The given equation is:

$\log_3(2x - 8) = 0$

Let's solve this step by step:

Step 1: Convert the logarithmic equation into an exponential equation.

The logarithmic equation can be rewritten using the property of logarithms: $\log_b(a) = c \quad \text{is equivalent to} \quad b^c = a$ In this case: $3^0 = 2x - 8$

Since $3^0 = 1$, the equation becomes: $1 = 2x - 8$

Step 2: Solve for $x$.

Add 8 to both sides: $1 + 8 = 2x$ $9 = 2x$

Now, divide by 2: $x = \frac{9}{2} = 4.5$

Final Solution:

The value of $x$ is $x = 4.5$.

Would you like a detailed explanation or have any questions about the steps? Here are a few related questions you might find useful:

1. What are the properties of logarithms and how are they used in equations like this?
2. How can you handle more complex logarithmic equations with different bases?
3. How do you check if the solution of a logarithmic equation is valid in its domain?
4. Can you explain the relationship between exponential and logarithmic functions?
5. How do you solve logarithmic equations that involve addition or subtraction of logarithms?

Tip: Always ensure that the argument inside a logarithmic function is positive, as logarithms are not defined for zero or negative values.