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in Logarithm by (24.0k points)
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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)

1 Answer

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by (23.5k points)
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Best answer

(b) loga (loge a) – loga (loge b) 

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

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

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

⇒ x = loga (logba)

= log a = \(\frac{\text{log}_e\,a}{\text{log}_e\,b}\) = loga (loge a) – loga (loge b) 

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