Let `f` be a real-valued invertible function such that `f((2x-3)/(x-2))=5x-2, x!=2.` Then value of `f^(-1)(13)` is________

Explanation

To solve the problem, we need to find the value of \( f^{-1}(13) \) given the function \( f\left(\frac{2x-3}{x-2}\right) = 5x - 2 \).

### Step-by-Step Solution:

1. **Understand the Given Function**:
We have the function defined as:
\[ f\left(\frac{2x-3}{x-2}\right) = 5x - 2 \]
Our goal is to find \( f^{-1}(13) \).

2. **Set Up the Equation**:
To find \( f^{-1}(13) \), we need to find \( x \) such that:
\[ f(x) = 13 \]
From the given function, we can set:
\[ 5x - 2 = 13 \]

3. **Solve for \( x \)**:
Rearranging the equation:
\[ 5x - 2 = 13 \implies 5x = 13 + 2 \implies 5x = 15 \implies x = \frac{15}{5} = 3 \]

4. **Find \( f^{-1}(13) \)**:
Now that we have \( x = 3 \), we can substitute this back into the expression we derived from the function:
\[ f^{-1}(13) = \frac{2(3) - 3}{3 - 2} = \frac{6 - 3}{1} = \frac{3}{1} = 3 \]

5. **Conclusion**:
Therefore, the value of \( f^{-1}(13) \) is:
\[ \boxed{3} \]