1972 AHSME Problems/Problem 8

Revision as of 21:46, 28 January 2021 by Coolmath34 (talk | contribs) (Created page with "== Problem == If <math>|x-\log y|=x+\log y</math> where <math>x</math> and <math>\log y</math> are real, then <math>\textbf{(A) }x=0\qquad \textbf{(B) }y=1\qquad \textbf{(C)...")
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If $|x-\log y|=x+\log y$ where $x$ and $\log y$ are real, then

$\textbf{(A) }x=0\qquad \textbf{(B) }y=1\qquad \textbf{(C) }x=0\text{ and }y=1\qquad\\ \textbf{(D) }x(y-1)=0\qquad  \textbf{(E) }\text{None of these}$


We have two cases: $x$ is positive and $x$ is negative. \[\text{Positive} \qquad x - \log y = x + \log y \qquad \rightarrow \qquad 2\log y = 0 \rightarrow y = 1\] \[\text{Negative} \qquad -x + \log y = x + \log y \qquad \rightarrow \qquad 2x = 0 \rightarrow x = 0\]

We can write $x(y-1)=0$ to show that $x=0, y=1,$ or both.

The answer is $\textbf{(D)}.$

-edited by coolmath34

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