If the equation `2^x(1-2^x)+4^y=2^y` is solved for `y` in terms of `x` where `x<0,` then the sum of the solution is `x(log)_2(1-2^x)` (b) `x+(log)_2(1-2^x)` `(log)_2(1-2^x)` (d) `x(log)_2(2^x+1)`

Options

A. ` x log_(2)(1-2^(x))`
B. `x+log_(2)(1-2^(x))`
C. `log_(2)(1-2^(x))`
D. `x log_(2)(2^(x)+1)`

Explanation

To solve the equation \( 2^x(1-2^x) + 4^y = 2^y \) for \( y \) in terms of \( x \), we will follow these steps:

### Step 1: Rearranging the Equation
Start with the original equation:
\[ 2^x(1 - 2^x) + 4^y = 2^y \]
We can rewrite \( 4^y \) as \( (2^2)^y = 2^{2y} \). Thus, the equation becomes:
\[ 2^x(1 - 2^x) + 2^{2y} = 2^y \]

### Step 2: Moving Terms
Rearranging gives us:
\[ 2^{2y} - 2^y + 2^x(1 - 2^x) = 0 \]
This is a quadratic equation in terms of \( 2^y \).

### Step 3: Letting \( t = 2^y \)
Let \( t = 2^y \). Then the equation becomes:
\[ t^2 - t + 2^x(1 - 2^x) = 0 \]

### Step 4: Identifying Coefficients
In the quadratic equation \( at^2 + bt + c = 0 \), we have:
- \( a = 1 \)
- \( b = -1 \)
- \( c = 2^x(1 - 2^x) \)

### Step 5: Using the Quadratic Formula
The sum of the roots \( \alpha \) and \( \beta \) (where \( \alpha \) and \( \beta \) are the roots corresponding to \( t_1 = 2^{\alpha} \) and \( t_2 = 2^{\beta} \)) can be found using the formula:
\[ \alpha + \beta = -\frac{b}{a} = -\frac{-1}{1} = 1 \]

### Step 6: Product of the Roots
The product of the roots is given by:
\[ t_1 t_2 = \frac{c}{a} = 2^x(1 - 2^x) \]
Since \( t_1 = 2^{\alpha} \) and \( t_2 = 2^{\beta} \), we have:
\[ 2^{\alpha + \beta} = 2^x(1 - 2^x) \]

### Step 7: Taking Logarithm
Taking logarithm base 2 on both sides gives:
\[ \alpha + \beta = x + \log_2(1 - 2^x) \]

### Conclusion
Thus, the sum of the solutions \( y \) in terms of \( x \) is:
\[ \alpha + \beta = x + \log_2(1 - 2^x) \]

### Final Answer
The correct option is:
(b) \( x + \log_2(1 - 2^x) \)