What is the Value of log (e(e(e(---------)^1/2)^1/2)^1/2)^1/2 ?????

Here log is base e.

• Let z=(e(e(e.........)^1/2)^1/2)^1/2)
  now squaring both side we get
  z^2=e(e(e(e.........)^1/2)^1/2)^1/2)
  i.e., z^2=e*z [since z=(e(e(e.........)^1/2)^1/2)^1/2)]
  therefore z=e
  taking log, log(z)=log(e)=1(asz=e^log(e)=1 & base is e)
  
  So answer is 1
• 9 years agoHelpfull: Yes(22) No(2)
• ans is 1..........
  log(e)^1/2+1/4+1/8...
  log(e)^1
  =1
• 9 years agoHelpfull: Yes(2) No(0)
• ans 1 but equation is
  
  e^y2=ey for y=1 is satisfy
  
• 9 years agoHelpfull: Yes(0) No(4)
• suppose y=log(e(e(e.........)^1/2)^1/2)^1/2)^1/2
  then y=1/2(log(e(e(e.........)^1/2)^1/2)^1/2)
  y=1/2(y)
  2y=y
  2y-y=0
  y=0
  so the result is 0.
• 9 years agoHelpfull: Yes(0) No(2)
• since :
  log e base e =1;
  let z=log(e(e(e(...)^1/2)^1/2)^1/2
  z^2=loge(e(e..)^1/2)^1/2
  =1/2+ 1/4+1/8.....up to infinity
  now this is a GP series
  
  so, sum=1/2/(1-1/2)=1
  z^2=1
  z=+1 or -1;
• 9 years agoHelpfull: Yes(0) No(0)