# Question #b91f5

##### 1 Answer

Jan 29, 2015

The solution is

The argument of the first absolute value (i.e.

The argument of the second absolute value (i.e.

So there are three intervals:

- For
#x>=0# the arguments are both negative, so the absolute value of them are their opposite:

(remember that

- For
#0<x<=1# the first argument is negative and the second is positive.

(not acceptable, out of interval)

- For
#x>1# the arguments are both positive:

#2^(x-1)-2^x=2^x-1+1rArr2^(x-1)=2.2^xrArr2^(x-1)=2^(x+1)rArrx-1=x+1rArr-1=1#

Impossible.