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Question

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


Solution

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

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

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

$$\Rightarrow \log_{10} \left(\dfrac{y}{\left|x\right|}\right)  = \log_{10} 2$$            ($$\because \log x^m=m\log x$$)
$$\Rightarrow \dfrac{y}{\left|x\right|} = 2 $$    
$$\Rightarrow  y = 2 \left|x\right|$$         ....(2)  

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

If $$x > 0$$ then
$$3x = \pm 10$$
$$x = \pm \dfrac{10}{3}$$
Since, $$x>0$$ . So, $$x \neq \dfrac{-10}{3}$$        
$$\therefore x = \dfrac{10}{3}$$
$$\Rightarrow  y = \dfrac{20}{3}$$   
So, the ordered pair is $$\left(\cfrac{10}{3}, \cfrac{20}{3} \right)$$
                        
If $$x < 0$$, then 
$$x - 2x = \pm 10$$   
$$\therefore x = \pm 10$$
Since $$x<0$$. So, $$x \neq 10, $$  
$$\therefore  x = - 10$$                                
$$\therefore  y= 20$$ 
So, the ordered pair is (-10,20)
Solutions are $$(-10, 20), (\cfrac{10}{3}, \cfrac{20}{3})$$
Hence, 2 ordered pairs are possible.


Maths

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