log2(x-1)-log2(2x-11)=log2 2

To solve the equation \( \log_2(x - 1) - \log_2(2x - 11) = \log_2 2 \), we can use the properties of logarithms.

1. Use the quotient rule of logarithms: \[ \log_2 \left(\frac{x - 1}{2x - 11}\right) = \log_2 2. \]

2. Since the logarithm on both sides has the same base (base 2), we can exponentiate to eliminate the logs: \[ \frac{x - 1}{2x - 11} = 2. \]

3. Cross-multiply to solve for \( x \): \[ x - 1 = 2(2x - 11). \]

4. Expand the right side: \[ x - 1 = 4x - 22. \]

5. Rearranging the equation to isolate \( x \): \[ x - 4x = -22 + 1, \] \[ -3x = -21, \] \[ x = 7. \]

6. Finally, we should check if \( x = 7 \) is valid by substituting back into the original logarithms:

   • Check \( x - 1 = 7 - 1 = 6 > 0 \)
   • Check \( 2x - 11 = 2(7) - 11 = 14 - 11 = 3 > 0 \)

Both conditions are satisfied.

Thus, the solution is: \[ \boxed{7}. \]