18^n ————————— 3^2n-1*2^n-2

To simplify the expression \(\frac{18^n}{3^{2n-1} \cdot 2^{n-2}}\), let's first rewrite \(18^n\) in terms of its prime factors:

\[ 18 = 2 \cdot 3^2 \quad \implies \quad 18^n = (2 \cdot 3^2)^n = 2^n \cdot 3^{2n} \]

Now substituting this into the original expression:

\[ \frac{18^n}{3^{2n-1} \cdot 2^{n-2}} = \frac{2^n \cdot 3^{2n}}{3^{2n-1} \cdot 2^{n-2}} \]

Now, we can separate this into the fraction of the bases \(2\) and \(3\):

\[ = \frac{2^n}{2^{n-2}} \cdot \frac{3^{2n}}{3^{2n-1}} \]

Now we can simplify each part:

For the base \(2\):

\[ \frac{2^n}{2^{n-2}} = 2^{n - (n-2)} = 2^2 = 4 \]

For the base \(3\):

\[ \frac{3^{2n}}{3^{2n-1}} = 3^{2n - (2n-1)} = 3^1 = 3 \]

Now, putting both the simplified parts together:

\[ = 4 \cdot 3 = 12 \]

So, the simplified version of the original expression is:

\[ \frac{18^n}{3^{2n-1} \cdot 2^{n-2}} = 12 \]