The given expression is:

$\frac{6^{2n+1} \times 3^{n-2}}{18^{n+2}}$

Let's simplify this step by step.

Step 1: Express $18$ as a product of primes

Since $18 = 2 \times 3^2$, we can rewrite the denominator:

$18^{n+2} = (2 \times 3^2)^{n+2} = 2^{n+2} \times 3^{2(n+2)}$

Thus, the expression becomes:

$\frac{6^{2n+1} \times 3^{n-2}}{2^{n+2} \times 3^{2(n+2)}}$

Step 2: Express $6$ as a product of primes

Since $6 = 2 \times 3$, we can rewrite $6^{2n+1}$:

$6^{2n+1} = (2 \times 3)^{2n+1} = 2^{2n+1} \times 3^{2n+1}$

Now the expression becomes:

$\frac{2^{2n+1} \times 3^{2n+1} \times 3^{n-2}}{2^{n+2} \times 3^{2n+4}}$

Step 3: Combine powers of 2 and 3

Let's combine the powers of 2:

$\frac{2^{2n+1}}{2^{n+2}} = 2^{(2n+1) - (n+2)} = 2^{n-1}$

Now for the powers of 3:

$\frac{3^{2n+1} \times 3^{n-2}}{3^{2n+4}} = 3^{(2n+1 + n - 2) - (2n+4)} = 3^{n-5}$

Step 4: Final simplified expression

The simplified expression is:

$2^{n-1} \times 3^{n-5}$

This is the final simplified result.

Would you like more details, or do you have any questions?

Related questions:

1. How do you simplify exponents with different bases?
2. What is the prime factorization of a number and how does it help in solving such problems?
3. Can you provide more examples of simplifying expressions with exponents?
4. How do you apply exponent laws when dividing powers of the same base?
5. How do prime factorizations affect the outcome of simplifying fractions?

Tip:

When simplifying expressions with exponents, always look for opportunities to break numbers into their prime factors for easier manipulation.