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 properties and formulas:** - Change of base formula: $$\log_a b = \frac{\log_c b}{\log_c a}$$ for any positive base $c \neq 1$. - We know $$\log_x 16 = \frac{\log_2 16}{\log_2 x}$$ since base 2 is convenient here. - Also, $$\log_2 16 = 4$$ because $2^4 = 16$. 3. **Rewrite the equation using these properties:** $$\log_2 \left( \log_2 (7x - 10) \times \frac{4}{\log_2 x} \right) = 3$$ 4. **Simplify inside the logarithm:** Let $$A = \log_2 (7x - 10)$$ and $$B = \log_2 x$$. Then the inside becomes $$A \times \frac{4}{B} = \frac{4A}{B}$$. So the equation is: $$\log_2 \left( \frac{4A}{B} \right) = 3$$ 5. **Rewrite the logarithmic equation in exponential form:** $$\frac{4A}{B} = 2^3 = 8$$ 6. **Express $A$ and $B$ back in terms of $x$:** $$A = \log_2 (7x - 10), \quad B = \log_2 x$$ So: $$\frac{4 \log_2 (7x - 10)}{\log_2 x} = 8$$ 7. **Divide both sides by 4:** $$\frac{\log_2 (7x - 10)}{\log_2 x} = \frac{8}{4} = 2$$ 8. **Rewrite the left side as a single logarithm:** $$\frac{\log_2 (7x - 10)}{\log_2 x} = \log_x (7x - 10)$$ by the change of base formula. So: $$\log_x (7x - 10) = 2$$ 9. **Rewrite in exponential form:** $$7x - 10 = x^2$$ 10. **Rearrange to form a quadratic equation:** $$x^2 - 7x + 10 = 0$$ 11. **Factor the quadratic:** $$x^2 - 7x + 10 = (x - 5)(x - 2) = 0$$ 12. **Solve for $x$:** $$x = 5 \quad \text{or} \quad x = 2$$ 13. **Check domain restrictions:** - Inside the logarithms, arguments must be positive: - $7x - 10 > 0 \Rightarrow x > \frac{10}{7} \approx 1.428$ - $x > 0$ - Also, bases of logarithms must be positive and not 1: - $x > 0$ and $x \neq 1$ Both $x=2$ and $x=5$ satisfy these conditions. 14. **Find the sum of all solutions:** $$2 + 5 = 7$$ **Final answer:** The sum of all solutions is $$\boxed{7}$$.