# Difference between revisions of "2018 AIME I Problems/Problem 5"

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Expilncalc (talk | contribs) (Added solution and problem.) |
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+ | For each ordered pair of real numbers <math>(x,y)</math> satisfying <cmath>\log_2(2x+y) = \log_4(x^2+xy+7y^2)</cmath>there is a real number <math>K</math> such that <cmath>\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).</cmath>Find the product of all possible values of <math>K</math>. | ||

+ | ==Solutions== | ||

+ | |||

+ | ==Straightforward Solution== | ||

+ | Note that <math>(2x+y)^2 = x^2+xy+7y^2</math>. DO NOT SQUARE THE WRONG SIDE! | ||

+ | That gives <math>x^2+xy-2y^2=0</math> upon simplification and division by <math>3</math>. Then, <math>x=y</math> or <math>x=-2y</math>. | ||

+ | From the second equation, <math>9x^2+6xy+y^2=3x^2+4xy+Ky^2</math>. If we take <math>x=y</math>, we see that <math>K=9</math>. If we take <math>x=-2y</math>, we see that <math>K=21</math>. The product is <math>\boxed{189}</math>. |

## Revision as of 18:34, 7 March 2018

For each ordered pair of real numbers satisfying there is a real number such that Find the product of all possible values of .

## Solutions

## Straightforward Solution

Note that . DO NOT SQUARE THE WRONG SIDE! That gives upon simplification and division by . Then, or . From the second equation, . If we take , we see that . If we take , we see that . The product is .