what is value of log(e(e(e....)^1/2)^1/2)^1/2? Base of the log is e
(a)0(b)1/3(c)1/2(d)1

• d) 1
  
  let log(e(e(e....)^1/2)^1/2)^1/2 = x
  => x = (1/2)*[(loge) + log(e(e(e....)^1/2)^1/2)^1/2 ]
  => x = (1/2)*(1 + x)
  => 2x = 1 + x
  => x = 1
• 11 years agoHelpfull: Yes(34) No(4)