If the equation `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)`

Answer

B

Explanation

`2^(2y) - 2^(y) + 2^(x) (1-2)^(x)) = 0`
Putting ` 2^(y) = t`, we get
` t^(2) - t+2^(x) (1-2^(x)) = 0 ," where "t_(1) = 2^(y_(1)) and t_(2) = 2^(y_(2))`
` t_(1)t_(2)=2^(x)(1-2^(x))`
` 2^(y_(1)+y_(2))=2^(x) (1-2^(x))`
` y_(1)+y_(2) = x + log _(2) (1-2^(x))`