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

Options

A. `log_(a) (log_(b)a)`
B. `(log(log a) - log(log b))/(log a)`
C. `log_(b) (log_(a)b)`
D. none of these

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**:
If three terms \( a \), \( b \), and \( c \) are in G.P., then the square of the middle term \( b \) is equal to the product of the other two terms:
\[ b^2 = a \cdot c \]
In our case, let:
- \( a = \log_x a \)
- \( b = a^{(x/2)} \)
- \( c = \log_b x \)

Thus, we have:
\[ (a^{(x/2)})^2 = \log_x a \cdot \log_b x \]

2. **Expressing Logarithms**:
We can rewrite \( \log_x a \) and \( \log_b x \) using the change of base formula:
\[ \log_x a = \frac{\log a}{\log x} \]
\[ \log_b x = \frac{\log x}{\log b} \]

3. **Substituting into the G.P. Condition**:
Substitute these expressions into the G.P. condition:
\[ \left(a^{(x/2)}\right)^2 = \left(\frac{\log a}{\log x}\right) \cdot \left(\frac{\log x}{\log b}\right) \]
Simplifying the right-hand side:
\[ a^x = \frac{\log a}{\log b} \]

4. **Taking Logarithms**:
Now, we take logarithms on both sides:
\[ \log(a^x) = \log\left(\frac{\log a}{\log b}\right) \]
This simplifies to:
\[ x \log a = \log(\log a) - \log(\log b) \]

5. **Solving for \( x \)**:
Rearranging gives us:
\[ x = \frac{\log(\log a) - \log(\log b)}{\log a} \]

6. **Final Result**:
Thus, the value of \( x \) is:
\[ x = \frac{\log(\log a) - \log(\log b)}{\log a} \]