AM-GM equality is unique, there is exactly one solution pair: (x, y) = (-1/4, -1/4).</p><p><strong>∴ Answer:</strong> b</p>
Correct Answer: b