If 2lg (x-2y) = lgx + lgY, then log2xy=______ ．

If 2lg (x-2y) = lgx + lgY, then log2xy=______ ．

∵ 2lg (x-2y) = lgx + lgY, ∵ x − 2Y ＞ 0x ＞ 0, y ＞ 0 (x − 2Y) 2 = XY, the solution is xy = 4. ∵ log2xy = log24 = 2

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

∵ lgx + lgY = 2lg (x-2y) ∵ x > 0 Y > 0x-2y > 0 ∵ x > 2Y > 0 ∵ lgx + lgY = 2lg (x-2y) ∵ lgxy = 2lg (x-2y) ∵ lgxy = LG (x-2y) ^ 2 ∵ xy = (x-2y) ^ 2 ∵ x ^ 2 + 4Y ^ 2-4xy = xYx ^ 2 + 4Y ^ 2-5xy = 0 (X-Y) (x-4y) = 0 ∵ x = y or x = 4Y ∵ x > 2; 2Y > 0 ∵ x = y

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

xy=(x-2y)^2
(x-y)(x-4y)=0
x/y=1 or 4
Original formula = 0 or 2
{0,2}