If the interval x satisfying the equation
`|x| +|-x|=(log_(3)(x-2))/(|log_(3)(x-2)|) " is " (a,b), " then " a+b=` _______.

Explanation

To solve the equation \( |x| + |-x| = \frac{\log_3(x-2)}{|\log_3(x-2)|} \), we will break it down step by step.

### Step 1: Simplify the Left-Hand Side
The left-hand side of the equation is \( |x| + |-x| \). We know that:
\[ |-x| = |x| \]
Thus, we can simplify:
\[ |x| + |-x| = |x| + |x| = 2|x| \]
So, the equation becomes:
\[ 2|x| = \frac{\log_3(x-2)}{|\log_3(x-2)|} \]

### Step 2: Analyze the Right-Hand Side
The right-hand side \( \frac{\log_3(x-2)}{|\log_3(x-2)|} \) can take two values:
- It is \( 1 \) when \( \log_3(x-2) > 0 \) (i.e., \( x - 2 > 1 \) or \( x > 3 \))
- It is \( -1 \) when \( \log_3(x-2) < 0 \) (i.e., \( x - 2 < 1 \) or \( x < 3 \))

### Step 3: Set Up Cases
We will consider two cases based on the value of the right-hand side.

#### Case 1: \( \log_3(x-2) > 0 \)
In this case, we have:
\[ 2|x| = 1 \]
This implies:
\[ |x| = \frac{1}{2} \]
Thus, \( x = \frac{1}{2} \) or \( x = -\frac{1}{2} \). However, for \( x > 3 \), neither of these values is valid.

#### Case 2: \( \log_3(x-2) < 0 \)
In this case, we have:
\[ 2|x| = -1 \]
This is impossible since \( |x| \) cannot be negative.

### Step 4: Determine Valid Interval
From the analysis, we need to ensure that \( x - 2 < 1 \) for the logarithm to be negative, which gives us:
\[ x < 3 \]
Also, \( x - 2 > 0 \) implies:
\[ x > 2 \]
Thus, we have the interval:
\[ 2 < x < 3 \]

### Step 5: Find \( a + b \)
Here, \( a = 2 \) and \( b = 3 \). Therefore:
\[ a + b = 2 + 3 = 5 \]

### Final Answer
The value of \( a + b \) is \( \boxed{5} \).