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 $x=2,y=6$

${\log }_2(x+y)=3$

Or, $(x+y)=2^3=8$ .............(1)

And, ${\log }_{2}x+{\log }_{2}y=2+{\log }_{2}3$

${\log }_2(x.y)={\log }_{2}4+{\log }_{2}3$ .........(2) ($\because 2={\log }_{2}4$)

Then, we have,

$x+y=8$............(3)

$xy=12$ ...........(4)

$x=8-y$

Putting it in Eq. 4 we get, $(8-y)y=12$

Or, $-y^2+8y-12=0$

$y^2-8y+12=0$

Or, $(y-6)(y-2)=0$

Hence, $y=2$ or $y=6$

And using y in Eq 3 for values of $y$ we get $x=6$ or $x=2$