Question

If ${\log }_2(x+y)=3$ and ${\log }_{2}x+{\log }_{2}y=2+{\log }_{2}3$, then the values of $x$ and $y$ are

• A. $x=1$, $y=8$
• B. $x=4$, $y=4$
• C. $x=4$, $y=8$
• D. $x=2$, $y=6$

Answer 1

The correct option is D $x=2$, $y=6$

${\log }_2(x+y)=3$
and ${\log }_{2}x+{\log }_{2}y=2+{\log }_{2}3$
since ${\log }_2(x+y)=3={\log }_2{2}^3$
$={\log }_{2}8$
$\Rightarrow x+y=8$ ...(1)
Again ${\log }_{2}x+{\log }_{2}y=2+{\log }_{2}3={\log }_{2}4+{\log }_{2}3$
$\Rightarrow {\log }_2\left(xy\right)={\log }_{2}12$
$\Rightarrow xy=12$ ...(2)
By solving (1) and (2), we get $x=2$ or $x=6$ and $y=6$ or $y=2$

Then, $x=2,y=6$