## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{10}{3}$
The domain of the variable requires that $3x-1>0$. Thus, $3x>1\\ x>\dfrac{1}{3}$ Recall: $y = \log_a b \text{ is equivalent to } b=a^y$. Therefore, $\log_3 (3x-1) = 2 \text{ is equivalent to } 3x-1 = 3^2$ Solve the equation to obtain: $3x-1 = 3^2$ $3x-1 = 9$ $3x = 9+1$ $3x=10$ $x=\boxed{\frac{10}{3}}$