Let a = 3 sqrt{2} and b = frac{1}{5^{frac{1}{ | vedclass

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(C) Given $a = 3\sqrt{2} = \sqrt{18}$. Thus,$3x + 2y = \log_{\sqrt{18}}(18)^{\frac{5}{4}} = \log_{18^{\frac{1}{2}}}(18)^{\frac{5}{4}} = \frac{5/4}{1/2

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$8$

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(C) Given $a = 3\sqrt{2} = \sqrt{18}$. Thus,$3x + 2y = \log_{\sqrt{18}}(18)^{\frac{5}{4}} = \log_{18^{\frac{1}{2}}}(18)^{\frac{5}{4}} = \frac{5/4}{1/2} = \frac{5}{2}$.Multiplying by $2$,we get $6x + 4y = 5$ ... $(1)$.Given $b = \frac{1}{5^{\frac{1}{6}} \sqrt{6}} = (5^{\frac{1}{6}} \cdot 6^{\frac{1}{2}})^{-1} = (5^{\frac{1}{6}} \cdot (6^3)^{\frac{1}{6}})^{-1} = (30^{\frac{1}{6}})^{-1} = 30^{-\frac{1}{6}}$.Also,$\sqrt{1080} = \sqrt{36 	imes 30} = 6\sqrt{30} = 6^1 \cdot 30^{\frac{1}{2}}$. This does not simplify easily to base $b$. Let us re-evaluate $2x - y = \log_b(\sqrt{1080})$.Note $1080 = 216 	imes 5 = 6^3 	imes 5$. So $\sqrt{1080} = 6^{\frac{3}{2}} \cdot 5^{\frac{1}{2}}$.Since $b = (5^{\frac{1}{6}} \cdot 6^{\frac{1}{2}})^{-1}$,then $b^{-3} = (5^{\frac{1}{6}} \cdot 6^{\frac{1}{2}})^3 = 5^{\frac{1}{2}} \cdot 6^{\frac{3}{2}} = \sqrt{1080}$.Thus,$2x - y = \log_b(b^{-3}) = -3$ ... $(2)$.From $(1)$ and $(2)$,we have $6x + 4y = 5$ and $2x - y = -3$. Multiplying $(2)$ by $4$ gives $8x - 4y = -12$.Adding this to $(1)$: $(6x + 4y) + (8x - 4y) = 5 - 12$ $\Rightarrow 14x = -7$ $\Rightarrow x = -0.5$.Substituting $x = -0.5$ into $(2)$: $2(-0.5) - y = -3$ $\Rightarrow -1 - y = -3$ $\Rightarrow y = 2$.Then $4x + 5y = 4(-0.5) + 5(2) = -2 + 10 = 8$.

Let $a = 3 \sqrt{2}$ and $b = \frac{1}{5^{\frac{1}{6}} \sqrt{6}}$. 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:

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