Q. If `log_x a, a^(x/2)` and `log_b x` are in G.P. then x is equal to (1) `log_a(log_b a)` (2) `log_a(log_e a)+log_a log_b b` (3)`-log_a(log_a b)` (4) none of these

Options

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

Answer

To solve the problem, we need to determine the value of \( x \) given that \( \log_x a \), \( a^{(x/2)} \), and \( \log_b x \) 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 condition is:
\[ B^2 = A \cdot C \]
Here, let \( A = \log_x a \), \( B = a^{(x/2)} \), and \( C = \log_b x \).

2. **Setting up the Equation**:
According to the G.P. condition:
\[ (a^{(x/2)})^2 = \log_x a \cdot \log_b x \]
This simplifies to:
\[ a^x = \log_x a \cdot \log_b x \]

3. **Using Logarithm Properties**:
We can express \( \log_x a \) using the change of base formula:
\[ \log_x a = \frac{\log_e a}{\log_e x} \]
and \( \log_b x \) as:
\[ \log_b x = \frac{\log_e x}{\log_e b} \]
Substituting these into our equation gives:
\[ a^x = \left(\frac{\log_e a}{\log_e x}\right) \cdot \left(\frac{\log_e x}{\log_e b}\right) \]
This simplifies to:
\[ a^x = \frac{\log_e a}{\log_e b} \]

4. **Taking Logarithm on Both Sides**:
Taking the logarithm of both sides, we have:
\[ \log_e(a^x) = \log_e\left(\frac{\log_e a}{\log_e b}\right) \]
Using the property of logarithms, this becomes:
\[ x \log_e a = \log_e(\log_e a) - \log_e(\log_e b) \]

5. **Solving for \( x \)**:
Rearranging gives:
\[ x = \frac{\log_e(\log_e a) - \log_e(\log_e b)}{\log_e a} \]
This can be rewritten using the change of base formula:
\[ x = \log_a\left(\frac{\log_e a}{\log_e b}\right) \]
This is equivalent to:
\[ x = \log_a(\log_b a) \]

### Conclusion:
Thus, the value of \( x \) is:
\[ x = \log_a(\log_b a) \]