Question

If $\left|{\log }_2(x+y)\right|+\left|{\log }_2(x-y)\right|=3$ and $xy=3$ then

• A. $x=3,y=1$
• B. $x=1,y=3$
• C. $x=3\sqrt{\frac{3}{7}},y=\sqrt{\frac{7}{3}}$
• D. $x=\sqrt{\frac{7}{3}},y=3\sqrt{\frac{3}{7}}$

Answer 1

Answer: (A), (C)

$x=3,y=1$
$x=3\sqrt{\frac{3}{7}},y=\sqrt{\frac{7}{3}}$

Given equations
$xy=3$ .....(2)
$\left|{\log }_2(x+y)\right|+\left|{\log }_2(x-y)\right|=3$ ....(2)
Case I: When both terms are positive
$\Rightarrow {\log }_2(x+y)+{\log }_2(x-y)=3$
$\Rightarrow {\log }_2(x^2-y^2)=3$
$\Rightarrow x^2-y^2=8$ ....(3)
Solving (1) and (3), we get
$x^4-8x^2-9=0$
$\Rightarrow (x^2-9)(x^2+1)=0$
$\Rightarrow x^2=9$ ($x^2=-1$ is not possible)
$\Rightarrow x=\pm 3$
$\Rightarrow y=\pm 1$
From eqn (2), it follows that
$x+y>0,x-y>0$
So, permissible values of x and y are 3,1
Case II : When first term of eqn (1) is positive and second term is negative
${\log }_2(x+y)-{\log }_2(x-y)=3$
$\Rightarrow {\log }_2\left(\frac{x+y}{x-y}\right)=3$
$\Rightarrow 8=\frac{x+y}{x-y}$
$\Rightarrow 7x-9y=0$ ....(4)
Solving (1) and (4), we get
$7x^2-27=0$
$\Rightarrow x=\pm 3\sqrt{\frac{3}{7}}$
$\Rightarrow x=3\sqrt{\frac{3}{7}}$
$\Rightarrow y=\sqrt{\frac{7}{3}}$
Case III : When first term of eqn (1) is negative and second term is positive
$-{\log }_2(x+y)+{\log }_2(x-y)=3$
$\Rightarrow {\log }_2\left(\frac{x-y}{x+y}\right)=3$
$\Rightarrow 8=\frac{x-y}{x+y}$
$\Rightarrow 7x+9y=0$ ....(5)
Solving (1) and (5), we get
$7x^2=-27$
No real value of x.
Hence, in this case no solution
Case IV : When both terms of eqn (1) is negative
$-{\log }_2(x+y)-{\log }_2(x-y)=3$
$\Rightarrow -{\log }_2(x^2-y^2)=3$
$\Rightarrow x^2-y^2=\frac{1}{8}$ ....(6)
Solving (1) and (6), we get
$8x^4-x^2-72=0$
$\Rightarrow 8t^2-t-72=0$ where $t=x^2$
$\Rightarrow t=\frac{1\pm \sqrt{577}}{2}$
$\Rightarrow x^2=\frac{1\pm \sqrt{577}}{2}$
So, it is not a solution