What is the value of loge(e(e(e....)1/2)1/2)1/2 ?

A. 0

B. 1/3

C. 1/2

D. 1

• Let y=(e(e(e.........)^1/2)^1/2)^1/2)
  now squaring both side we get
  y^2=e(e(e(e.........)^1/2)^1/2)^1/2)
  i.e., y^2=e*y [since y=(e(e(e.........)^1/2)^1/2)^1/2)]
  therefore y=e
  taking log, log(y)=log(e)=1
  
  So answer is 1
• 10 years agoHelpfull: Yes(25) No(3)
• loge(e(e(e....)1/2)1/2)1/2
  
  (e(e(e....)1/2)1/2)1/2= e^(1/2 + 1/4 +1/8 +….) = e^1
  
  Log(e^1) = log e = 1
• 10 years agoHelpfull: Yes(10) No(2)
• 1,log(y)=log(e)=1
• 10 years agoHelpfull: Yes(1) No(1)
• loge(e(e(e....)1/2)1/2)1/2
  (e(e(e....)1/2)1/2)1/2= e^(1/2 + 1/4 +1/8 +….) = e^1
  Log(e^1) = log e = 1
• 10 years agoHelpfull: Yes(1) No(0)
• loge(e(e(e....)1/2)1/2)1/2
  (e(e(e....)1/2)1/2)1/2= e^(1/2 + 1/4 +1/8 +….) = e^1
  Log(e^1) = log e = 1
• 8 years agoHelpfull: Yes(0) No(0)