loge(e(e(e....)^1/2)^1/2)^1/2?

• log(e(e(e...)^1/2)^1/2)^1/2
  =log(e^1/2(e(e...)^1/2)^1/4
  =log(e^1/2.e^1/4(e...)^1/8
  =log(e^1/2.e^1/4.e^1/8(e...)^1/16)
  =1/2loge+1/4loge+1/8loge+....
  since loge =1,so
  =1/2+1/4+1/8+....
  this expression is in G.P.
  so here,a=1/2 and r=1/2 and n->infinite,so using formula
  s=a/(1-r)
  s=(1/2)/(1-(1/2)
  s=1
  so the answer is 1.
  
• 11 years agoHelpfull: Yes(22) No(0)
• I think mp is right.
  
  let loge(e(e(e....)^1/2)^1/2)^1/2 = x
  =>(1/2) log (e) + log(e(e(e....)^1/2)^1/2)^1/2 = x
  => 1/2 + (1/2)loge(e(e(e....)^1/2)^1/2)^1/2 = x
  => 1/2 + (x/2) = x
  => x = 1
• 11 years agoHelpfull: Yes(13) No(2)
• loge(e(e(e....)^1/2)^1/2)^1/2 =1
• 11 years agoHelpfull: Yes(6) No(0)
• ans this question ASAP......
• 11 years agoHelpfull: Yes(0) No(8)
• @MP-can you solve it plz????//
• 11 years agoHelpfull: Yes(0) No(7)
• @mp please explain it.
• 11 years agoHelpfull: Yes(0) No(8)
• let loge(e(e(e....)^1/2)^1/2)^1/2 = x
  => log (e) + log(e(e(e....)^1/2)^1/2)^1/2 = x
  => 1 + (1/2)loge(e(e(e....)^1/2)^1/2)^1/2 = x
  => 1 + (x/2) = x
  => x = 2
• 11 years agoHelpfull: Yes(0) No(28)
• log e base e is 1 ....so 1^anything is 1..
  
• 8 years agoHelpfull: Yes(0) No(0)