Given lgx + lgY = 2lg (x-2y), the value of log2 ^ (x / y) is

Given lgx + lgY = 2lg (x-2y), the value of log2 ^ (x / y) is

The original formula is reduced to 2lgy-2
That is, xy = (x-2y) ^ 2 is reduced to x = 4Y or x = y
Because x-2y is greater than zero, x = 4Y
So log2 (x / y) = log2 4 = 2

Lgx + lgY = 2lg (x-2y), then what is the set of values of log2 (x / y)

lgxy=lg(x-2y)^2
xy=x^2-4xy+4y^2
x^2-5xy+4y^2=0
(x-y)(x-4y)=0
x=y,x=4y
The true number is greater than 0
So x > 0, Y > 0, x-2y > 0
Then when x = Y > 0, x-2y

2lg (X-Y / 2) = lgx + lgY, find the value of X / y,

2lg(x-2y)=lgx+lgy
2lg(x-2y)=lg(x-2y)^2
lgx+lgy=lgxy
(x-2y)^2=xy
x^2-5xy+4y^2=0
x=4y or x=y
X-2y > 0 x > 0 Y > 0 (true number is positive)
x=4y
x/y=4