We are given the logarithmic equation:

$\log_6 (2x - 9) = 2$

Step 1: Rewrite the equation in exponential form

Recall that a logarithmic equation $\log_b(y) = x$ can be rewritten in exponential form as $b^x = y$. Applying this to our equation:

$\log_6 (2x - 9) = 2 \quad \Rightarrow \quad 6^2 = 2x - 9$

Step 2: Simplify

We know that $6^2 = 36$, so the equation becomes:

$36 = 2x - 9$

Step 3: Solve for $x$

Now, solve for $x$ by isolating it on one side of the equation.

$36 + 9 = 2x$ $45 = 2x$ $x = \frac{45}{2} = 22.5$

Step 4: Verify the solution

Substitute $x = 22.5$ back into the original logarithmic equation to ensure it is valid:

$\log_6(2(22.5) - 9) = \log_6(45 - 9) = \log_6(36)$

Since $\log_6(36) = 2$, the solution is correct.

Thus, the solution is: $x = 22.5$

Would you like more details or have any questions?

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Here are 5 related questions to further explore:

1. How do you convert logarithmic equations to exponential form?
2. What is the general method to solve equations involving logarithms?
3. How would the solution change if the base of the logarithm was different?
4. Can logarithmic equations have multiple solutions?
5. How would you solve an equation like $\log_6(2x - 9) = \log_6(3x - 5)$?

Tip: Always check the domain when solving logarithmic equations to avoid extraneous solutions.