Art of Problem Solving

1991 AHSME Problems/Problem 24: Difference between revisions

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== Problem ==
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>
== Solution ==
<math>\fbox{}</math>
== See also ==
{{AHSME box|year=1991|num-b=23|num-a=25}} 
[[Category: Intermediate Algebra Problems]]
{{MAA Notice}}
{{MAA Notice}}

Revision as of 01:58, 28 September 2014

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

$\fbox{}$

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

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