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Difference between revisions of "2010 AIME II Problems/Problem 5"
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However, we want to find <math>\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}</math>, so we take the square root of <math>\ 5625</math>, or <math>\boxed{075}</math>.
 
However, we want to find <math>\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}</math>, so we take the square root of <math>\ 5625</math>, or <math>\boxed{075}</math>.
  
 +
==Solution 2==
 +
<math>a = \log{x}</math>
 +
 +
<math>b = \log{y}</math>
 +
 +
<math>c = \log{z}</math>
 +
 +
<math>xyz = 10^81</math>
 +
 +
<math>\log{xyz} = 81</math>
 +
 +
<math>\log{x} + \log{y} + \log{z} = 81</math>
 +
 +
<math>a + b + c = 81</math>
 +
 +
<math>\log{x}(\log{yz}) + \log{y}\log{z} = \log{x}(\log{y} + \log{z}) + \log{y}\log{z} = ab + ac + bc = 468</math>
 +
 +
<math>a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 6561</math>
 +
 +
<math>a^2 + b^2 + c^2 = 5621 = 75^2</math>
 +
 +
<math>\sqrt{\log{x^2} + \log{y^2} + \log{z^2}} = \sqrt{a^2 + b^2 + c^2} = \boxed{075}</math>
 +
~MathleteMA
 
== See also ==
 
== See also ==
 
{{AIME box|year=2010|num-b=4|num-a=6|n=II}}
 
{{AIME box|year=2010|num-b=4|num-a=6|n=II}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:54, 2 March 2017

Problem

Positive numbers $x$, $y$, and $z$ satisfy $xyz = 10^{81}$ and $(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468$. Find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$.

Solution

Using the properties of logarithms, $\log_{10}xyz = 81$ by taking the log base 10 of both sides, and $(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}z)= 468$ by using the fact that $\log_{10}ab = \log_{10}a + \log_{10}b$.

Through further simplification, we find that $\log_{10}x+\log_{10}y+\log_{10}z = 81$. It can be seen that there is enough information to use the formula $\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$, as we have both $\ a+b+c$ and $\ 2ab+2ac+2bc$, and we want to find $\sqrt {a^2 + b^2 + c^2}$.

After plugging in the values into the equation, we find that $\ (\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2$ is equal to $\ 6561 - 936 = 5625$.

However, we want to find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$, so we take the square root of $\ 5625$, or $\boxed{075}$.

Solution 2

$a = \log{x}$

$b = \log{y}$

$c = \log{z}$

$xyz = 10^81$

$\log{xyz} = 81$

$\log{x} + \log{y} + \log{z} = 81$

$a + b + c = 81$

$\log{x}(\log{yz}) + \log{y}\log{z} = \log{x}(\log{y} + \log{z}) + \log{y}\log{z} = ab + ac + bc = 468$

$a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 6561$

$a^2 + b^2 + c^2 = 5621 = 75^2$

$\sqrt{\log{x^2} + \log{y^2} + \log{z^2}} = \sqrt{a^2 + b^2 + c^2} = \boxed{075}$ ~MathleteMA

See also

2010 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AIME Problems and Solutions

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