1.

If logxa, ax/2 and logbx are in GP, then x is(a) loga (logba) (b) loga (loge a) + loga(loge b) (c) – loga (logab) (d) loga (loge b) – loga (loge a)

Answer»

(a) loga (logba) 

If logx a, ax/2 and logbx are in GP, then (ax/2)2  = (logbx) × (logx a) 

⇒ ax = logba ⇒ log ax = log (logba) ⇒ x log a = log (logba) ⇒ x loga a = loga (logba) 

x = loga (logba).



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