There are a,b real numbers which satisfy this \(\displaystyle 2lg(x-2y)=lg(x)+lg(y)\).
I need to find the values of this expression: \(\displaystyle \frac{x}{y}\)
My try: lg(x^2-4xy+4y^2)=lgxy
x^2-4xy+4y^2=xy / : y^2
(x/y)^2 -5(x/y)+4=0
I noted x/y=a and I got a^2-5a+4=0 so a=4 or a=1 so the values of x/y are 4 and 1 but the right answer is just 4.Why 1 is not good?
I need to find the values of this expression: \(\displaystyle \frac{x}{y}\)
My try: lg(x^2-4xy+4y^2)=lgxy
x^2-4xy+4y^2=xy / : y^2
(x/y)^2 -5(x/y)+4=0
I noted x/y=a and I got a^2-5a+4=0 so a=4 or a=1 so the values of x/y are 4 and 1 but the right answer is just 4.Why 1 is not good?