1991 AHSME Problems/Problem 24

Revision as of 16:47, 23 February 2018 by Hapaxoromenon (talk | contribs) (Added a solution with explanation)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $y=10^{-x}.$

See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS