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Algebra Level 2

{ log ⁡ 100 ∣ x + y ∣ = 1 2 log ⁡ 10 y − log ⁡ 10 ∣ x ∣ = log ⁡ 100 4 \large{\begin{cases}{\log_{100}|x+y|=\frac{1}{2}} \\ { \log_{10}y - \log_{10}|x|=\log_{100}4}\end{cases}} The solution set of the equations above is?

For more questions on logarithms try this set Logs of logs .
{ − 10 , − 20 } , { − 10 3 , 20 3 } \{-10,-20 \},\left \{\frac{-10}{3},\frac{20}{3} \right \} { 10 , 20 } , { − 10 3 , − 20 3 } \{10,20\},\left \{\frac{-10}{3},\frac{-20}{3} \right \} { − 10 , 20 } , { 10 3 , 20 3 } \{-10,20\}, \left \{\frac{10}{3},\frac{20}{3} \right \} { − 10 , − 20 } , { − 10 3 , − 20 3 } \{-10,-20\}, \left \{\frac{-10}{3},\frac{-20}{3} \right \}

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Gautam Sharma by Gautam Sharma , 23, India

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2 solutions

The first equation requires that ∣ x + y ∣ = 10 , |x + y| = 10, which is represented by the lines x + y = 10 ⟹ y = − x + 10 x + y = 10 \Longrightarrow y = -x + 10 and x + y = − 10 ⟹ y = − x − 10. x + y = -10 \Longrightarrow y = -x - 10.

The second equation requires that

log ⁡ 10 ( y ∣ x ∣ ) = log ⁡ 10 ( 4 ) log ⁡ 10 ( 100 ) = log ⁡ 10 ( 2 ) ⟹ y = 2 ∣ x ∣ . \log_{10}\left(\dfrac{y}{|x|}\right) = \dfrac{\log_{10}(4)}{\log_{10}(100)} = \log_{10}(2) \Longrightarrow y = 2|x|.

In the first quadrant the solution will be where y = − x + 10 y = -x + 10 intersects y = 2 x , y = 2x, , i.e.,

2 x = − x + 10 ⟹ x = 10 3 ⟹ y = 20 3 . 2x = -x + 10 \Longrightarrow x = \dfrac{10}{3} \Longrightarrow y = \dfrac{20}{3}.

In the second quadrant we require that

− x + 10 = − 2 x ⟹ x = − 10 ⟹ y = 20. -x + 10 = -2x \Longrightarrow x = -10 \Longrightarrow y = 20.

Thus the solution set is ( − 10 , 20 ) , ( 10 3 , 20 3 ) . \boxed{(-10,20), \left(\dfrac{10}{3}, \dfrac{20}{3}\right)}.

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JEE style : for 1st condition to be true , ∣ x + y ∣ |x+y| should be 10 which is given only in the one of the given options. So that's the correct answer provided that the question is framed with correct answer. :P

Sandeep Bhardwaj - 6 years, 2 months ago
Aman Gautam
Apr 5, 2015

a lazy approach...!
as y cannot be negative ,eliminate the other options which have y as negative..!!!

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