1. **Stating the problem:** Simplify the expression $\log_2\left(\sqrt{x+1} - \sqrt{x-1}\right)$. 2. **Recall the logarithm and radical properties:** The logarithm base 2 is defined as $\log_2(y)$, and the difference of square roots can be rationalized using the conjugate. 3. **Rationalize the expression inside the logarithm:** Multiply numerator and denominator by the conjugate $\sqrt{x+1} + \sqrt{x-1}$ to simplify the difference of roots: $$\sqrt{x+1} - \sqrt{x-1} = \frac{(\sqrt{x+1} - \sqrt{x-1})(\sqrt{x+1} + \sqrt{x-1})}{\sqrt{x+1} + \sqrt{x-1}} = \frac{(x+1) - (x-1)}{\sqrt{x+1} + \sqrt{x-1}} = \frac{2}{\sqrt{x+1} + \sqrt{x-1}}.$$ 4. **Rewrite the logarithm:** $$\log_2\left(\sqrt{x+1} - \sqrt{x-1}\right) = \log_2\left(\frac{2}{\sqrt{x+1} + \sqrt{x-1}}\right).$$ 5. **Use logarithm properties:** $$\log_2\left(\frac{2}{\sqrt{x+1} + \sqrt{x-1}}\right) = \log_2(2) - \log_2\left(\sqrt{x+1} + \sqrt{x-1}\right).$$ 6. **Evaluate $\log_2(2)$:** Since $2$ is the base, $\log_2(2) = 1$. 7. **Final simplified form:** $$1 - \log_2\left(\sqrt{x+1} + \sqrt{x-1}\right).$$ **Answer:** $$\boxed{1 - \log_2\left(\sqrt{x+1} + \sqrt{x-1}\right)}$$