1991 AHSME Problems/Problem 24

Problem

The graph, $G$ of $y=\log_{10}x$ is rotated $90^{\circ}$ counter-clockwise about the origin to obtain a new graph $G'$ . Which of the following is an equation for $G'$ ?

(A) $y=\log_{10}\left(\frac{x+90}{9}\right)$ (B) $y=\log_{x}10$ (C) $y=\frac{1}{x+1}$ (D) $y=10^{-x}$ (E) $y=10^x$

Solution

$\fbox{D}$ Rotating a point $(x,y)$ $90^{\circ}$ anticlockwise about the origin maps it to $(-y,x).$ (You can prove this geometrically, or using matrices if you aren't convinced). Thus $(x, \log_{10}x)$ maps to $(-\log_{10}x, x)$ , so new $y =$ old $x = 10^{\log_{10}x} = 10^{-(-\log_{10}x)} = 10^{-\text{new} x}$ , so the new equation is $y=10^{-x}.$

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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1991 AHSME Problems/Problem 24

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