Question

If logx a, ax/2 and logb x are in GP, ​​then x is

• A. loga(logb a)
• B. loga (loge a) + loga (loge b)
• C. -loga (loga b)
• D. loga (loge b)-loga (loge a)

Answer 1

Answer: (A)

loga(logb a)

Answer 2

The correct option is (A): loga(logb 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:

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

$x=\log_a(\log_ba)$