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Difference between revisions of "2010 AIME II Problems/Problem 5"

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== Solution ==  
 
== Solution ==  
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Using the properties of logarithms, <math>\log_{10}xyz = 81</math> by taking the log base 10 of both sides, and <math>(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}x)= 468</math> by using the fact that <math>\log_{10}ab = \log_{10}a + \log_{10}b</math>. Through further simplification, we find that <math>\log_{10}x+\log_{10}y+\log_{10}z = 81</math>. It can be seen that there is enough information to use the formula <math>\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc</math>, as we have both <math>\ a+b+c</math> and <math>\ 2ab+2ac+2bc</math>, and we want to find <math>\sqrt {a^2 + b^2 + c^2}</math>. After plugging in the values into the equation, we find that <math>\(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2</math> is equal to <math>\ 6561 - 936 = 5625</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{75}</math>.
+
Using the properties of logarithms, <math>\log_{10}xyz = 81</math> by taking the log base 10 of both sides, and <math>(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}x)= 468</math> by using the fact that <math>\log_{10}ab = \log_{10}a + \log_{10}b</math>.  
 +
 
 +
Through further simplification, we find that <math>\log_{10}x+\log_{10}y+\log_{10}z = 81</math>. It can be seen that there is enough information to use the formula <math>\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc</math>, as we have both <math>\ a+b+c</math> and <math>\ 2ab+2ac+2bc</math>, and we want to find <math>\sqrt {a^2 + b^2 + c^2}</math>.  
 +
 
 +
After plugging in the values into the equation, we find that <math>\ (\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2</math> is equal to <math>\ 6561 - 936 = 5625</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{75}</math>.
  
 
== 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}}

Revision as of 16:48, 3 April 2010

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 $(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}x)= 468$ 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 $\boxed{75}$.

See also

2010 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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