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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
Linear Inequalities
Solution:
$2^{2y}-2^{y}+2^{x}\left(1-2^{x}\right)=0$
put $2^{y}=t$
$t^{2}-t+2^{x}\left(1-2^{x}\right)=0$ where $t_{1}=2^{y_{1}}$ and $t_{2}=2^{y_{2}}$
$t_{1}t_{2}=2^{x}\left(1-2^{x}\right)$
$2^{y_{1}+y_{2}}=2^{x}\left(1-2^{x}\right)$
$y_{1}+y_{2}=x+log_{2}\left(1-2^{x}\right)$