WHAT WILL BE THE VALUE OF
log e(e(e....)^1/2)^1/2)^1/2).....log base is e

• e(e(e....)^1/2)^1/2)^1/2 = e ^ (1/2 + 1/4 + 1/8 +....)
  it is in G.P(Geometric progression) then sum to infinite terms = (a)/(1-r)
  (1/2 + 1/4 + 1/8 +....) = (1/2)/(1-1/2) = 1
  so e ^ 1
  log(e^1) base e = 1
  answer is 1;
• 12 years agoHelpfull: Yes(10) No(2)
• e(e(e....)^1/2)^1/2)^1/2)=x
  or,ex=x^2
  0r,x=e
  so ans will be 1
• 13 years agoHelpfull: Yes(8) No(16)
• let (e(e(e....)^1/2)^1/2)^1/2)=x then
  we can write
  x=(ex)^1/2 why....?
  x=e^1/2*x^1/2
  or
  x^1/2=e^1/2
  or
  x=e
  so rewrite log (e(e(e....)^1/2)^1/2)^1/2) as log e with base e
  
  which will be 1.
• 10 years agoHelpfull: Yes(1) No(1)