If the equation 2x + 4y = 2y + 4x is solved for y in terms of x where x

Question

If the equation 2x + 4y = 2y + 4x is solved for y in terms of x where x < 0, then the sum of the solutions is (A) x log2 (1 - 2x) (B) x+ log2 (1 - 2x) (C) log2(1 - 2x) (D) x log2(2x + 1)

Tardigrade Admin · Accepted Answer

22y-2y+2x(1-2x)=0 put 2y=t t2-t+2x(1-2x)=0 where t1=2y1 and t2=2y2 t1t2=2x(1-2x) 2y1+y2=2x(1-2x) y1+y2=x+log2(1-2x)

Q. 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 solutions is

$x l o g_{2} (1 - 2^{x})$

$x + l o g_{2} (1 - 2^{x})$

$l o g_{2} (1 - 2^{x})$

$x l o g_{2} (2^{x} + 1)$

$2^{2 y} - 2^{y} + 2^{x} (1 - 2^{x}) = 0$
put $2^{y} = t$
$t^{2} - t + 2^{x} (1 - 2^{x}) = 0$ where $t_{1} = 2^{y_{1}}$ and $t_{2} = 2^{y_{2}}$
$t_{1} t_{2} = 2^{x} (1 - 2^{x})$
$2^{y_{1} + y_{2}} = 2^{x} (1 - 2^{x})$
$y_{1} + y_{2} = x + l o g_{2} (1 - 2^{x})$

Q. If the equation 2x+4y=2y+4x is solved for y in terms of x where x<0, then the sum of the solutions is

xlog2​(1−2x)

x+log2​(1−2x)

log2​(1−2x)

xlog2​(2x+1)