0\ \wedge\ 3x-5 > 0\\\\x > \frac{1}{3}\ \wedge\ x > \frac{5}{3}\\\\x\in\left(\frac{5}{3};\ \infty\right)" alt="log_3(3x-1)+log_3(3x-5)=1\\\\D:3x-1 > 0\ \wedge\ 3x-5 > 0\\\\x > \frac{1}{3}\ \wedge\ x > \frac{5}{3}\\\\x\in\left(\frac{5}{3};\ \infty\right)" align="absmiddle" class="latex-formula">