Find the set of all solutions of the equation `2^|y| -|2^(y -1) -1| = 2^(y -1) +1 `

Answer

To solve the equation \( 2^{|y|} - |2^{y-1} - 1| = 2^{y-1} + 1 \), we will analyze different cases based on the value of \( y \).

### Step 1: Case 1 - When \( y < 0 \)

In this case, \( |y| = -y \) and \( 2^{y-1} < 1 \) implies \( |2^{y-1} - 1| = 1 - 2^{y-1} \).

Substituting into the equation, we have:
\[ 2^{-y} - (1 - 2^{y-1}) = 2^{y-1} + 1 \]
This simplifies to:
\[ 2^{-y} + 2^{y-1} - 1 = 2^{y-1} + 1 \]
Cancelling \( 2^{y-1} \) from both sides:
\[ 2^{-y} - 1 = 1 \]
Thus:
\[ 2^{-y} = 2 \]
Taking logarithm base 2:
\[ -y = 1 \implies y = -1 \]
Since \( y = -1 \) is in the interval \( (-\infty, 0) \), it is a valid solution.

### Step 2: Case 2 - When \( 0 \leq y < 1 \)

In this case, \( |y| = y \) and \( 2^{y-1} < 1 \) implies \( |2^{y-1} - 1| = 1 - 2^{y-1} \).

Substituting into the equation gives:
\[ 2^{y} - (1 - 2^{y-1}) = 2^{y-1} + 1 \]
This simplifies to:
\[ 2^{y} + 2^{y-1} - 1 = 2^{y-1} + 1 \]
Cancelling \( 2^{y-1} \) from both sides:
\[ 2^{y} - 1 = 1 \]
Thus:
\[ 2^{y} = 2 \]
Taking logarithm base 2:
\[ y = 1 \]
Since \( y = 1 \) is not included in the interval \( [0, 1) \), we do not accept this solution.

### Step 3: Case 3 - When \( y \geq 1 \)

In this case, \( |y| = y \) and \( 2^{y-1} \geq 1 \) implies \( |2^{y-1} - 1| = 2^{y-1} - 1 \).

Substituting into the equation gives:
\[ 2^{y} - (2^{y-1} - 1) = 2^{y-1} + 1 \]
This simplifies to:
\[ 2^{y} - 2^{y-1} + 1 = 2^{y-1} + 1 \]
Cancelling \( 2^{y-1} \) from both sides:
\[ 2^{y} = 2 \cdot 2^{y-1} \]
This simplifies to:
\[ 2^{y} - 2^{y} = 0 \]
This is true for all \( y \geq 1 \).

### Final Solution

From the three cases, we find that the solutions are:
- From Case 1: \( y = -1 \)
- From Case 3: All \( y \geq 1 \)

Thus, the set of all solutions is:
\[ \{ -1 \} \cup [1, \infty) \]