1. Interpret the equations carefully

We are given $a=3\sqrt{2},\qquad b=\frac{1}{5^{1/6}\sqrt{6}}$ and $3x+2y=\log_a\left(18^{5/4}\right),$ $2x-y=\log_b(\sqrt{1080}).$

We need to find $4x+5y.$

2. Simplify the first logarithm

Since $a=3\sqrt{2}=\sqrt{18}=18^{1/2},$ we get $\log_a\left(18^{5/4}\right)=\log_{18^{1/2}}\left(18^{5/4}\right).$

Using $\log_{m^p}(m^q)=\frac{q}{p},$ we have $\log_{18^{1/2}}\left(18^{5/4}\right)=\frac{5/4}{1/2}=\frac{5}{2}.$

So the first equation becomes 3x+2y=\frac{5}{2}.\tag{1}

3. Simplify the second logarithm

First simplify $b$: $b=\frac{1}{5^{1/6}\sqrt{6}}=5^{-1/6}\cdot 6^{-1/2}.$

Now simplify $\sqrt{1080}$: $1080=36\cdot 30,$ so $\sqrt{1080}=6\sqrt{30}=6\sqrt{5\cdot 6}=6\cdot 5^{1/2}6^{1/2}=5^{1/2}6^{1}.$

Also note $b^{-6}=(5^{-1/6}6^{-1/2})^{-6}=5^1 6^3=5\cdot 216=1080.$ Hence $b^{-3}=\sqrt{1080}.$

Therefore, $\log_b(\sqrt{1080})=\log_b(b^{-3})=-3.$

So the second equation becomes 2x-y=-3.\tag{2}

4. Solve the linear system

From (2), $y=2x+3.$

Substitute into (1): $3x+2(2x+3)=\frac{5}{2}$ $3x+4x+6=\frac{5}{2}$ $7x=\frac{5}{2}-6=\frac{5-12}{2}=-\frac{7}{2}$ $x=-\frac{1}{2}.$

Then $y=2\left(-\frac12\right)+3=-1+3=2.$

5. Compute $4x+5y$

$4x+5y=4\left(-\frac12\right)+5(2)=-2+10=8.$

6. Compare with stored answer

Derived answer is $8,$ which matches the stored correct answer.