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>,
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<cmath>\begin{align*}
<math>\log_3(xyz-3+\log_5 y)=4</math>,
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\log_2(xyz-3+\log_5 x)&=5,\\
<math>\log_4(xyz-3+\log_5 z)=4</math>,
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\log_3(xyz-3+\log_5 y)&=4,\\
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\log_4(xyz-3+\log_5 z)&=4.\\
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\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>.
  
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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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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}}
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{{MAA Notice}}

Latest revision as of 20:33, 8 November 2022

Problem

Let $x,y,$ and $z$ be real numbers satisfying the system \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*} 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 $90+41+134=\boxed{265}$.

Note: $xyz=125$ because we know $xyz$ has to be a power of $5$, and so it is not hard to test values in the equation $3xyz+\log_5{xyz} = 378$ in order to achieve desired value for $xyz$.

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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