1. The problem states: Solve the equation $$\log_4 x + \log_2 y = 5$$ given the conditions from the previous system. 2. Recall the change of base formula and properties of logarithms: - $$\log_a b = \frac{\log_c b}{\log_c a}$$ for any positive base $c$. - Here, $$\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}$$ since $4 = 2^2$. 3. Rewrite the equation using base 2 logarithms: $$\log_4 x + \log_2 y = 5 \implies \frac{\log_2 x}{2} + \log_2 y = 5$$ 4. Multiply both sides by 2 to clear the denominator: $$\log_2 x + 2 \log_2 y = 10$$ 5. Use the logarithm power rule on the second term: $$\log_2 x + \log_2 y^2 = 10$$ 6. Combine the logarithms: $$\log_2 (x y^2) = 10$$ 7. Convert from logarithmic to exponential form: $$x y^2 = 2^{10} = 1024$$ 8. The original system also includes $x + y = 34$ and $\log_2 x + \log_2 y = 6$ (from the first equation in the system), which implies: $$\log_2 (xy) = 6 \implies xy = 2^6 = 64$$ 9. Now we have two equations: $$x + y = 34$$ $$xy = 64$$ $$x y^2 = 1024$$ 10. From $xy = 64$, express $x = \frac{64}{y}$. 11. Substitute into $x y^2 = 1024$: $$\frac{64}{y} \cdot y^2 = 1024 \implies 64 y = 1024 \implies y = \frac{1024}{64} = 16$$ 12. Substitute $y=16$ into $x + y = 34$: $$x + 16 = 34 \implies x = 18$$ 13. Check $xy = 18 \times 16 = 288$, which contradicts $xy=64$ from step 8. 14. Re-examine step 8: The original problem states $\log_2 x + \lg_2 y = 6$; assuming $\lg_2 y$ means $\log_2 y$, so $\log_2 x + \log_2 y = 6$ implies $xy=64$. 15. But from step 7, $x y^2 = 1024$. 16. Substitute $x = \frac{64}{y}$ into $x y^2 = 1024$: $$\frac{64}{y} y^2 = 1024 \implies 64 y = 1024 \implies y = 16$$ 17. Then $x = \frac{64}{16} = 4$. 18. Check $x + y = 4 + 16 = 20$, which contradicts $x + y = 34$. 19. The problem states $x + y = 34$ and $\log_2 x + \log_2 y = 6$ from the first system, but the user only asked to solve the last equation $\log_4 x + \log_2 y = 5$. 20. Without the other equations, the solution is: $$\log_4 x + \log_2 y = 5 \implies x y^2 = 1024$$ 21. The problem is underdetermined with one equation and two variables. 22. Express $x$ in terms of $y$: $$x = \frac{1024}{y^2}$$ 23. This is the general solution for the given equation. Final answer: $$x = \frac{1024}{y^2}$$ where $x > 0$, $y > 0$.