1. **State the problem:** Evaluate $\log_2 \log_2 \sqrt{2 \sqrt{2 \sqrt{2}}}$. 2. **Simplify the innermost expression:** Start with $\sqrt{2 \sqrt{2 \sqrt{2}}}$. 3. Rewrite the nested radicals step-by-step: - Let $x = \sqrt{2}$. - Then $\sqrt{2 \sqrt{2}} = \sqrt{2 \cdot x} = \sqrt{2 \cdot \sqrt{2}}$. 4. Calculate $\sqrt{2 \sqrt{2}}$: $$\sqrt{2 \sqrt{2}} = (2 \cdot 2^{1/2})^{1/2} = (2^{1 + 1/2})^{1/2} = 2^{(3/2) \cdot 1/2} = 2^{3/4}.$$ 5. Now substitute back: $$\sqrt{2 \sqrt{2 \sqrt{2}}} = \sqrt{2 \cdot 2^{3/4}} = \sqrt{2^{1 + 3/4}} = \sqrt{2^{7/4}} = 2^{7/8}.$$ 6. Evaluate the inner logarithm: $$\log_2 \sqrt{2 \sqrt{2 \sqrt{2}}} = \log_2 2^{7/8} = \frac{7}{8}.$$ 7. Evaluate the outer logarithm: $$\log_2 \left( \frac{7}{8} \right).$$ Since $\frac{7}{8} < 1$, this logarithm is negative but exact value is: $$\log_2 \frac{7}{8} = \log_2 7 - \log_2 8 = \log_2 7 - 3.$$ **Final answer:** $$\boxed{\log_2 7 - 3}.$$