Difference between revisions of "1991 AHSME Problems/Problem 24"
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| + | == Problem == | ||
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The graph, <math>G</math> of <math>y=\log_{10}x</math> is rotated <math>90^{\circ}</math> counter-clockwise about the origin to obtain a new graph <math>G'</math>. Which of the following is an equation for <math>G'</math>? | The graph, <math>G</math> of <math>y=\log_{10}x</math> is rotated <math>90^{\circ}</math> counter-clockwise about the origin to obtain a new graph <math>G'</math>. Which of the following is an equation for <math>G'</math>? | ||
(A) <math>y=\log_{10}\left(\frac{x+90}{9}\right)</math> (B) <math>y=\log_{x}10</math> (C) <math>y=\frac{1}{x+1}</math> (D) <math>y=10^{-x}</math> (E) <math>y=10^x</math> | (A) <math>y=\log_{10}\left(\frac{x+90}{9}\right)</math> (B) <math>y=\log_{x}10</math> (C) <math>y=\frac{1}{x+1}</math> (D) <math>y=10^{-x}</math> (E) <math>y=10^x</math> | ||
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| + | == Solution == | ||
| + | <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> | ||
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| + | == See also == | ||
| + | {{AHSME box|year=1991|num-b=23|num-a=25}} | ||
| + | |||
| + | [[Category: Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:47, 23 February 2018
Problem
The graph, 
of 
is rotated 
counter-clockwise about the origin to obtain a new graph 
. Which of the following is an equation for ?
(A) 
(B) 
(C) 
(D) 
(E) 
Solution
Rotating a point 
anticlockwise about the origin maps it to 
(You can prove this geometrically, or using matrices if you aren't convinced). Thus 
maps to 
, so new 
old 
, so the new equation is 
See also
| 1991 AHSME (Problems • Answer Key • Resources) | ||
| 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. 
