Multiply by e x at the same time

2e^x<； e^2x+ 1<； 2e^x(b- 1)

Let e x = t

2t & ltt？ + 1 & lt； 2(b- 1) tons

T on the left? -2t+ 1 & gt； 0 (t- 1)？ & gt0 So t≠ 1 is e x ≠ 1, x ≠ 0.

T on the right? -2(b- 1)t+ 1 & lt； 0

△=4(b- 1)？ -4=4b？ -8b①

This problem is transformed into the solution of quadratic function about t under t≥0.

Formula ① is discussed below.

∫4b(b-2)= 0, b=0 or 2

When 0

When b=0, or 2, delta = 0, which has an intersection with the X axis. B=0, the intersection point is t=- 1 (excluding), b = 2, t = 1. E x = 1, x = 0 and the left formula x≦ are excluded. So b≠0 and 2.

When b>2 or b<0, △ > 0, which has two unequal roots with the x axis. And because its intersection with the Y axis is (0, 1), its intersection with the X axis can only be on one side of the X axis. The symmetry axis is (b- 1),

When b>, it is on the right side of the x axis. Explain that these two intersections are on the right side of the X axis. at present

T∈((b- 1)-√(4b？ -8b)/2，(b- 1)+√(4b？ -8b)/2)=(b- 1-√(b？ -2b)、b- 1+√(b？ -2b))。

X belongs to (ln[b- 1-√(b? -2b)]，ln[b- 1+√(b？ -2b)])

When b<0 is on the left side of the X axis, it means that the two intersections are on the left side of the X axis. Thus, this quadratic function is >: 0, so it is.

To sum up, we can draw a conclusion. b & gt2，x∈(ln[b- 1-√(b？ -2b)]，ln[b- 1+√(b？ -2b)])