Question

If $2\log (x+y)=\log x+\log y+\log 8$ and $(x-y{)}^2=kxy$ then find the value of $k$.

Answer 1

Given,

$2\log (x+y)=\log x+\log y+\log 8$

or, $\log (x+y{)}^2=\log (8xy)$

or, $(x+y{)}^2=8xy$

or, $(x+y{)}^2-4xy=4xy$

or, $(x-y{)}^2=4xy$

Comparing this with given equation we get, $k=4$.