<s>[INST] Logarithm-Based Sum Find the value of $x$ satisfying 
$18^{4x-3}=(54\sqrt{2})^{3x-4}$.
The given options are $2,6,3,4$.
I don't know how to do this.
 [/INST] Notice that, following Math Lover's hint, $\displaystyle 54\sqrt{2}=18^\frac{3}{2}$, because $54=(3\sqrt{2})^3$ and $18=(3\sqrt{2})^2$. 
Therefore, $18^{4x-3}=(18^\frac{3}{2})^{3x-4}$, so $18^{4x-3}=18^{\frac{3}{2}(3x-4)}$. 
Taking the base $18$ logarithm of each side gives 
$ 4x-3=\frac{3}{2}(3x-4)$
$8x-6=9x-12$
$\boxed{x=6}$.
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