Difference between revisions of "2016 AIME II Problems/Problem 3"
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− | ==Problem | + | ==Problem== |
Let <math>x,y,</math> and <math>z</math> be real numbers satisfying the system | Let <math>x,y,</math> and <math>z</math> be real numbers satisfying the system | ||
− | <math>\log_2(xyz-3+\log_5 x)=5</math> | + | <math>\log_2(xyz-3+\log_5 x)=5</math>, |
− | <math>\log_3(xyz-3+\log_5 y)=4</math> | + | <math>\log_3(xyz-3+\log_5 y)=4</math>, |
− | <math>\log_4(xyz-3+\log_5 z)=4</math> | + | <math>\log_4(xyz-3+\log_5 z)=4</math>, |
Find the value of <math>|\log_5 x|+|\log_5 y|+|\log_5 z|</math>. | Find the value of <math>|\log_5 x|+|\log_5 y|+|\log_5 z|</math>. | ||
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First, we get rid of logs by taking powers: <math>xyz-3+\log_5 x=2^{5}=32</math>, <math>xyz-3+\log_5 y=3^{4}=81</math>, and <math>(xyz-3+\log_5 z)=4^{4}=256</math>. Adding all the equations up and using the <math>\log {xy}=\log {x}+\log{y}</math> property, we have <math>3xyz+\log_5{xyz} = 378</math>, so we have <math>xyz=125</math>. Solving for <math>x,y,z</math> by substituting <math>125</math> for <math>xyz</math> in each equation, we get <math>\log_5 x=-90, \log_5 y=-41, \log_5 z=134</math>, so adding all the absolute values we have <math>90+41+134=\boxed{265}</math>. | First, we get rid of logs by taking powers: <math>xyz-3+\log_5 x=2^{5}=32</math>, <math>xyz-3+\log_5 y=3^{4}=81</math>, and <math>(xyz-3+\log_5 z)=4^{4}=256</math>. Adding all the equations up and using the <math>\log {xy}=\log {x}+\log{y}</math> property, we have <math>3xyz+\log_5{xyz} = 378</math>, so we have <math>xyz=125</math>. Solving for <math>x,y,z</math> by substituting <math>125</math> for <math>xyz</math> in each equation, we get <math>\log_5 x=-90, \log_5 y=-41, \log_5 z=134</math>, so adding all the absolute values we have <math>90+41+134=\boxed{265}</math>. | ||
− | + | Note: <math>xyz=125</math> because we know <math>xyz</math> has to be a power of <math>5</math>, and so it is not hard to test values in the equation <math>3xyz+\log_5{xyz} = 378</math> in order to achieve desired value for <math>xyz</math>. | |
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2016|n=II|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:40, 15 March 2021
Problem
Let and be real numbers satisfying the system , , , Find the value of .
Solution
First, we get rid of logs by taking powers: , , and . Adding all the equations up and using the property, we have , so we have . Solving for by substituting for in each equation, we get , so adding all the absolute values we have .
Note: because we know has to be a power of , and so it is not hard to test values in the equation in order to achieve desired value for .
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.