If `log_(x)a,a^(x//2)` and `log_(a)x` are in G.P, then x is equal to:

Options

A. `log_(a)(log_(b)a)`
B. `log_(a)(log_(e)a)-log_(e)(log_(e)b)`
C. `-log_(e)(log_(a)b)`
D. both (a) and (b)

Answer

To solve the problem, we need to determine the value of \( x \) given that \( \log_x a \), \( a^{(x/2)} \), and \( \log_a x \) are in geometric progression (G.P.).

### Step-by-Step Solution:

1. **Understanding G.P. Condition**:
For three terms \( A, B, C \) to be in G.P., the square of the middle term \( B \) must equal the product of the other two terms \( A \) and \( C \). Therefore, we can write:
\[ (a^{(x/2)})^2 = \log_x a \cdot \log_a x \]

2. **Simplifying the Left Side**:
The left side simplifies to:
\[ a^{x} \]

3. **Using Change of Base Formula**:
We can express \( \log_x a \) and \( \log_a x \) using the change of base formula:
\[ \log_x a = \frac{1}{\log_a x} \]
Thus, we can rewrite the right side:
\[ \log_x a \cdot \log_a x = \frac{1}{\log_a x} \cdot \log_a x = 1 \]

4. **Setting Up the Equation**:
Now we have:
\[ a^x = 1 \]

5. **Finding the Value of \( x \)**:
The equation \( a^x = 1 \) holds true when \( x = 0 \) (assuming \( a \neq 0 \) and \( a \neq 1 \)). Thus:
\[ x = 0 \]

### Final Answer:
\[ x = 0 \]