calculate (e(e(e...)^1/3)^1/3)^1/3...??? this was the question asked in 2nd august exam

• let p=e(e(e...)^1/3)^1/3)^1/3......
  then,
  p=(e*p)^1/3 (becoz it is uncountable series so we can write like that)
  =>p=e^1/2 .............ans
• 9 years agoHelpfull: Yes(8) No(0)
• (1/3)/(1-1/3)=1/2
  
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• sorry i forgot to mention log in front of the expression
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• let p=(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3)^1/3
  two side log
  logp=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
  (e(e(e(e.........)^1/3)^1/3)^1/3)^1/3)^1/3=e^1/2
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• the question was log(e(e(e...)^1/3)^1/3)^1/3) base e was there...so ans is 1...anything to log e to base e is 1
  
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• Take total sum as x and apply log base e both sides
  ln(x)=1/3+(1/3)^2 ....==1/2
  x=e^1/2
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