Question

${\log }_{100}|x+y|=\frac{1}{2}$
${\log }_{10}y-{\log }_{10}\left|x\right|={\log }_{100}4$
The possible number of ordered sets of $(x,y)$ is

Answer 1

Given ${\log }_{100}|x+y|=\frac{1}{2}$

Converting into exponential form
$\Rightarrow |x+y|=(100{)}^{1/2}$
$\Rightarrow |x+y|=10$
$\Rightarrow x+y=\pm 10$ ...(1)

Also given ${\log }_{10}y-{\log }_{10}\left|x\right|={\log }_{100}4$
$\Rightarrow {\log }_{10}\left(\frac{y}{\left|x\right|}\right)={\log }_{{10}^2}{2}^2$ ($\because \log \left(\frac{m}{n}\right)=\log m-\log n)$)

$\Rightarrow {\log }_{10}\left(\frac{y}{\left|x\right|}\right)=\frac{1}{2}{\log }_{10}{2}^2$ $(\because {\log }_{a^b}m=\frac{1}{b}{\log }_{a}m)$

$\Rightarrow {\log }_{10}\left(\frac{y}{\left|x\right|}\right)={\log }_{10}2$ ($\because \log x^m=m\log x$)
$\Rightarrow \frac{y}{\left|x\right|}=2$

$\Rightarrow y=2\left|x\right|$ ....(2)

From (1) & (2)
$x+2\left|x\right|=\pm 10$

If $x>0$ then

$3x=\pm 10$

$x=\pm \frac{10}{3}$

Since, $x>0$ . So, $x\ne \frac{-10}{3}$

$\therefore x=\frac{10}{3}$

$\Rightarrow y=\frac{20}{3}$

So, the ordered pair is $(\frac{10}{3},\frac{20}{3})$

If $x<0$, then

$x-2x=\pm 10$
$\therefore x=\pm 10$

Since $x<0$. So, $x\ne 10,$

$\therefore x=-10$
$\therefore y=20$

So, the ordered pair is (-10,20)
Solutions are $(-10,20),(\frac{10}{3},\frac{20}{3})$

Hence, 2 ordered pairs are possible.