logX(base e) + logYcube(base e) = a
logXsquare(base e) + logY(base e) = b
Find the value of Y.

• y=e^((2a-b)/5)
• 10 years agoHelpfull: Yes(16) No(2)
• logX(base e) + logYcube(base e) = a
  or, log{(X*y^3)base e}=a
  which can be rewritten as e^a =X*Y^3 ----- (1)
  
  logXsquare(base e) + logY(base e) = b
  or, log{(X^2*y)base e}=b
  which can be rewritten as E^b = X^2*Y----- (2)
  
  now from expression 1:-
  X = e^a / Y^3
  Substituting this value of X in expression 2:-
  e^b = ((e^a/Y^3)^2)*Y
  = e^2a / Y^5
  or Y^5 = e^2a/e^b
  therefore, Y = e ^((2a-b)/5)
  
  
  
  
  
• 10 years agoHelpfull: Yes(8) No(2)
• i think it should be the ans, y=e^((2a-b))
  can u explain @mukesh nayal how is it y=e^((2a-b)/5)
• 10 years agoHelpfull: Yes(1) No(0)
• i think it should be
  y=e^(6b-3a)/5
• 10 years agoHelpfull: Yes(0) No(2)
• this question is not able for elitmus
• 10 years agoHelpfull: Yes(0) No(0)