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

find the solution

• G.P. series
  ans-1/2
• 10 years agoHelpfull: Yes(8) No(1)
• Let z = (e(e(e...)^1/3)^1/3)^1/3)
  Now cubing both sides:-
  z^3 = e * (e(e...)^1/3)1/3
  or, z^3 = e * z
  or, z^3 - ez = 0
  or, z(z^2-e)=0
  
  Therefore either z=0 or z = e^1/2
  Putting z=e^1/2 :-
  log z base e => log e^1/2 base e => 1/2 *log e base e=> 1/2 (ANS)
• 10 years agoHelpfull: Yes(6) No(0)
• it will form a G.P. (1/3)+(1/9)+(1/27)...
  so sum=(1/3)/(1 - 1/3)=1/2
  loge^1/2=1/2.
  
• 10 years agoHelpfull: Yes(3) No(1)
• Let log(e(e(e(e(e(......)1/3)1/3)1/3)1/3)1/3)1/3=X
  Base is e.So 1/3[ln e+ ln(e(e(e(e(......)1/3)1/3)1/3)1/3)1/3)]=X
  1/3[1+X]=x
  3X=X+1
  2X=1
  X=1/2,ANS
• 10 years agoHelpfull: Yes(3) No(0)