1. **State the problem:** Calculate the value of $$\log_{\frac{1}{2}} 8 + \log_2 \frac{1}{8} + \log_3 \frac{1}{9}.$$ 2. **Rewrite each logarithm individually:** - For $$\log_{\frac{1}{2}} 8$$, recall that $$\frac{1}{2} = 2^{-1}$$ and 8 = $$2^3$$. - For $$\log_2 \frac{1}{8}$$, note that $$\frac{1}{8} = 2^{-3}$$. - For $$\log_3 \frac{1}{9}$$, note that $$\frac{1}{9} = 3^{-2}$$. 3. **Evaluate each logarithm:** - Use the change of base rule: $$\log_{a} b = \frac{\log b}{\log a}$$, or recognize exponents directly. 4. **Calculate $$\log_{\frac{1}{2}} 8$$:** $$\log_{2^{-1}} 2^{3} = \frac{3}{-1} = -3.$$ Because changing base from $$2^{-1}$$ to $$2$$ flips the sign. 5. **Calculate $$\log_2 \frac{1}{8}$$:** $$\log_2 2^{-3} = -3.$$ 6. **Calculate $$\log_3 \frac{1}{9}$$:** $$\log_3 3^{-2} = -2.$$ 7. **Sum the results:** $$-3 + (-3) + (-2) = -8.$$ **Final answer:** $$\boxed{-8}$$