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

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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