d) \( \frac{1}{2} \log _{2}(3 x+3)-\frac{1}{2} \log _{2}(2 x-3)=\log _{2} 2 \)

To simplify the equation, we start by combining the logarithmic terms on the left side: \[ \frac{1}{2} \log _{2}(3 x+3)-\frac{1}{2} \log _{2}(2 x-3) = \frac{1}{2} \left( \log_{2}(3x+3) - \log_{2}(2x-3) \right) \] Using the property of logarithms that states \(\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)\), we have: \[ \frac{1}{2} \log_{2}\left(\frac{3x+3}{2x-3}\right) = \log_{2} 2 \] Since \(\log_{2} 2 = 1\), we can rewrite the equation: \[ \frac{1}{2} \log_{2}\left(\frac{3x+3}{2x-3}\right) = 1 \] Now, multiplying both sides by 2 gives us: \[ \log_{2}\left(\frac{3x+3}{2x-3}\right) = 2 \] Converting from logarithmic to exponential form, we get: \[ \frac{3x+3}{2x-3} = 2^2 \] This simplifies to: \[ \frac{3x+3}{2x-3} = 4 \] Next, we cross-multiply: \[ 3x + 3 = 4(2x - 3) \] Expanding the right side: \[ 3x + 3 = 8x - 12 \] To isolate \(x\), we move the \(x\) terms to one side and the constant terms to the other: \[ 3 + 12 = 8x - 3x \] This simplifies to: \[ 15 = 5x \] Dividing both sides by 5 gives us: \[ x = 3 \] Finally, don't forget to check if this solution is valid! Substituting \(x = 3\) back into the original logarithmic expressions, we must ensure that all arguments of your logarithms remain positive. In this case: For \(3x + 3\): \[ 3(3) + 3 = 12 \quad (\text{valid since } > 0) \] For \(2x - 3\): \[ 2(3) - 3 = 3 \quad (\text{valid since } > 0) \] Since both expressions are valid, the solution to the equation is: \[ \boxed{3} \]