How do you prove log_10 x +log_10(3x-5)=log_10 2?

Not proof but calculation. $\left\{x = - \frac{1}{3} , x = 2\right\}$
${\log}_{a} x + {\log}_{a} \left(3 x - 5\right) - {\log}_{a} 2 = {\log}_{a} \left(\frac{x \left(3 x - 5\right)}{2}\right) = 0$
But ${\log}_{a} y = 0 \to y = 1$ So we get as a result
$\left(\frac{x \left(3 x - 5\right)}{2}\right) = 1$. Now solving for $x$ we get
$\left\{x = - \frac{1}{3} , x = 2\right\}$ two solutions.