If logxa, \(a^{\frac{x}{2}}\) and logbx are in GP, then x is equal to

1.	If logxa, \(a^{\frac{x}{2}}\) and logbx are in GP, then x is equal to :(a) loga (logba) (b) loga (loge a) – loga (loge b) (c) –loga (logab) (d) both (a) and (b)
Answer» (b) log_a (log_e a) – log_a (log_e b) log_xa, \(a^{\frac{x}{2}}\), log_bx are in GP ⇒ \(\big[a^{\frac{x}{2}}\big]^2\) = log_xa . log_bx ⇒ a^x = \(\frac{\text{log}\,a}{\text{log}\,x}.\frac{\text{log}\,x}{\text{log}\,b}\) ⇒ a^x = \(\frac{\text{log}\,a}{\text{log}\,b} = \text{log}_b\,a\) ⇒ x = log_a (log_ba) = log a = \(\frac{\text{log}_e\,a}{\text{log}_e\,b}\) = log_a (log_e a) – log_a (log_e b)

If logxa, \(a^{\frac{x}{2}}\) and logbx are in GP, then x is equal to :(a) loga (logba) (b) loga (loge a) – loga (loge b) (c) –loga (logab) (d) both (a) and (b)

Answer»

(b) log_a (log_e a) – log_a (log_e b)

log_xa, \(a^{\frac{x}{2}}\), log_bx are in GP ⇒ \(\big[a^{\frac{x}{2}}\big]^2\) = log_xa . log_bx

⇒ a^x = \(\frac{\text{log}\,a}{\text{log}\,x}.\frac{\text{log}\,x}{\text{log}\,b}\)

⇒ a^x = \(\frac{\text{log}\,a}{\text{log}\,b} = \text{log}_b\,a\)

⇒ x = log_a (log_ba)

= log a = \(\frac{\text{log}_e\,a}{\text{log}_e\,b}\) = log_a (log_e a) – log_a (log_e b)