If the equation 2^x+4^y=2^y + 4^x is so...

If the equation `2^x+4^y=2^y + 4^x` 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)`

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