1. **State the problem:** Simplify the expression $\log 2\sqrt{2x+2\sqrt{x^2-1}} + \log 2(\sqrt{x+1} - \sqrt{x-1})$. 2. **Recall logarithm properties:** - $\log a + \log b = \log(ab)$ - Simplify inside the logarithms before combining. 3. **Combine the logs:** $$\log \left(2\sqrt{2x+2\sqrt{x^2-1}} \times 2(\sqrt{x+1} - \sqrt{x-1})\right) = \log \left(4(\sqrt{2x+2\sqrt{x^2-1}})(\sqrt{x+1} - \sqrt{x-1})\right)$$ 4. **Simplify inside the square root:** Note that $$2x + 2\sqrt{x^2 - 1} = 2\left(x + \sqrt{x^2 - 1}\right)$$ 5. **Recognize that $x + \sqrt{x^2 - 1} = (\sqrt{x+1} + \sqrt{x-1})^2 / 2$:** Calculate: $$(\sqrt{x+1} + \sqrt{x-1})^2 = (x+1) + (x-1) + 2\sqrt{(x+1)(x-1)} = 2x + 2\sqrt{x^2 - 1}$$ So, $$\sqrt{2x + 2\sqrt{x^2 - 1}} = \sqrt{(\sqrt{x+1} + \sqrt{x-1})^2} = \sqrt{x+1} + \sqrt{x-1}$$ 6. **Substitute back:** $$4(\sqrt{x+1} + \sqrt{x-1})(\sqrt{x+1} - \sqrt{x-1})$$ 7. **Use difference of squares:** $$(\sqrt{x+1} + \sqrt{x-1})(\sqrt{x+1} - \sqrt{x-1}) = (x+1) - (x-1) = 2$$ 8. **Final simplification:** $$4 \times 2 = 8$$ 9. **Answer:** $$\log 8$$ Since $8 = 2^3$, we can write: $$\log 8 = \log 2^3 = 3 \log 2$$ **Final answer:** $3 \log 2$