Question

Solve the following for $x$ and $y$

${\log }_{100}|x+y|=\frac{1}{2},{\log }_{10}y-{\log }_{10}\left|x\right|={\log }_{100}4$

Answer 1

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

$\Rightarrow {\log }_{{10}^2}|x+y|=\frac{1}{2}$

$\Rightarrow |x+y|=10$ ..............$\left(1\right)$

${\log }_{10}y-{\log }_{10}\left|x\right|={\log }_{100}4,y>0$

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

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

From $\left(1\right)$ and $\left(2\right)$ we get

$|x+2|\left|x\right|=10$ ..........$\left(3\right)$

Case$I$: If $x>0$ then $\left|x\right|=x$

From eqn$\left(3\right)$, $|x+2x|=10\Rightarrow \left|x\right|=\frac{10}{3}$

$\therefore x=\frac{10}{3}$ and $y=\frac{20}{3}$ from eqn$\left(2\right)$

Case$II$: If $x<0$ then $\left|x\right|=-x$

$\Rightarrow |x-2x|=10\Rightarrow |-x|=10$ or $x=-10$

$\therefore x=-10,y=2|-10|=20$

Hence, the solutions are $\{\frac{10}{3},\frac{20}{3}\}$ and $\{-10,20\}$