log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3)^1/3

opt
a)0 b)1 c)1/3 d) not remember

• ans-1/2
  log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3)^1/3
  loge^1/3+1/9+1/27+....
  since 1/3+1/9+1/27+.... is in GP so
  sum= (1/3)/(1-1/3)
  1/2
  so
  loge^1/2
  =1/2
• 9 years agoHelpfull: Yes(17) No(1)
• let e(e(e(e(e...)^1/3)^1/3)^1/3....=x
  then (e*x)^1/3=x
  e^1/3=x^2/3
  e=x^2
  e^1/2=x
  taking log both side then log x=1/2log e
  then log x=1/2
  answer is 1/2.
  
• 9 years agoHelpfull: Yes(8) No(0)
• log(e(e(e(e....)^1/3)^1/3)^1/3)^1/3
  lets take from inner most 3 values as
  (e(e(e^1/3)^1/3)^1/3)
  e^1/3(e(e^1/3)^1/9)
  e^1/3*e^1/9*e^1/27.....
  e^(1/3+1/9+1/27+......)
  e^((1/3)/(1-1/3)) as from GP infinite sum....
  e^(1/2)
  now loge^1/2 with base e
  1/2 is answer.......
  if log(e(e(e...)^1/2)^1/2)^1/2 with base e will give.. 1
  e^(1/2)(1-(1/2))
  e^1
  loge^1 with base e gives 1
  
• 9 years agoHelpfull: Yes(5) No(0)
• let log(e(e(e....)^1/3)^1/3)^1/3 = x
  => x = (1/3)*[(loge) + log(e(e(e....)^1/3)^1/3)^1/3 ]
  => x = (1/3)*(1 + x)
  => 3x = 1 + x
  => x = 1/2
• 9 years agoHelpfull: Yes(4) No(0)
• Let, loge(e(e(e...)1/3)1/3))1/3 = x
  => (1/3)loge(e(e(e...)1/3)1/3)) = x
  => (1/3)[logee + loge(e(e(e...)1/3)1/3))1/3] = x
  => (1/3)[1 + x] = x
  => 1 + x = 3x
  => x = 1/2
• 9 years agoHelpfull: Yes(3) No(0)
• ans = 1/d
  log(e(e(e.....)^1/3)^1/3)^1/3) = z
  1/3 log(e(e(e(.....)^1/3^1/3) = z
  1/3 [loge + log(e(e(e.....)^1/3)^1/3)^1/3)] = z
  1/3 [1 + z] = z
  z = 1/2
• 9 years agoHelpfull: Yes(1) No(0)
• Ans is b)1.
  bcoz log e^anything = 1.
• 9 years agoHelpfull: Yes(0) No(8)
• answer is 2 ...this question is very common and repeated many a time in elitmus
• 9 years agoHelpfull: Yes(0) No(0)
• log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3=1/3(loge+1/3(log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3))
  let log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3=p
  logp=1/3(loge+1/3(logp))
  logp -1/9logp=1/3*1
  8/9logp=1/3
  p=e^(3/8)
  log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3=logp=3/8
  
• 9 years agoHelpfull: Yes(0) No(2)
• let logp=log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3)^1/3
  logp=1/3(1+log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3)^1/3))
  logp=1/3(1+logp)
  logp=1/2
  log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3)^1/3=1/2
  
• 9 years agoHelpfull: Yes(0) No(0)