log(e(e(e...)^1/3)^1/3)^1/3 what will be its value?

• let us assume [log(e(e(e..)^1/3)^1/3)^1/3] = x
  now as log m^n = n log(m)
  we can write it as 1/3[log(e(e(e...)^1/3)^1/3)] = x
  now 1/3[log(e)+log(e(e..)^1/3)^1/3)] = x
  now we can see that it is again our question so we can replace it as x
  so 1/3(1+x)= x
  => 3x = 1+x
  => x= 1/2
• 10 years agoHelpfull: Yes(21) No(4)
• e(e(e..)^1/3)^1/3)^1/3=e^(1/3+1/9++1/27+......infinte series)=e^(1/3)/(1-1/3)=e^1/2
  now,
  loge^1/2=1/2loge=1/2*1=1/2 since loge base e=1
• 10 years agoHelpfull: Yes(13) No(0)
• Let x=(e(e(e.........)^1/3)^1/3)^1/3)
  now squaring both side we get
  x^3=e(e(e(e.........)^1/3)^1/3)^1/3)
  i.e., x^3=e*x [since x=(e(e(e.........)^1/3)^1/3)^1/3)]
  therefore x=e [since x can not be zero]
  Finally, log(x)=log(e^1/2)=1/2
  
• 10 years agoHelpfull: Yes(3) No(0)