Question #b91f5

1 Answer
Jan 29, 2015

The solution is #0#, that is the only solution.

The argument of the first absolute value (i.e. #x-1#) changes the signum in #1#, in fact:
#x-1>=0rArrx>=1#

The argument of the second absolute value (i.e. #2^x-1#), change the signum in #0#, in fact:

#2^x-1>=0rArr2^x>=1rArr2^x>=2^0rArrx>=0#.

So there are three intervals: #x<=0#; #0<x<=1#; #x>1#.

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

#2^(-x+1)-2^x=-2^x+1+1rArr2^(-x+1)=2rArr-x+1=1rArrx=0#

(remember that #2=2^1#).

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

#2^(-x+1)-2^x=2^x-1+1rArr2^(-x+1)=2^x+2^xrArr2^(-x+1)=2.2^xrArr2^(-x+1)=2^(x+1)rArr-x+1=x+1rArrx=0#

(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.