Difference between revisions of "1991 AHSME Problems/Problem 24"

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

Latest revision as of 16:47, 23 February 2018

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
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