Solveeit Logo
Login

Question

Question: If \(\log _ { x } a , a ^ { x / 2 }\) and \(\log _ { b } x\) are in G.P., then \(x =\)...

If log⁡xa,ax/2\log _ { x } a , a ^ { x / 2 } and log⁡bx\log _ { b } x are in G.P., then x=x =

A

−log⁡(log⁡ba)- \log \left( \log _ { b } a \right)

B

−log⁡a(log⁡ab)- \log _ { a } \left( \log _ { a } b \right)

C

log⁡a(log⁡ea)−log⁡a(log⁡eb)\log _ { a } \left( \log _ { e } a \right) - \log _ { a } \left( \log _ { e } b \right)

D

log⁡a(log⁡eb)−log⁡a(log⁡ea)\log _ { a } \left( \log _ { e } b \right) - \log _ { a } \left( \log _ { e } a \right)

Answer

log⁡a(log⁡ea)−log⁡a(log⁡eb)\log _ { a } \left( \log _ { e } a \right) - \log _ { a } \left( \log _ { e } b \right)

Explanation

Solution

Obviously (ax/2)2=log⁡xa⋅log⁡bx=log⁡ba\left( a ^ { x / 2 } \right) ^ { 2 } = \log _ { x } a \cdot \log _ { b } x = \log _ { b } a

⇒\Rightarrow ax=log⁡baa ^ { x } = \log _ { b } a ⇒\Rightarrow x=log⁡a(log⁡ba)x = \log _ { a } \left( \log _ { b } a \right)

⇒\Rightarrow x=log⁡a(log⁡ealog⁡eb)=log⁡a(log⁡ea)−log⁡a(log⁡eb)x = \log _ { a } \left( \frac { \log _ { e } a } { \log _ { e } b } \right) = \log _ { a } \left( \log _ { e } a \right) - \log _ { a } \left( \log _ { e } b \right).