If `log_(10) (x^3+y^3)-log_(10) (x^2+y^2-xy) <=2` then the minimum value of `xy` for all `x>=0, y>=0` is

If log_(10)(x^(3)+y^(3))-log_(10)(x^(2)+y^(2)-xy)<=2, where x,y are positive real number,then find the maximum value of xy.

If log_(10)|x^3 + y^3|-log_(10) |x^2-xy + y^2|+log_(10)|x^3-y^3|-log_(10)|x^2+xy+y^2|=log_(10)221 . wherex, y are integers , then (i) if x=111 then y can be: (ii) if y=2 then value of x can be:

If log_(10)|x^3 + y^3|-log_(10) |x^2-xy + y^2|+log_(10)|x^3-y^3|-log_(10)|x^2+xy+y^2|=log_(10)221. wherex, y are integers , then (i) if x=111 then y can be: (ii) if y=2then value of x can be:

If log_(10)|x^(3)+y^(3)|-log_(10)|x^(2)-xy+y^(2)|+log_(10)|x^(3)-y^(3)|-log_(10)|x^(2)+xy+y^(2)|=log_(10)221 . Where x, y are integers, then Q. If y=2 , then value of x can be :

lf_(log_(10)(x^(3)+y^(3))-log_(10)(x^(2)-xy+y^(2))+log_(10)(x^(3)-y^(3))-log_(10)(x^(2)+xy+y^(2)))=log_(10)221 then

If 3 log_(10) (x^(2) y) = 4 + 2 log_(10) x - log_(10) y , where x and y are both + ve, and x - y = 2 sqrt(6) , then the value of x is

If log_(10)x+log_(10)y=2, x-y=15 then :

If log_(10)x+log_(10)y=2, x-y=15 then :

If log_(10) x+log_(10) y= 3 and log_(10) x-log_(10) y=1, then x and y are respectively a. 10 and 100 b. 100 and 10 c. 1000 and 100 d. 100 and 1000

If "log"_(10) x =y, " then log"_(10^(3))x^(2) equals