1. **State the problem:** Simplify the expression $$6\log_2 \frac{2}{3} + 2\log_2 \frac{1}{6} - 4\log_2 \frac{2}{9}$$. 2. **Recall the logarithm power rule:** $$a \log_b x = \log_b x^a$$. We will apply this to each term. 3. Apply the power rule: $$6\log_2 \frac{2}{3} = \log_2 \left(\frac{2}{3}\right)^6$$ $$2\log_2 \frac{1}{6} = \log_2 \left(\frac{1}{6}\right)^2$$ $$-4\log_2 \frac{2}{9} = \log_2 \left(\frac{2}{9}\right)^{-4}$$ 4. Rewrite the expression as a single logarithm using the product and quotient rules: $$\log_2 \left(\frac{2}{3}\right)^6 + \log_2 \left(\frac{1}{6}\right)^2 + \log_2 \left(\frac{2}{9}\right)^{-4} = \log_2 \left[ \left(\frac{2}{3}\right)^6 \times \left(\frac{1}{6}\right)^2 \times \left(\frac{2}{9}\right)^{-4} \right]$$ 5. Simplify each term: $$\left(\frac{2}{3}\right)^6 = \frac{2^6}{3^6} = \frac{64}{729}$$ $$\left(\frac{1}{6}\right)^2 = \frac{1}{36}$$ $$\left(\frac{2}{9}\right)^{-4} = \left(\frac{9}{2}\right)^4 = \frac{9^4}{2^4} = \frac{6561}{16}$$ 6. Multiply the terms inside the logarithm: $$\frac{64}{729} \times \frac{1}{36} \times \frac{6561}{16} = \frac{64 \times 1 \times 6561}{729 \times 36 \times 16}$$ 7. Calculate numerator and denominator: $$64 \times 6561 = 419904$$ $$729 \times 36 = 26244$$ $$26244 \times 16 = 419904$$ 8. So the fraction is: $$\frac{419904}{419904} = 1$$ 9. Therefore, the expression simplifies to: $$\log_2 1 = 0$$ **Final answer:** $$0$$