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(base x){(x^1/2)/y}+log(base y){y2/x} that was the question
  on solving this expression we will be having,
  =1/2[log(base X){X}]-log(base X){Y}+2[log(base Y){Y}]-log(base Y){X} since [log(base y){y}]=1 simlilarly [log(base x){x}]=1
  =1/2-log(base X){Y}+2-log(base Y){X}=3/2-[log(base X){Y}+log(base Y){x}]
  [log(base X){Y}+log(base Y){x}] this part can be solved by using property AM>=GM
  ([log(base X){Y}+log(base Y){x}])/2>=([log(base X){Y}+log(base Y){x}])^1/2 SO
  ([log(base X){Y}+log(base Y){x}])>=2([log(base X){Y}+log(base Y){x}])^1/2 OR WE CAN SAY ([log(base X){Y}+log(base Y){x}])>=2 WHERE 2 is it's minimum value and other values are 3,4,5...so on but in quetion we were asked to get maximum value of that expression we will have to put ([log(base X){Y}+log(base Y){x}])=2 in the expression 3/2-[log(base X){Y}+log(base Y){x}]=3/2-2=2.5-2=0.5
  if we put ([log(base X){Y}+log(base Y){x}])=3
  the we will have value=3/2-3=2.5-3=-0.5 so the maximum value of the expression would be=0.5
  
  
• 10 years agoHelpfull: Yes(8) No(8)
• log(base x){(x^1/2)/y} + log(base y){(y^2)/x)
  = 1/2[log(base X){X}]-log(base X){Y}+2[log(base Y){Y}]-log(base Y){X}
  =1/2+2 [since log(base X){Y}= -log(base Y){X}]
  =5/2 Ans..
  
• 10 years agoHelpfull: Yes(3) No(1)
• 0.5 is the ans
  
• 10 years agoHelpfull: Yes(1) No(0)
• question is z= log x ((x^0.5)/y) +log y (y^2/x)
  
  take x=y, then Z= log x (X^-1/2) + log y (y) = -1/2 +1= 0.5 (Minimum value)
  
  take x>y, let x=4 and y=2
  then Z=log 2 ( 2/2) + log4 (4) = 0+ 1= 1 (Maximum value)
  
  Hence maximum value = 1
• 9 years agoHelpfull: Yes(1) No(1)