1. **State the problem:** Solve the equation $$\log_2 \left( \log_2 (7x - 10) \times \log_x 16 \right) = 3$$ and find the sum of all solutions. 2. **Recall logarithm properties:** - $$\log_a b \times \log_b c = \log_a c$$ - Change of base formula: $$\log_x 16 = \frac{\log_2 16}{\log_2 x}$$ - Since $$16 = 2^4$$, $$\log_2 16 = 4$$. 3. **Rewrite the inner expression:** $$\log_2 (7x - 10) \times \log_x 16 = \log_2 (7x - 10) \times \frac{4}{\log_2 x}$$ 4. **Substitute into the original equation:** $$\log_2 \left( \log_2 (7x - 10) \times \frac{4}{\log_2 x} \right) = 3$$ 5. **Let $$a = \log_2 (7x - 10)$$ and $$b = \log_2 x$$, then:** $$\log_2 \left( \frac{4a}{b} \right) = 3$$ 6. **Rewrite the equation:** $$\log_2 \left( \frac{4a}{b} \right) = 3 \implies \frac{4a}{b} = 2^3 = 8$$ 7. **Solve for $$a$$:** $$4a = 8b \implies a = 2b$$ 8. **Recall definitions:** $$a = \log_2 (7x - 10), \quad b = \log_2 x$$ So, $$\log_2 (7x - 10) = 2 \log_2 x$$ 9. **Use logarithm power rule:** $$\log_2 (7x - 10) = \log_2 (x^2)$$ 10. **Equate arguments:** $$7x - 10 = x^2$$ 11. **Rewrite as quadratic:** $$x^2 - 7x + 10 = 0$$ 12. **Factor the quadratic:** $$(x - 5)(x - 2) = 0$$ 13. **Solutions:** $$x = 5 \quad \text{or} \quad x = 2$$ 14. **Check domain restrictions:** - Inside $$\log_2 (7x - 10)$$, argument must be positive: $$7x - 10 > 0 \implies x > \frac{10}{7} \approx 1.428$$ - Inside $$\log_x 16$$, base $$x > 0$$ and $$x \neq 1$$ Both $$x=2$$ and $$x=5$$ satisfy these conditions. 15. **Find sum of solutions:** $$2 + 5 = 7$$ **Final answer:** $$\boxed{7}$$