Let $a = 3\sqrt{2}$ and $b = \frac{1}{5^{1/16}\sqrt{6}}$. If $x, y \in \mathbb{R}$ are such that $3x + 2y = \log_a (18)^{\frac{5}{4}}$ and $2x - y = \log_b (\sqrt{1080})$, then $4x + 5y$ is equal to ..............

Let a = 3√2 and b = 1/(5^(1/16)√6). If x, y ∈ R are such that 3x + 2y = logₐ (18)^(5/4) and 2x - y = log_b (√1080), then 4x + 5y is equal to .............. $\text{Given: } a = 3\sqrt{2}, \quad b = \dfrac{1}{5^{1/16}\sqrt{6}}$ $3x + 2y = \log_a (18)^{5/4}$ $2x - y = \log_b (\sqrt{1080})$ $\text{To find: } 4x + 5y$ $1.\ \text{Simplify } a,\ b \text{ and get both logs}$ $2.\ \text{Put the values into the equations}$ $3.\ \text{Find } x,\ y \text{ then } 4x+5y$ $a = 3\sqrt{2}$ $a^2 = (3\sqrt{2})^2 = 9 \cdot 2 = 18$ $a^2 = 18 \ \Longrightarrow \ \log_a 18 = 2$ $\log_a (18)^{5/4} = \frac{5}{4}\log_a 18$ $= \frac{5}{4} \cdot 2 = \frac{5}{2}$ $b = \dfrac{1}{5^{1/6}\sqrt{6}} = 5^{-1/6}\cdot 6^{-1/2}$ $1080 = 6^3 \cdot 5$ $\sqrt{1080} = 6^{3/2}\cdot 5^{1/2}$ $b^{-3} = \left(5^{-1/6}\cdot 6^{-1/2}\right)^{-3} = 5^{1/2}\cdot 6^{3/2}$ $b^{-3} = \sqrt{1080}$ $b^{-3} = \sqrt{1080} \ \Longrightarrow \ \log_b(\sqrt{1080}) = -3$ $3x + 2y = \frac{5}{2}$ $2x - y = -3$ $2x - y = -3 \ \Longrightarrow \ y = 2x + 3$ $3x + 2(2x+3) = \frac{5}{2}$ $7x + 6 = \frac{5}{2}$ $7x = \frac{5}{2} - 6 = \frac{5-12}{2} = -\frac{7}{2}$ $x = -\frac{1}{2}$ $y = 2x + 3 = 2\left(-\frac{1}{2}\right) + 3 = -1 + 3 = 2$ $4x + 5y = 4\left(-\frac{1}{2}\right) + 5(2) = -2 + 10$ $4x + 5y = 8$ $4x + 5y = 2(3x+2y) - (2x-y)$ $= 2\left(\frac{5}{2}\right) - (-3) = 5 + 3 = 8$ $\boxed{4x + 5y = 8}$