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 | ||
| − | < | + | <cmath>\begin{align*} |
| − | + | \log_2(xyz-3+\log_5 x)&=5,\\ | |
| − | + | \log_3(xyz-3+\log_5 y)&=4,\\ | |
| + | \log_4(xyz-3+\log_5 z)&=4.\\ | ||
| + | \end{align*}</cmath> | ||
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>. | ||
| Line 9: | Line 11: | ||
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 == | == See also == | ||
{{AIME box|year=2016|n=II|num-b=2|num-a=4}} | {{AIME box|year=2016|n=II|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 19:33, 8 November 2022
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. 
