Question

Solve the system $\{\begin{array}{c}{4log}_2^{2}x+1=2{\log }_{2}y\\ {\log }_2{x}^2\ge {\log }_{2}y\end{array}$ . The square of product of x and y coordinate of the solution is

Answer 1

The system of the equation is defined when $x>0$ and $y>0$
From second equation,
${\log }_2{x}^2\ge {\log }_{2}y\Rightarrow 2{\log }_{2}x\ge {\log }_{2}y\Rightarrow 4{\log }_{2}x\ge 2{\log }_{2}y$
Replacing $2{\log }_{2}y$ by $4{log}_2^{2}x+1$, (see first equation)
then we get,
$4{\log }_{2}x\ge 4{log}_2^{2}x+1\Rightarrow 4{log}_2^{2}x-4{log}_{2}x+1\le 0\Rightarrow {(2{\log }_{2}x-1)}^2\le 0$
which is possible only when
${(2{\log }_{2}x-1)}^2=0\therefore 2{\log }_{2}x-1=0\therefore {\log }_{2}x=\frac{1}{2}\Rightarrow x=2^{1/2}x=\sqrt{2}$
We find from the equation of the system that $y=2$.
Hence, pair ($\sqrt{2},2$) is the solution of the given system.
So, product of $xy=2\sqrt{2}$.
So, the square$=8$.