If 2lg (X-Y / 2 = lgx + lgY), then x / y? If 2lg (X-Y) / 2 = lgx + lgY, then x / y?

If 2lg (X-Y / 2 = lgx + lgY), then x / y? If 2lg (X-Y) / 2 = lgx + lgY, then x / y?

2lg (x-2y) = lgx + lgylg (x-2y) ^ 2 = lgxy (x-2y) ^ 2 = xYx ^ 2 + 4Y ^ 2-4xy-xy = 0x ^ 2-5xy + 4Y ^ 2 = 0 (x / y) ^ 2-5 (x / y) + 4 = 0 (x / y-4) (x / Y-1) = 0, so: X / y = 4 or X / y = 1

Given 2lg (x-2y) = lgx + lgY, find the value of X / y
I think it should be on the left side to change the square of 2 to LG (x-2y) ^ 2, and on the right side to LG (XY) with the formula. Then (x-2y) ^ 2 = XY. But at the beginning, I can't go on. I hope someone can help me point out my mistakes.)
The answer is "x-2y > 0, x > 0, Y > 0", but why? Why can't I see that?
But if you want to give up 1, you still need to use x-2y > 0, x > 0, Y > 0.

If we expand the above formula, we can get x square - 5xy + 4Y square = 0, and then factorize it into (X-Y) (x-4y) = 0, so x = y or x = 4Y, then x / y = 1 or 4. Yes, the problem is to set a trap in it. Most people think that they have finished the previous part only. In fact, it is not the case of x = y because x-2y > 0, that is, x > 2Y

Domain of function y = log2 (16-4 ^ x)

The definition field of function y = log2 (16-4 ^ x) is 16-4 ^ x > 0, and the solution is X