If the equation $2^{\mathrm{x}}+4^{\mathrm{y}}=2^{\mathrm{y}}+4^{\mathrm{x}}$ is solved for y in terms of x, is

$2^{2\mathrm{y}}-2^{\mathrm{y}}+2^{\mathrm{x}}\left(1-2^{\mathrm{x}}\right)=0$

$\text{ Putting }{2}^y=t,\text{ we get }$

${\mathrm{t}}^2-\mathrm{t}+2^{\mathrm{x}}\left(1-2^{\mathrm{x}}\right)=0$,$\text{ where }{\mathrm{t}}_1=2^{{\mathrm{y}}_1}\text{ and }{\mathrm{t}}_2=2^{{\mathrm{y}}_2}$

$\begin{array}{l}{\mathrm{t}}_1{\mathrm{t}}_2=2^{\mathrm{x}}\left(1-2^{\mathrm{x}}\right)\\ 2^{{\mathrm{y}}_1+{\mathrm{y}}_2}=2^{\mathrm{x}}\left(1-2^{\mathrm{x}}\right)\\ {\mathrm{y}}_1+{\mathrm{y}}_2=\mathrm{x}+{\log }_2\left(1-2^{\mathrm{x}}\right)\end{array}$