# Question #b91f5

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:
$x - 1 \ge 0 \Rightarrow x \ge 1$

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

${2}^{x} - 1 \ge 0 \Rightarrow {2}^{x} \ge 1 \Rightarrow {2}^{x} \ge {2}^{0} \Rightarrow x \ge 0$.

So there are three intervals: $x \le 0$; $0 < x \le 1$; $x > 1$.

• For $x \ge 0$ the arguments are both negative, so the absolute value of them are their opposite:

${2}^{- x + 1} - {2}^{x} = - {2}^{x} + 1 + 1 \Rightarrow {2}^{- x + 1} = 2 \Rightarrow - x + 1 = 1 \Rightarrow x = 0$

(remember that $2 = {2}^{1}$).

• For $0 < x \le 1$ the first argument is negative and the second is positive.

${2}^{- x + 1} - {2}^{x} = {2}^{x} - 1 + 1 \Rightarrow {2}^{- x + 1} = {2}^{x} + {2}^{x} \Rightarrow {2}^{- x + 1} = {2.2}^{x} \Rightarrow {2}^{- x + 1} = {2}^{x + 1} \Rightarrow - x + 1 = x + 1 \Rightarrow x = 0$

(not acceptable, out of interval)

• For $x > 1$ the arguments are both positive:
${2}^{x - 1} - {2}^{x} = {2}^{x} - 1 + 1 \Rightarrow {2}^{x - 1} = {2.2}^{x} \Rightarrow {2}^{x - 1} = {2}^{x + 1} \Rightarrow x - 1 = x + 1 \Rightarrow - 1 = 1$

Impossible.