log(e(e(e...)^1/2)^1/2)^1/2 what will be its value?

• Let x=e(e(e...)^1/2)^1/2)^1/2
  squaring both side we get
  x^2=e(e(e...)^1/2)^1/2)
  x^2=e*x (since x=e(e(e...)^1/2)^1/2)^1/2)
  x(x-e)=0
  x=e
  so logx=1
  
• 11 years agoHelpfull: Yes(33) No(3)
• y=log(e(e(e...)^1/2)^1/2)^1/2
  y=1/2 log(e(e(e...)^1/2)^1/2)^1/2
  2y=loge+y
  y=loge
  y=1
• 11 years agoHelpfull: Yes(19) No(4)
• it will come like 1/2+1/2+1/2+..... and so on..so sum of ths series up to unfinit series will be 1.
• 11 years agoHelpfull: Yes(2) No(1)
• let A=log(e(e(e.....)^1/2)^1/2)^1/2
  take log both side
  log A=1/2 log (log(e(e(e.....)^1/2)^1/2)^1/2)
  logA=1/2log(A)(Since log(e(e(e.....)^1/2)^1/2)^1/2 can be rewrite as A )
  1/2logA=0
  A=1
  
• 9 years agoHelpfull: Yes(0) No(0)