∴a<； =-1, and the number of solutions is 0.

When a>| x | = log at-1? (1+a), namely x = log? ( 1+a)

So, there are two situations.

Second, 2 | x |- 1-a = 0, that is, 2 | x |-( 1+a) = 0, 1+a > 0, so it is equivalent to shifting the image of 2 | x downward (1+a). Incremental from 0, incremental from X.