Let `agt1` be a real number and `f(x)=log_(a)x^(2)" for "xgt 0.` If `f^(-1)` is the inverse function fo f and b and c are real numbers then `f^(-1)(b+c)` is equal to

Options

A. `f^(-1)(b).f^(-1)(c)`
B. `f^(-1)(b)+f^(-1)(c)`
C. `(1)/(f(b+c))`
D. `(1)/(f^(-1)(b)+f^(-1)(c))`

Explanation

To solve the problem, we need to find the expression for \( f^{-1}(b+c) \) given the function \( f(x) = \log_a(x^2) \) for \( x > 0 \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \log_a(x^2) \]

2. **Simplify the function**:
Using the logarithmic identity \( \log_a(x^m) = m \cdot \log_a(x) \), we can rewrite:
\[ f(x) = 2 \cdot \log_a(x) \]

3. **Set \( y = f(x) \)**:
Let \( y = f(x) \). Then we have:
\[ y = 2 \cdot \log_a(x) \]

4. **Express \( x \) in terms of \( y \)**:
To find the inverse function, we need to express \( x \) in terms of \( y \):
\[ \frac{y}{2} = \log_a(x) \]
Exponentiating both sides gives:
\[ x = a^{y/2} \]

5. **Write the inverse function**:
Thus, the inverse function \( f^{-1}(y) \) is:
\[ f^{-1}(y) = a^{y/2} \]

6. **Find \( f^{-1}(b+c) \)**:
Now, we need to find \( f^{-1}(b+c) \):
\[ f^{-1}(b+c) = a^{(b+c)/2} \]

7. **Express \( f^{-1}(b+c) \)**:
We can rewrite this as:
\[ f^{-1}(b+c) = a^{b/2} \cdot a^{c/2} \]

8. **Relate to \( f^{-1}(b) \) and \( f^{-1}(c) \)**:
Recognizing that:
\[ f^{-1}(b) = a^{b/2} \quad \text{and} \quad f^{-1}(c) = a^{c/2} \]
We can express:
\[ f^{-1}(b+c) = f^{-1}(b) \cdot f^{-1}(c) \]

### Final Answer:
Thus, we conclude that:
\[ f^{-1}(b+c) = f^{-1}(b) \cdot f^{-1}(c) \]