Question

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

• A. $x{\log }_2(1-2^x)$
• B. $x+{\log }_2(1-2^x)$
• C. ${\log }_2(1-2^x)$
• D. $x+{\log }_2(1+2^x)$

Answer 1

The correct option is B $x+{\log }_2(1-2^x)$

$2^x+4^y=2^y+4^x\Rightarrow 2^x+2^{2y}=2^y+2^{2x}\Rightarrow 2^{2x}-2^x=2^{2y}-2^y$

Let $2^y=t$
$\therefore 2^{2x}-2^x=t^2-t$
$\Rightarrow t^2-t-(2^{2x}-2^x)=0$
If $t_1=2^{y_1},t_2=2^{y_2}$ are the roots of the equation,
then $t_1{t}_2=2^x-2^{2x},t_1+t_2=1$
$\Rightarrow 2^{y_1+y_2}=2^x-2^{2x}=2^x(1-2^x)$
Taking ${\log }_2$ both sides, we get
$y_1+y_2=x+{\log }_2(1-2^x)$

Alternate Solution:
$2^x+4^y=2^y+4^x$
Let $2^x=a,2^y=b$
$\Rightarrow a+b^2=b+a^2$
$\Rightarrow (a-b)=(a-b)(a+b)$
$\Rightarrow (a-b)(a+b-1)=0$
$\Rightarrow a=b$ or $a+b=1$

If $a=b$
$\Rightarrow 2^x=2^y$
$\Rightarrow x=y$

If $a+b=1$
$\Rightarrow 2^x+2^y=1$
$\Rightarrow 2^y=1-2^x$
Taking $\log$ with base $2$ on both sides
$y={\log }_2(1-2^x)$

Sum of roots $=x+{\log }_2(1-2^x)$