If the zeros of the equation INX = 8-2x x0 ∈ (k, K + 1) k ∈ Z, then K=

On the one hand, the images of y = LNX and y = 8-2x were observed,
On the other hand, LNX - (8-2x) = f (x), f (1) = - 6, f (2) = ln2-4

Let x0 be the solution of the equation INX + x = 4, then x0 is in which of the following intervals
A （3,4）
B （2,3）
C （1,2）
D （0,1）

LNX + x = 4lnx = - x + 4, Let f (x) = LNX g (x) = - x + 4, so as long as we get the abscissa of the intersection of F (x) g (x), that is, x0. From the image, we can see that x0 must be larger than 1. The following is the comparison. When x0 = 2, f (2) = LN2 g (2) = 2 = lne ^ 2. Obviously, 2 is smaller than e ^ 2, so LN2 is smaller than 2, so x0 is larger than 2, and then go down to 3

Let x0 be the approximate solution of the equation LNX + 2x-6 = 0, and X belong to (a, b), B-A = 1, then find the value of a and B

Note that E = 2.73, LN2 < 1, Ln3 > 1, we have: F (2) = LN2 - 2 < 0, f (3) = Ln3 > 0. Because f (x) is continuous, there must be a root in the interval (2,3). So we can take a = 2, B = 3

If the solution of equation lnx-6 + 2x = 0 is x0, then the largest integer solution of inequality x ≤ x0 is______ ．

∵ equation lnx-6 + 2x = 0, ∵ equation LNX = 6-2x. Draw the images of two functions y = 6-2x, y = LNX respectively: the abscissa of the intersection point of two function images is known from the graph, that is, the solution x0 ∈ (2, 3) of equation lnx-6 + 2x = 0. The largest integer solution of inequality x ≤ x0 is 2, so the answer is 2

If the zeros of the equation INX = 8-2x x0 ∈ (k, K + 1) k ∈ Z, then K=