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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