Question

If $A$ is the solution set of the equation ${\log }_{x}2\cdot {\log }_{2x}2={\log }_{4x}2$ and $B$ is the solution set of the equation $x^{{\log }_x(3-x{)}^2}=25,$ then $n(A\cup B)$ is equal to

• A. $n(A\times B)$
• B. $n(A-B)+n(B-A)$
• C. $n((A-B)\times (B-A))$
• D. $n(A\cup B)-n(A\cap B)$

Answer 1

Answer: (B), (D)

The correct options are
B $n(A-B)+n(B-A)$
D $n(A\cup B)-n(A\cap B)$

${\log }_{x}2\cdot {\log }_{2x}2={\log }_{4x}2$
$\Rightarrow \frac{{\log }_{2}2}{{\log }_{2}x}\cdot \frac{{\log }_{2}2}{{\log }_{2}2x}=\frac{{\log }_{2}2}{{\log }_{2}4x}$
$\Rightarrow \frac{1}{{\log }_{2}x(1+{\log }_{2}x)}=\frac{1}{2+{\log }_{2}x}$
$\Rightarrow {\log }_{2}x+({\log }_{2}x{)}^2=2+{\log }_{2}x$
$\Rightarrow ({\log }_{2}x{)}^2=2$
$\Rightarrow {\log }_{2}x=\pm \sqrt{2}$
$\Rightarrow x=2^{-\sqrt{2}},2^{\sqrt{2}}$
$\therefore A=\{2^{-\sqrt{2}},2^{\sqrt{2}}\}$

$x^{{\log }_x(3-x{)}^2}=25$
Clearly, $x>0$
$(x-3{)}^2=25$
$\Rightarrow x=3\pm 5=-2,8$
$B=\left\{8\right\}$

$\therefore n(A\cup B)=3;n(A\cap B)=0$
$n(A-B)+n(B-A)=3;n(A\times B)=2$