Basic Mathematics & Logarithm
Logarithmic expressions
Grade 11
Question:
<p>Value of \(\log_6\!\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)\) is:</p>
<p>negative integer</p>
<p>rational but not integer</p>
<p>irrational</p>
<p>prime</p>
Step-by-Step Solution
Key Concept: Recognize that the expression under the logarithm is a sum of nested radicals that can be simplified by squaring to find a perfect form, then apply logarithm properties.
<p><strong>Step 1:</strong> Let $x = \sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}$</p><p><strong>Step 2:</strong> Square both sides:</p><p>$x^2 = (2-\sqrt{3}) + 2\sqrt{(2-\sqrt{3})(2+\sqrt{3})} + (2+\sqrt{3})$</p><p><strong>Step 3:</strong> Simplify the product under the radical:</p><p>$(2-\sqrt{3})(2+\sqrt{3}) = 4 - 3 = 1$</p><p><strong>Step 4:</strong> Therefore:</p><p>$x^2 = 4 + 2\sqrt{1} = 4 + 2 = 6$</p><p>So $x = \sqrt{6}$ (taking positive root since both terms are positive)</p><p><strong>Step 5:</strong> Apply logarithm:</p><p>$\log_6(\sqrt{6}) = \log_6(6^{1/2}) = \frac{1}{2}\log_6(6) = \frac{1}{2}$</p><p>∴ Answer: $\boxed{\frac{1}{2}}$ (A)</p>
Correct Answer: A