Question

Find the number of pairs for (x, y) from the following equations:
$lo{g}_{100}|x+y|=\frac{1}{2}lo{g}_{10}y-lo{g}_{10}|x|=lo{g}_{100}4$

• A. 1
• B. 2
• C. none of these

Answer 1

The correct option is B 2

$lo{g}_{100}|x+y|=\frac{1}{2}\Rightarrow (100{)}^{\frac{1}{2}}=|x+y|$
$\Rightarrow$ |x + y| = 10 . . . (1)
Again, $lo{g}_{10}y-lo{g}_{10}|x|=lo{g}_{100}4$
$lo{g}_{10}y-lo{g}_{10}|x|=lo{g}_{10}2\Rightarrow lo{g}_{10}\frac{y}{|x|}=lo{g}_{10}2$
$\Rightarrow$ y = 2|x| . . .(2)
From eq. (2) we can conclude that y is always positive.
Now, when x > 0 and y > 0 (always)
|x + y| = 10 $\Rightarrow$ |x + 2|x|| = 10
$\Rightarrow$ x + 2 |x| = 10 ($\because$ x > 0)
$\Rightarrow$ x + 2x = 10
$\Rightarrow$ $x=\frac{10}{3}$
$\therefore$ $y=\frac{20}{3}$
Again, x < 0 and y > 0 (always positive)
|-x + 2| - x|| = 10
$\Rightarrow$ |- x + 2x| = 10
$\Rightarrow$ |x| = 10
$\Rightarrow$ x = -10 ($\because$ x < 0)
$\therefore$ y = 20
Hence, x = -10, y = 20 and $x=\frac{10}{3}andy=\frac{20}{3}$.