Question

If log_x a, a^x/2 and log_b x are in GP, ​​then x is

• 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 1

The Correct Option is A

The correct option is (A): log_a(log_b a)
To explain why the correct option is \( \log_a(\log_b a) \):

### Given:
We have that \( \log_x a, a^{x/2}, \log_b x \) are in geometric progression (GP). By the property of GP, we can express it as:

\[(a^{x/2})^2 = \log_x a \cdot \log_b x\]

### Step 1: Using Logarithmic Identities

Using the change of base formula:
\( \log_x a = \frac{\log a}{\log x} \)
\( \log_b x = \frac{\log x}{\log b} \)

Substituting these into the GP condition gives:

\[(a^{x/2})^2 = \frac{\log a}{\log x} \cdot \frac{\log x}{\log b}\]
Simplifying further, we have:

\[(a^{x}) = \frac{\log a}{\log b}\]
\[(a^{x})=\log_ba\]

\[x=\log_a(\log_ba)\]