X>0, y = | a x- 1 | = a x- 1∑(0, +∞) and the function is strictly monotonous;

x=0，y = | a^x- 1| = 0；

x & lt0，y=| a^x- 1| = 1 - a^x ∈(0， 1)

Therefore, a> is at 1, and y=2a and y = | a x- 1 | have 1 in common.

When 0

X<0, y = | a x- 1 | = a x- 1∑(0, +∞) and the function is strict and simple;

x=0，y = | a^x- 1| = 0；

x & gt0，y=| a^x- 1| = 1 - a^x ∈(0， 1)

If y=2a and y = | a x- 1 | have two things in common, 2a∈(0, 1), then a∈(0, 1/2).

Namely 0