d) \( \log _{5} x+\log _{5}(2 x-1)-\log _{5}(2 x+2)=1 \) \( \log _{3}(x+2)+\log _{3}(x+1)=\log _{3}\left(x^{2}-1\right. \) \( \log _{3} \frac{1}{x}=6-2 \cdot \log _{3} x \)

Let's dive into solving these logarithmic equations step by step! Starting with the first equation: \( \log_{5} x + \log_{5} (2x - 1) - \log_{5} (2x + 2) = 1 \). Using the properties of logarithms, we can combine the logs on the left: \[ \log_{5} \left( \frac{x(2x - 1)}{2x + 2} \right) = 1 \] This implies: \[ \frac{x(2x - 1)}{2x + 2} = 5 \] Cross-multiplying and simplifying gives you a polynomial which you can solve for x. Next, for the second equation: \(\log_{3}(x+2) + \log_{3}(x+1) = \log_{3}(x^{2}-1)\). Again, we can combine the logs: \[ \log_{3} \left((x + 2)(x + 1)\right) = \log_{3}(x^{2} - 1) \] Setting the insides equal: \[ (x + 2)(x + 1) = x^{2} - 1 \] This again leads to a polynomial that you can solve to find x. Lastly, for the equation: \(\log_{3} \frac{1}{x} = 6 - 2 \cdot \log_{3} x\). We can rewrite \(\log_{3} \frac{1}{x}\) as \(-\log_{3} x\): \[ -\log_{3} x = 6 - 2 \cdot \log_{3} x \] Rearranging gives you: \[ -\log_{3} x + 2 \cdot \log_{3} x = 6 \] This simplifies to: \[ \log_{3} x = 6 \] So \(x = 3^6\). Now go ahead and solve those polynomials to find your specific values for \(x\)! Happy solving!