Question:

Step 1: Define the expression inside the logarithm. Let $S = \sqrt{2-\sqrt{3}} + \sqrt{2+\sqrt{3}}$. We need to find the value of $\log_6(S)$. Step 2: Square both sides to simplify the expression. We calculate $S^2$ by expanding: $$S^2 = \left(\sqrt{2-\sqrt{3}} + \sqrt{2+\sqrt{3}}\right)^2$$ Using the formula $(a+b)^2 = a^2 + 2ab + b^2$: $$S^2 = (2-\sqrt{3}) + 2\sqrt{(2-\sqrt{3})(2+\sqrt{3})} + (2+\sqrt{3})$$ Step 3: Simplify the product under the square root. Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$: $$(2-\sqrt{3})(2+\sqrt{3}) = 4 - 3 = 1$$ Therefore: $$S^2 = (2-\sqrt{3}) + 2\sqrt{1} + (2+\sqrt{3})$$ Step 4: Combine like terms. $$S^2 = 2 - \sqrt{3} + 2 + \sqrt{3} + 2(1)$$ $$S^2 = 4 + 2 = 6$$ Step 5: Find the value of $S$. Since $S > 0$ (as it is a sum of positive square roots): $$S = \sqrt{6}$$ Step 6: Evaluate the logarithm. $$\log_6(S) = \log_6(\sqrt{6}) = \log_6(6^{1/2}) = \frac{1}{2}$$ Step 7: Identify the nature of the answer. The value $\frac{1}{2}$ is rational (it can be expressed as a ratio of integers) but not an integer. **Final Answer:** The value of $\log_6(\sqrt{2-\sqrt{3}} + \sqrt{2+\sqrt{3}}) = \frac{1}{2}$, which is **rational but not integer**. This corresponds to **Option 2**.