The equation provided is:

$\log_2 (3x - 1) = 3$

To solve for $x$, follow these steps:

1. Recall that the logarithmic equation $\log_b(a) = c$ can be rewritten in its exponential form as $a = b^c$.

   For the given equation:

   $3x - 1 = 2^3$

2. Calculate $2^3$:

   $3x - 1 = 8$

3. Solve for $x$:

   $3x = 8 + 1$ $3x = 9$ $x = \frac{9}{3} = 3$

So, the solution is:

$x = 3$

Would you like more details or have any other questions?

Here are some related questions you might find useful:

1. How do you solve logarithmic equations with different bases?
2. What are the properties of logarithms and how are they used in equations?
3. Can logarithmic functions have negative arguments?
4. How do you graph logarithmic functions?
5. How does the change of base formula work in logarithms?

Tip: When dealing with logarithmic equations, always check if the argument of the logarithm is positive, as logarithms are only defined for positive numbers.