The equation you've provided is:

$\log_3 (x - a) = b$

This is a logarithmic equation where the base of the logarithm is 3. To solve for $x$, we can rewrite the logarithmic equation in its exponential form. The general form of the logarithmic equation:

$\log_b y = x$

is equivalent to:

$y = b^x$

Applying this to your equation:

$\log_3 (x - a) = b \quad \Rightarrow \quad x - a = 3^b$

Finally, solve for $x$:

$x = 3^b + a$

This is the solution for $x$ in terms of $a$ and $b$.

Would you like further details on logarithmic properties or have any questions on this topic? Here are 5 related questions to expand your understanding:

1. How do you handle logarithmic equations with different bases?
2. Can you solve a similar equation if $b$ is negative or fractional?
3. What happens if the argument of the logarithm (i.e., $x - a$) is negative?
4. How do we solve equations with logarithms on both sides?
5. What are some applications of logarithmic equations in real-world problems?

Tip: When solving logarithmic equations, always check that the argument of the logarithm is positive, as logarithms are only defined for positive arguments.