Question Number 114597 by O Predador last updated on 19/Sep/20

      What  is  the  number  value  of   f[((log((( (√(x  )) −   1 )/(x   −   1)))  ))^(1/3) ]  =  (√((√(x ))   +   x ))  for  f(−1)?        a) 0,1     b) 27     c) 81     d) 10     e) 12

Answered by Olaf last updated on 19/Sep/20

((log((((√x)−1)/(x−1)))))^(1/3)  = −1  log((((√x)−1)/(x−1))) = −1  (((√x)−1)/(x−1)) = 10^(−1)  =(1/(10))  (√x)−1 = (1/(10))(x−1)  x−10(√x)+9 = 0  ((√x)−1)((√x)−9) = 0  (√x) = 1 : impossible  otherwise (((√x)−1)/(x−1)) is not defined  (√x) = 9 ⇒ (√((√x)+x)) = (√(90)) = 3(√(10))  Sorry but I find 3(√(10))  I supposed log is log_(10)  and not ln

Commented byO Predador last updated on 20/Sep/20

 Thank you!

Answered by floor(10²Eta[1]) last updated on 20/Sep/20

f(((log((((√x)−1)/(x−1)))))^(1/3) )=(√((√x)+x))  ((log((((√x)−1)/(x−1)))))^(1/3) =−1⇒log((((√x)−1)/(x−1)))=−1  (((√x)−1)/(x−1))=10^(−1) =(1/(10))⇒10(√x)−10=x−1  10(√x)=x+9  100x=x^2 +18x+81  x^2 −82x+81=0  x=((82±80)/2)⇒x=1 or 81⇒x=81 (because x≠1)  x=81⇒  f(−1)=(√((√(81))+81))=(√(90))  now if log is ln the logic is the same and  the answer is (√(e^2 −e))

Answered by 1549442205PVT last updated on 20/Sep/20

We need find x>0, x ≠1such that   log(((√x)−1)/(x−1))=−1⇔log(((√x)−1)/(((√x)−1)((√x)+1)))=−1  ⇔log(1/( (√x)+1))=−1⇔−log((√x)+1)=−1  •If logx=log_(10) x then  ⇔log((√x)+1)=1⇒(√x)+1=10  ⇔(√x)=9⇔x=81  ⇒f(−1)=(√(9+81))=(√(90))  •If logx = lnx then (√x)+1=e⇒(√x)=e−1  ⇒x=e^2 −2e+1⇒f(−1)=(√( e^2 −e))  Available answer is false!