1. **State the problem:** Evaluate $$\log_2\left(\log_2 82 - \log_{\sqrt{3}} 27 + \log_{\sqrt{10}} 10\right)$$. 2. **Recall logarithm change of base formula:** $$\log_a b = \frac{\log_c b}{\log_c a}$$ for any positive base $c \neq 1$. 3. **Evaluate each logarithm inside the parentheses:** - $$\log_2 82$$ is left as is for now. - $$\log_{\sqrt{3}} 27 = \frac{\log_3 27}{\log_3 \sqrt{3}}$$. Since $$27 = 3^3$$, $$\log_3 27 = 3$$. Also, $$\sqrt{3} = 3^{1/2}$$, so $$\log_3 \sqrt{3} = \frac{1}{2}$$. Therefore, $$\log_{\sqrt{3}} 27 = \frac{3}{\frac{1}{2}} = 3 \times 2 = 6$$. - $$\log_{\sqrt{10}} 10 = \frac{\log_{10} 10}{\log_{10} \sqrt{10}}$$. Since $$\log_{10} 10 = 1$$ and $$\sqrt{10} = 10^{1/2}$$, so $$\log_{10} \sqrt{10} = \frac{1}{2}$$. Therefore, $$\log_{\sqrt{10}} 10 = \frac{1}{\frac{1}{2}} = 2$$. 4. **Substitute back into the expression:** $$\log_2\left(\log_2 82 - 6 + 2\right) = \log_2\left(\log_2 82 - 4\right)$$. 5. **Approximate $$\log_2 82$$:** Since $$2^6 = 64$$ and $$2^7 = 128$$, $$\log_2 82$$ is between 6 and 7. Calculate: $$\log_2 82 = \frac{\ln 82}{\ln 2} \approx \frac{4.407}{0.693} \approx 6.36$$. 6. **Evaluate inside the parentheses:** $$6.36 - 4 = 2.36$$. 7. **Finally, evaluate $$\log_2 2.36$$:** $$\log_2 2.36 = \frac{\ln 2.36}{\ln 2} \approx \frac{0.859}{0.693} \approx 1.24$$. **Final answer:** $$\boxed{1.24}$$