what is the value of log(e(e(e..)^1/2)^1/2)^1/2?
option
a) 0
b) 1
c) 1/2
d) 1/3

• Question was to find.......log(e(e(e...)^1/3)^1/3)^1/3 = ??
  Sol: Let y=log(e(e(e...)^1/3)^1/3)^1/3
  y=1/3 log e.(e(e..)^1/3)^1/3
  3y=log e + log(e(e(e...)^1/3)^1/3)^1/3
  3y= 1 + y
  2y = 1
  y = 1/2......answer......was easy naaaaa
  
• 11 years agoHelpfull: Yes(28) No(0)
• ans is a)o
• 11 years agoHelpfull: Yes(4) No(1)
• right answer is 1.
• 11 years agoHelpfull: Yes(2) No(0)
• you write a wrong question my frd..
  ques is..log(e(e(e...)^1/3)^1/3)^1/3..
  so thw write answer is..1/2
• 11 years agoHelpfull: Yes(1) No(2)
• @subhajit, saurabh, akshay : Approach plz
• 11 years agoHelpfull: Yes(1) No(0)
• b> 1 sure...
• 11 years agoHelpfull: Yes(0) No(1)
• if the base is e then ans is :1
  and if base is 10 then ans is:0
• 11 years agoHelpfull: Yes(0) No(0)