<p>Value of \(\log_6\!\left(\sqrt{2-\sqrt{3}} + \sqrt{2+\sqrt{3}}\right)\) is:</p>
Step-by-Step Solution
Key Concept: Recognize that the expression under the logarithm can be simplified by squaring: if we let $x = \sqrt{2-\sqrt{3}} + \sqrt{2+\sqrt{3}}$, then $x^2 = (2-\sqrt{3}) + (2+\sqrt{3}) + 2\sqrt{(2-\sqrt{3})(2+\sqrt{3})} = 4 + 2\sqrt{4-3} = 6$, so $x = \sqrt{6}$.
<p><strong>Step 1:</strong> Let $x = \sqrt{2-\sqrt{3}} + \sqrt{2+\sqrt{3}}$</p><p><strong>Step 2:</strong> Square both sides: $x^2 = (2-\sqrt{3}) + (2+\sqrt{3}) + 2\sqrt{(2-\sqrt{3})(2+\sqrt{3})}$</p><p><strong>Step 3:</strong> Simplify: $x^2 = 4 + 2\sqrt{(2-\sqrt{3})(2+\sqrt{3})}$</p><p><strong>Step 4:</strong> Calculate $(2-\sqrt{3})(2+\sqrt{3}) = 4 - 3 = 1$</p><p><strong>Step 5:</strong> Therefore $x^2 = 4 + 2\sqrt{1} = 4 + 2 = 6$</p><p><strong>Step 6:</strong> So $x = \sqrt{6}$ (taking positive root since both terms are positive)</p><p><strong>Step 7:</strong> Now evaluate: $\log_6(\sqrt{6}) = \log_6(6^{1/2}) = \frac{1}{2}\log_6(6) = \frac{1}{2}$</p><p>∴ Answer: A ($\frac{1}{2}$)</p>
Correct Answer: A