1. **State the problem:** Evaluate $\log_2 \log_2 \sqrt{2\sqrt{2\sqrt{2}}}$ and verify if the answer $\frac{7}{16}$ is correct. 2. **Simplify the innermost expression:** $$\sqrt{2\sqrt{2\sqrt{2}}}$$ Start from the innermost square root: $$\sqrt{2} = 2^{\frac{1}{2}}$$ Then, $$\sqrt{2\sqrt{2}} = \sqrt{2 \times 2^{\frac{1}{2}}} = \sqrt{2^{1 + \frac{1}{2}}} = \sqrt{2^{\frac{3}{2}}} = 2^{\frac{3}{4}}$$ Now, $$\sqrt{2\sqrt{2\sqrt{2}}} = \sqrt{2 \times 2^{\frac{3}{4}}} = \sqrt{2^{1 + \frac{3}{4}}} = \sqrt{2^{\frac{7}{4}}} = 2^{\frac{7}{8}}$$ 3. **Evaluate the inner logarithm:** $$\log_2 \sqrt{2\sqrt{2\sqrt{2}}} = \log_2 2^{\frac{7}{8}} = \frac{7}{8}$$ 4. **Evaluate the outer logarithm:** $$\log_2 \left( \frac{7}{8} \right) = \log_2 7 - \log_2 8 = \log_2 7 - 3$$ 5. **Interpretation:** The value is $\log_2 7 - 3$, which is approximately $2.807 - 3 = -0.193$, not $\frac{7}{16}$. **Conclusion:** The answer $\frac{7}{16}$ is incorrect. **Final answer:** $$\log_2 \log_2 \sqrt{2\sqrt{2\sqrt{2}}} = \log_2 7 - 3$$