Solve For x:

Let $a=18=2^13^2$ and $b=54\sqrt 2 = 2^\frac 32 3^3$ . $\implies b^2 = 2^33^6=(2^13^2)^3=a^3$ . Therefore,

$\begin{aligned} 18^{4x-3} & = (54\sqrt 2)^{3x-4} \\ a^{4x-3} & = b^{3x-4} \\ a^{4x-3} & = a^{\frac 32(3x-4)} \\ \implies 4x-3 & = \frac 32(3x-4) \\ \frac 4x-3 & = \frac 92x -6 \\ \frac x2 & = 3 \\ \implies x & = \boxed 6 \end{aligned}$