If x>=y and y>1 , what is the maximum value of log (base x) -(x/y) + log (base y)x^2/y?
option
a) -0.5
b) 0
c) 0.5
d) 1

• log -(x/y) + log ( x^2)/y => - log x/y + 2log (x/y)
  x y x y
  
  By Formula (i) - log x/y = log y/x (ii) log y/x = log y - log x
  x x x x x
  
  log y - log x + 2{log x - log y }
  x x y y
  
  log y -1 + 2log x - 2 ==> log y + 2log x -3
  x y x y
  
  by Sub x &y as 2 by Question Condition---> Ans 1+2-3 = 0
• 8 years agoHelpfull: Yes(4) No(3)
• question wrong .
• 6 years agoHelpfull: Yes(1) No(0)
• option b
  log (base x) -(x/y) + log (base y)x^2/y
  => -{log (base x) x - log (base y) x}+ 2{log (base y) - 1}
  => -1 + log(base x) y + 2 log (base y) x - 2
  =>{log (base x) y+2 log (base y) x} - 3
  y>1 and x>=y thus if y =2 then x=y
  then
  {log (base x) y+2 log (base y) x} - 3= 1+2-3
  = 0
  thus opt (b)
  
  
  
• 8 years agoHelpfull: Yes(0) No(5)
• log(base x)(-x/y)+ log(base y)((x^2)/y)
  = -log(base x)(x/y)+ 2*log(base y)(x/y)
  = -( log(base x)(x)- log(base x)(y)) + 2*( log(base y)(x)- log(base y)(y))
  = -( 1- log(base x)(y)) + 2*( log(base y)(x)- 2)
  = -3 + log(base x)(y) + 2*log(base y)(x)
  = -3 + (log y/ log x) + (log (base y)(x^2))
  = -3 + (log y/ log x) + (log (x^2)/ log y)
  = -3 + ( log y - log x) + (2* log x - log Y)
  = -3 + ( log y/ log y) + 2*(log x/ log x)
  = -3 + 1 + 2
  = 0
• 6 years agoHelpfull: Yes(0) No(1)