Let's rewrite the expression using only prime numbers for the bases

Notice the following \(4^4=(2^2)^4=2^8\)
And \(9^n=(3^2)^n=3^{2n}\)
And: \(6^9=(2\times3)^9=(2^9)(3^9)\)

Now replace the given values with their equivalents to get:\(\frac{(3^7)(2^8)(3^{2n})}{(2^9)(3^9)}=40.5\)

Multiply both sides by 2: \(\frac{(2)(3^7)(2^8)(3^{2n})}{(2^9)(3^9)}=81\)

Simplify the numerator and rewrite 81: \(\frac{(3^7)(2^9)(3^{2n})}{(2^9)(3^9)}=3^4\)

Notice that \(2^9\) cancels out in numerator and denominator: \(\frac{(3^7)(3^{2n})}{(3^9)}=3^4\)

Multiply both sides of the equation by \(3^9\) to get: \((3^7)(3^{2n}) = (3^9)(3^4)\)

Simplify both sides: \(3^{7+2n} = 3^{13}\)

Since the bases are equal we can conclude that the exponents are equal to get: \(7+2n=13\)

Solve to get \(n=3\)

Answer: 3

Cheers,
Brent