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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.

This detailed solution explains how to solve the logarithmic equation log_3(x - a) = b by converting it to its exponential form and solving for x. Explore core logarithmic concepts, suitable for Grades 10-12.

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