Question

Question #b91f5

Answer 1

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.