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 inverse of the function \( f(x) = \log_a(x^2) \) and then evaluate \( f^{-1}(b+c) \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \log_a(x^2) \]
where \( a > 1 \) and \( x > 0 \).

2. **Set \( f(x) \) equal to \( y \)**:
\[ y = \log_a(x^2) \]

3. **Use the properties of logarithms**:
We can rewrite the logarithmic equation using the property \( \log_a(b) = c \) implies \( a^c = b \):
\[ x^2 = a^y \]

4. **Solve for \( x \)**:
Taking the square root of both sides, we get:
\[ x = \sqrt{a^y} = a^{y/2} \]
Since \( x > 0 \), we only consider the positive root.

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

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

7. **Express \( f^{-1}(b+c) \)** in terms of \( f^{-1}(b) \) and \( f^{-1}(c) \):
We can express \( f^{-1}(b) \) and \( f^{-1}(c) \):
\[ f^{-1}(b) = a^{b/2} \quad \text{and} \quad f^{-1}(c) = a^{c/2} \]
Hence,
\[ f^{-1}(b+c) = a^{(b+c)/2} = a^{b/2} \cdot a^{c/2} = f^{-1}(b) \cdot f^{-1}(c) \]

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