If log_ab = 2; log_bc = 2 and log_3c = 3+log_3a then (a+b+c) equals (A) 90 (B) 93 (C) 102 (D) 243

log_ab=2,  log_b c = 2,  log_3c = 3+log_3a

⇒log_b/log_a  ....1. = 2,  logc/logb ....2. = 2, logc = 3log3+loga

logb/loga × logc/logb = 2×2          (Multiply 1. & 2.)

⇒logc/loga = 4 ⇒logc = 4loga

⇒4loga = 3 log3+loga       (from 3)

⇒ 3 loga 3log3

⇒ a = 3

⇒ logb = 2loga = 2log3 = log9

⇒  b = 9

logc = 2logb = 2log9 = log81

⇒  c = 81

⇒  a+b+c = 3+9381 = 93

Option B