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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) |
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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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