Description : If logx a, a^(x/2) and logbx are in GP, then x is equal to : -Maths 9th
Last Answer : (b) loga (loge a) - loga (loge b) logxa, \(a^{rac{x}{2}}\), logbx are in GP ⇒ \(\big[a^{rac{x}{2}}\big]^2\) = logxa . logbx⇒ ax = \(rac{ ext{log}\,a}{ ext{log}\,x}.rac{ ext{log}\,x}{ ext{log}\,b ... ⇒ x = loga (logba)= log a = \(rac{ ext{log}_e\,a}{ ext{log}_e\,b}\) = loga (loge a) - loga (loge b)