Let's solve the equation \(\frac{9^{2x-1}}{27^{x+1}} = 1\) step by step:

1. Rewrite 9 and 27 as powers of 3: \(\frac{(3^2)^{2x-1}}{(3^3)^{x+1}} = 1\)

2. Simplify the exponents: \(\frac{3^{4x-2}}{3^{3x+3}} = 1\)

3. Subtract the exponents: \(3^{4x-2-(3x+3)} = 1\)

4. Simplify the exponent: \(3^{x-5} = 1\)

5. Take the logarithm of both sides: \(x-5 = \log_3(1)\)

6. Solve for x: \(x = 5 + \log_3(1) = 5 + 0 = 5\)

So, the value of x that satisfies the equation \(\frac{9^{2x-1}}{27^{x+1}} = 1\) is 5