Difference between revisions of "2016 AIME II Problems/Problem 3"

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==Problem 3==
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==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>
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<math>\log_2(xyz-3+\log_5 x)=5</math>,
<math>\log_3(xyz-3+\log_5 y)=4</math>
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<math>\log_3(xyz-3+\log_5 y)=4</math>,
<math>\log_4(xyz-3+\log_5 z)=4</math>
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<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>.
  
Solution by Shaddoll
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== See also ==
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{{AIME box|year=2016|n=II|num-b=2|num-a=4}}
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{{MAA Notice}}

Revision as of 23:29, 23 February 2021

Problem

Let $x,y,$ and $z$ be real numbers satisfying the system $\log_2(xyz-3+\log_5 x)=5$, $\log_3(xyz-3+\log_5 y)=4$, $\log_4(xyz-3+\log_5 z)=4$, Find the value of $|\log_5 x|+|\log_5 y|+|\log_5 z|$.

Solution

First, we get rid of logs by taking powers: $xyz-3+\log_5 x=2^{5}=32$, $xyz-3+\log_5 y=3^{4}=81$, and $(xyz-3+\log_5 z)=4^{4}=256$. Adding all the equations up and using the $\log {xy}=\log {x}+\log{y}$ property, we have $3xyz+\log_5{xyz} = 378$, so we have $xyz=125$. Solving for $x,y,z$ by substituting $125$ for $xyz$ in each equation, we get $\log_5 x=-90, \log_5 y=-41, \log_5 z=134$, so adding all the absolute values we have $90+41+134=\boxed{265}$.

See also

2016 AIME II (ProblemsAnswer KeyResources)
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

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