Question

Find the set of all solutions of the equation $2^{\left|y\right|}-|2^{y-1}-1|=2^{y-1}+1$, the solution includes

• A. $y=-1$
• B. $y>1$
• C. $y=1$
• D. $y<1$

Answer 1

The correct option is A $y=-1$

Here, $2^{\left|y\right|}-|2^{y-1}-1|=2^{y-1}+1$
We know to define modulus, we have three cases as
Case I: y< 0
$\Rightarrow$ $2^{-y}+(2^{y-1}-1)=2^{y-1}+1$
$\Rightarrow$ $2^{-y}=2^1$ (as when y< 0 |y|=-y and $|2^{y-1}-1|=-(2^{y-1}-1)$)
Hence, y=-1, which is true when y< 0 (i)
Case II: $0\le y<1$
$\Rightarrow$ $2^y+(2^{y-1}-1)=2^{y-1}+1$
$\Rightarrow$ $2^y=2$ (as when $0\le y<1$ |y|=-y and $|2^{y-1}-1|=-(2^{y-1}-1)$)
$\Rightarrow$ $y=1$, which shows no solution as,
$0\le y<1$ (ii)
Case III: $y\ge 1$
$\Rightarrow$ $2^y-(2^{y-1}-1)=2^{y-1}+1$
$\Rightarrow$ $2^y=2^{y-1}+2^{y-1}$
$\Rightarrow$ $2^y={2.2}^{y-1}$ (as when $y\ge 0$ |y|=y and $|2^{y-1}-1|=(2^{y-1}-1)$)
$\Rightarrow$ $2^y=2^y$, which is an identity therefor, it is true $\forall y\ge 1$ (iii)
Hence, from Eqs. (i), (ii), (iii) the solution of set is $\{y:y\ge 1\cup y=-1\}$.