What is the Value of log (e(e(e(---------)^1/3)^1/3)^1//3)^1/3 ?????

• Take z = e(e(e.....)^1/3)^1/3)^1/3
  Cube both side
  z^3 = e(e(e.....)^1/3)^1/3)
  z^3= e(z) (coz its infinite series)
  z^2 = e
  z = e^1/2
  
  now, log e(e(e.....)^1/3)^1/3)^1/3 = log z
  
  so, log e^1/2
  = 1/2 log e
  = 1/2 (coz base of log e is e hence = 1)
  
• 10 years agoHelpfull: Yes(10) No(2)
• Take z = e(e(e.....)^1/3)^1/3)^1/3
  z= e^(1/3+1/9+1/27+.....)
  z=e^((1/3)/(1-1/3))
  z=e^1/2
  log z=1/2 log e
  log z=1/2
• 10 years agoHelpfull: Yes(6) No(1)
• ans: 1
  log(e(e(e(-----)^1/3)^1/3)^1/3)^1/3
  (e(e(e(-----)^1/3)^1/3)^1/3)^1/3
  = e^(1/3+1/9+1/27+.....)=e^1= log(e^1)=1
  
• 10 years agoHelpfull: Yes(4) No(9)
• which answer is correct?
• 10 years agoHelpfull: Yes(3) No(0)